Mahesh

Students get through AP Inter 2nd Year Physics Important Questions 10th Lesson Alternating Current which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 10th Lesson Alternating Current

Very Short Answer Questions

Question 1.
A transformer converts 200 V ac into 2000 V ac. Calculate the number of turns in the ‘ secondary if the primary has 10 turns. [T.S. Mar. 16]
Answer:
\(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\)
V_p = 200V, V_s = 2000V, N_p = 10
N_s = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}\) × N_p = \(\frac{2000}{200}\) × 10
N_s = 100.

Question 2.
What type of transformer is used in a 6V bed lamp ? [A.P. Mar. 17]
Answer:
Step down transformer is used in 6V bed lamp.

Question 3.
What is the phenomenon involved in the working of a transformer ? [Mar. 16(A.P.) Mar. 14]
Answer:
Transformer works on the principle of mutual induction.

Question 4.
What is transformer ratio ?
Answer:
The ratio of secondary e.m.f to the primary e.m.f. (or) number of turns in secondary to the number of turns in the primary is called the transformer ratio.
Transformer ratio = \(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{V}_{\mathrm{p}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}\)

Question 5.
Write the expression for the reactance of i) an inductor and (ii) a capacitor.
Answer:

1. Inductive reactance (X_L) = ωL
2. Capacitive reactance (X_C) = \(\frac{1}{\omega C}\)

Question 6.
What is the phase difference between A.C emf and current in the following: Pure resistor, pure inductor and pure capacitor. [T.S. Mar. 15]
Answer:

1. In pure resistor A.C. e.m.f and current are in phase with each other.
2. In pure inductor, current lags behind the e.m.f. by an angle of \(\frac{\pi}{2}\) (or) 90°.
3. In pure capacitor, current leads the e.m.f by an angle \(\frac{\pi}{2}\) (or) 90°.

Question 7.
Define power factor. On which factors does power factor depend ?
Answer:
The ratio of true power and apparent power (virtual power) in an a.c circuit is called as power factor of the circuit.
Power factor (cosΦ) = \(\frac{\mathrm{P}}{\mathrm{P}_{\mathrm{rms}}}\) [∵ P_rms = V_rms I_rms]
Power factor depends on r.m.s voltage, r.m.s current and average power (P).

Question 8.
What is meant by wattless component of current ?
Answer:
Average power (P) = V_rms(I_rms sinΦ) cos\(\frac{\pi}{2}\)
The average power consumed in the circuit due to (I_rms sinΦ) component of current is zero. This component of current is known as wattless current. (I_rms sinΦ) is the wattless component of current.

Question 9.
When does a LCR series circuit have minimum impedance ?
Answer:
In LCR series circuit, Impendence (Z) = \(\sqrt{R^2+\left(\frac{1}{\omega C}-\omega L\right)^2}\)
At a particular frequency, ωL = \(\frac{1}{\omega C}\)
The impedance is minimum (Z = R)
This frequency is called resonant frequency.

Question 10.
What is the phase difference between voltage and current when the power factor in LCR series circuit is unity ?
Answer:
In LCR series circuit power factor (cosΦ) = 1
Phase difference between voltage and current is zero. (Φ = 0)

Short Answer Questions

Question 1.
State the principle on which a transformer works. Describe the working of a transformer with necessary theory.
Answer:
Transformer is a device to convert a low alternating current of high voltage into high alternating current of low voltage and vice versa.
Principle : It works on the principle of mutual induction between two coils.
Working : When an alternating emf is applied across the primary coil, the input voltage changes with time. Hence the magnetic flux through the primary also changes with time.

This changing magnetic flux will be linked with secondary through the core. An emf is induced in the secondary.

Theory: Let N₁ and N₂ be the number of turns in the primary and secondary. Let V_P and V_S be the emf s across the primary and secondary.
\(\frac{V_S}{V_p}=\frac{\text { Output emf }}{\text{Input emf}}=\frac{-N_2 \frac{d \phi}{d t}}{-N_1 \frac{d \phi}{d t}}=\frac{N_2}{N_1}\)
∴ \(\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}=\frac{\mathrm{N}_2}{\mathrm{~N}_1}\) = Transformer ratio
Efficiency of transformer :
It is the ratio of output power to the input power.
η = \(\frac{\text { Outputpower }}{\text { Input power }}\) × 100

Problems

Question 1.
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb;
(b) the peak voltage of the source; and
(c) the rms current through the bulb. [A.P. Mar. 15]
Solution:
(a) We are given P = 100 W and V = 220V The resistance of the bulb is
R = \(\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(220 \mathrm{~V})^2}{100 \mathrm{~W}}\) = 484 Ω

(b) The peak voltage of the source is υ_m = \(\sqrt{2}\)V = 311 V

(c) Since, P = 1 V
I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.450 A.

Question 2.
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Solution:
The inductive reactance,
X_L = 2πvL = 2 × 3.14 × 50 × 25 × 10^-3 = 7.85 Ω
The rms current in the circuit is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}=\frac{220 \mathrm{~V}}{7.85 \Omega}\) = 28 A

Question 3.
The instantaneous current and instantaneous voltage across a series circuit containing resistance and inductance are given by i = \(\sqrt{2}\) sin (100t – π/4)A and υ = 40 sin (100t) V. Calculate the resistance ?
Solution:
i = \(\sqrt{2}\) sin (100t – π/4)A (∵i = i₀sin(ωt – Φ))
υ = 40 sin(100t)V (∵ V = V₀sin(ωt ))
i₀ = \(\sqrt{2}\) , V₀ = 40, ω = 100, Φ = π/4
R = \(\frac{\mathrm{V}_0}{\mathrm{i}_0}\) cosΦ = \(\frac{40}{\sqrt{2}}\) cos\(\frac{\pi}{4}\), R = \(\frac{40}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\), R = 20 Ω

Question 4.
In an AC circuit, a condenser, a resistor and a pure inductor are connected in series across an alternator (AC generator). If the voltages across them are 20 V, 35 V and 20 V respectively, find the voltage supplied by the alternator.
Solution:
V_C = 20V, V_R = 35V, V_L = 20V
V = \(\sqrt{V_R^2+\left(V_L^2-V_C^2\right)}\) ; V = \(\sqrt{(35)^2+\left(20^2-20^2\right)}\) ; V = \(\sqrt{35^2}\); V = 35 Volt.

Question 5.
What is step up transformer ? How it differs from step down transformer ?
Solution:
The ratio of number of turns in the secondary coil to the number of turns in the primary coil is called transformer ratio.
T = \(\frac{N_S}{N_p}=\frac{\text {No. of turns in the secondary }}{\text {No. of turns in the primary }}\)
If N_S > N_P, then the transformer is called step up transformer.
If N_S < N_P, then the transformer is called step down transformer.

Textual Examples

Question 1.
A light bulb is rated at 100W for a 220 V supply. Find
(a) the resistance of the bulb;
(b) the peak voltage of the source; and
(c) the rms current through the bulb. [A.P. Mar. 15]
Solution:
(a) We are given P = 100 W and V = 220V.
The resistance of the bulb is
R = \(\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(220 \mathrm{~V})^2}{100 \mathrm{~W}}\) = 484 Ω

(b) The peak voltage of the source is υ_m = \(\sqrt{2}\)V = 311 V

(c) Since, P = 1 V
I = \(\frac{\mathrm{P}}{\mathrm{V}}=\frac{100 \mathrm{~W}}{220 \mathrm{~V}}\) = 0.450 A.

Question 2.
A pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.
Solution:
The inductive reactance,
X_L = 2πvL = 2 × 3.14 × 50 × 25 × 10^-3 = 7.85 Ω
The rms current in the circuit is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{L}}}=\frac{220 \mathrm{~V}}{7.85 \Omega}\) = 28 A

Question 3.
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced ?
Solution:
When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (1/ωC) and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.

Question 4.
A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current ?
Solution:
The capacitive reactance is
X_c = \(\frac{1}{2 \pi v C}=\frac{1}{2 \pi(50 \mathrm{~Hz})\left(15.0 \times 10^{-6} \mathrm{~F}\right)}\)
= 212 Ω
The rms current is I = \(\frac{\mathrm{V}}{\mathrm{X}_{\mathrm{C}}}=\frac{220 \mathrm{~V}}{212 \Omega}\)
= 1.04 A
The peak current is
i_m = \(\sqrt{2}\)I = (1.41)(1.04A) = 1.47A
This current oscillates between + 1.47A and – 1.47 A, and is ahead of the voltage by π/2.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Question 5.
A light bulb and an open coil inductor are connected to an ac source through a key as shown in the figure.

The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons.
Solution:
As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

Question 6.
A resistor of 200Ω and a capacitor of 15.0 μF are connected in series to a 220V, 50 Hz ac source,
(a) Calculate the current in the circuit;
(b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage ? If yes, resolve the paradox.
Solution:
Given
R = 200Ω. C = 15.0 μF = 15.0 × 10^-6F
V = 220V, v = 50Hz
(a) In order to calculate the current, we need the impedance of the circuit. It is
Z = \(\sqrt{\mathrm{R}^2+\mathrm{X}_{\mathrm{C}}^2}=\sqrt{\mathrm{R}^2+(2 \pi v C)^{-2}}\)
= \(\sqrt{(200 \Omega)^2+\left(2 \times 3.14 \times 50 \times 10^{-6} \mathrm{~F}\right)^{-2}}\)
= \(\sqrt{(200 \Omega)^2+(212 \Omega)^2}\) = 291.5Ω
Therefore, the current in the circuit is
I = \(\frac{\mathrm{V}}{\mathrm{Z}}=\frac{220 \mathrm{~V}}{291.5 \Omega}\) = 0.755A

(b) Since the current is the same throughout the circuit, we have
V_R = IR = (0.755 A) (200Ω) = 151V
V_C = IX_C = (0.755A) (212.3Ω) = 160.3V
The algebraic sum of the two voltages, V_R and V_C is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox ? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem:
V_R+C = \(\sqrt{\mathrm{V}_{\mathrm{R}}^2+\mathrm{V}_{\mathrm{C}}^2}\) = 220 V
Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

Question 7.
a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.
Solution:
a) We know that P = IV cosΦ where cosΦ is the power factor. To supply a given power at a given voltage, if cosΦ is small, we have to increase current accordingly. But this will lead to large power loss (I^R) in transmission.

b) Suppose in a circuit, current I lags the voltage by an angle Φ.
Then power factor cosΦ = R/Z
We can improve the power factor (tending to 1) by making Z tend to R. Let us understand, with the help of a phasor diagram in the figure how this can be achieved. Let us resolve I into two components, I_P

along the applied voltage V and I_q perpendicular to the applied voltage. I_q is called the wattless component since corresponding to this component of current, there is no power loss. I_P is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit.

It’s clear from this analysis that if we want to improve power factor, we must completely neutralize the lagging wattless current I_q by an equal leading wattless current I’_q. This can be done by connecting a capacitor of appropriate value in parallel so that I_q and I’_q cancel each other and P is effectively I_P V.

Question 8.
A sinusoidal voltage of peak value 283 V , and frequency 50 Hz is applied to a series LCR circuit in which R = 3Ω. L = 25.48 mH. and C = 796μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and
(d) the power factor.
Solution:
a) To find the impedance of the circuit, we first calculate X_L and X_C.
X_L = 2πvL
= 2 × 3.14 × 50 × 25.48 × 10^-3Ω = 8Ω
X_C = \(\frac{1}{2 \pi v \mathrm{C}}\)
= \(\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}\) = 4Ω
Therefore,
z = \(\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}=\sqrt{3^2+(8-4)^2}\)
= 5Ω

b) Phase difference, Φ = tan^-1\(\frac{\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}\)
= tan^-1\(\left(\frac{4-8}{3}\right)\) = -53.1°

c) The power dissipated in the circuit is
P = I²R

Therefore, P = (40A)² × 3Ω = 4800W
= 4.8 kW

d) Power factor = cos Φ = cos 53.1° = 0.6.

Question 9.
Suppose the frequency of the source in the previous example can be varied,
(a) What is the frequency of the source at which resonance occurs ?
(b) Calculate the impedance, the current, and the power dissipated at the resonant condition.
Solution:
(a) The frequency at which the resonance occurs is
ω₀ = \(\frac{1}{\sqrt{\mathrm{LC}}}=\frac{1}{\sqrt{25.48 \times 10^{-3} \times 796 \times 10^{-6}}}\)
= 222.1 rad/s
v_r = \(\frac{\omega_0}{2 \pi}=\frac{221.1}{2 \times 3.14}\) Hz = 35.4Hz

b) The impedance Z at resonant condition is equal to the resistance
Z = R = 3Ω
The rms current at resonance is ,
as V = \(\frac{v_{\mathrm{m}}}{\sqrt{2}}\)
I = \(\frac{\mathrm{V}}{\mathrm{Z}}=\frac{\mathrm{V}}{\mathrm{R}}=\left(\frac{283}{\sqrt{2}}\right) \frac{1}{3}\) = 66.7 A
The power dissipated at resonance is
P = I² × R = (66.7)² × 3 = 13.35 kW
You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 8.

Question 10.
At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work ?
Solution:
The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.

Question 11.
Show that in the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the inductor is constant in time.
Solution:
Let q₀ be the initial charge tin a capacitor. Let the charged capacitor be connected to an inductor of inductance L. this LC circuit will sustain an oscillation with frequency
\(\omega\left(2 \pi v=\frac{1}{\sqrt{\mathrm{LC}}}\right)\)
At an instant t, charge q on the capacitor and the current i are given by :
q(t) = q₀ cos ωt
i(t) = -q₀ co sin ωt
Energy stored in the capacitor at time ‘t’ is
U_E = \(\frac{1}{2}\) C V² = \(\frac{1}{2} \frac{\mathrm{q}^2}{\mathrm{C}}=\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\) cos² (ωt)
Energy stored in the inductor at time ‘t’ is
U_M = \(\frac{1}{2}\) L i²
= \(\frac{1}{2}\) L q₀² ω²sin² (ωt)
= \(\frac{Q_0^2}{2 C} \sin ^2(\omega t)\) [∵  ω = 1/ \(\sqrt{L C}\)]
Sum of energies
U_E + U_M = \(\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\) [cos² ωt + sin²ωt) = \(\frac{\mathrm{q}_0^2}{2 \mathrm{C}}\)
This sum is constant in time as q₀ and C, both are time-independent.

Students get through AP Inter 2nd Year Physics Important Questions 11th Lesson Electromagnetic Waves which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 11th Lesson Electromagnetic Waves

Very Short Answer Questions

Question 1.
Give any one use of infrared rays. [T.S. Mar. 17; A.P. Mar. 16]
Answer:

1. Infrared radiation plays an important role in maintaining the Earth warm.
2. Infrared lamps are used in physical therapy.
3. Infrared detectors are used in Earth Satellites.
4. These are used in taking photographs during the conditions of fog, smoke etc.

Question 2.
How are infrared rays produced ? How they can be detected ?
Answer:
Infrared rays can be produced by vibrations of atoms and molecules. These waves can be detected by Thermopile, Bolometer, IR photographic film.

Question 3.
How are radio waves produced ? How can they detected ?
Answer:
Radio waves can be produced by rapid acceleration and deceleration of electrons in aerials (conductors). These can be detected by receivers of aerials.

Question 4.
If the wave length of E.M radiation is doubled, what happens to the energy of photon ? [IPE 2016 (TS)]
Answer:
If the wave length of electromagnetic radiation is doubled, then energy will be halved because energy is inversely proportional to.wavelength of electromagnetic waves.
E = hυ = hc/λ ⇒ E ∝ 1/λ (∵ hc is a constant)

Question 5.
What is the principle of production of electromagnetic waves ?
Answer:
If the charge is accelerated both the magnetic field and electric field will change with Space and time, then electromagnetic waves are produced.

Question 6.
What is the ratio of speed of infrared rays and ultraviolet rays in vaccum ?
Answer:
The ratio of speed of infrared rays and ultraviolet rays in vacuum is 1 : 1.
All electromagnetic waves travel with same speed 3 × 10⁸ m /s in vaccum.

Question 7.
What is the relation between the amplitudes of the electric and magnetic fields in free space for an electromagnetic wave ?
Answer:
E₀ = CB₀
Where E₀ = Amplitude of electric field.
B₀ = Amplitude of magnetic field.
C = velocity of light.

Question 8.
What are the applications of microwaves ? [A.P. Mar. 17; T.S. Mar. 15]
Answer:

1. Microwaves are used in Radars.
2. Microwaves are used for cooking purposes.
3. A radar using microwave can help in detecting the speed of automobile while in motion.

Question 9.
Microwaves are used in Radars, why ? [Mar. 14]
Answer:
As microwaves are of smaller wavelengths, hence they can be transmitted as a beam signal in a particular direction. Microwaves do not bend around the comers of any obstacle coming in their path.

Question 10.
Give two uses of infrared rays.
Answer:

1. Infrared rays are used for producing dehydrated fruits.
2. They are used in the secret writings on the ancient walls.
3. They are used in green houses to keep the plants warm.

Question 11.
How are microwaves produced ? How can they detected ? [A.P. Mar. 16; IPE 15]
Answer:
Microwaves can be produced using Klystron valve or Magnetrons.
Microwaves can be detected using point contact diodes.

Question 12.
The charging current for a capacitor is 0.6 A. What is the displacement current across its plates ?
Answer:
i = charging current for a capacitor = 0.6 A
i = i_d = ε₀ = \(\frac{\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{dt}}\)
∴ i = i_d = 0.6 A
∴ Displacement current (i_d) = 0.6 A.

Question 13.
What physical quantity is the same for X-rays of wavelength 10^-10m, red light of wavelength 6800 Å and radiowaves of wavelength 500in ?
Answer:
The speed in vaccum is same for all the given wavelengths, which is 3 × 10⁸ m/s.

Question 14.
A radio can tune into any station in the 7.5 MHz to 12MHz band. What is the corresponding wavelength band ?
Answer:
λ₁ = \(\frac{3 \times 10^8}{7.5 \times 10^6}\) = 40 m
λ₂ = \(\frac{3 \times 10^8}{12 \times 10^6}\) = 25 m
Thus wavelength band is 40m to 25m.

Question 15.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vaccum is B₀ = 510 nT. What is the amplitude of the electric field part of the wave ?
Answer:
Here, B₀ = 510nT = 510 × 10^-9T
E₀ = CB₀ = 3 × 10⁸ × 510 × 10^-9 = 153 NC^-1.

Question 16.
Define displacement current
Answer:
Displacement current (I_d) is equal to ε₀ times to the rate of change of electric flux. Displacement current is not the current produced due to charge carried. But it is due to varying electric flux. It is the current in the sense that it produces a magnetic field.
I_d = ε₀ \(\frac{\mathrm{d} \phi_{\varepsilon}}{\mathrm{dt}}\)

Short Answer Questions

Question 1.
State six characteristics of electromagnetic waves.
Answer:
Characteristics of electromagnetic waves :

1. Electromagnetic waves are produced by accelerated charges.
2. Electromagnetic waves are transverse in nature.
3. Electromagnetic waves donot require material medium for their propagation.
4. Electromagnetic waves obey principle of superposition of waves.
5. Velocity of E.M waves in vaccum depends on permittivity and permeability of free space.
6. Electromagnetic waves carry energy and momentum.
7. Electromagnetic waves exert pressure when they strike a surface.

Question 2.
What is Greenhouse effect and its contribution towards the surface temperature of earth ?
Answer:
Green house effect: Temperature of the earth increases due to the radiation emitted by
the earth is trapped by atmospheric gases like CO₂, CH₄, N₂, chlorofluoro carbons etc., is called green house effect.

1. Radiation from the sun enters the atmosphere and heat the objects on the earth. These heated objects emit infrared rays.
2. These rays are reflected back to Earth’s surface and trapped in the Earth’s atmosphere. Due to this temperature of the earth increases.
3. The layers of carbon dioxide (CO₂) and low lying clouds prevent infrared rays to escape Earth’s atmosphere.
4. Since day-by-day the amount of carbondioxide in the atmosphere increases, more • infrared rays are entrapped in the atmosphere.
5. Hence the temperature of the Earth’s surface increases day by day.

Problems

Question 1.
A plane electromagnetic wave travels in vaccum along z-direction. What can you say about the directions of its electric and magnetic field vectors ? If the frequency of the wave is 30 MHz. What is its wavelength ?
Answer:
Electric and magnetic fields \(\overline{\mathrm{E}}\) and \(\overline{\mathrm{B}}\) of an electromagnetic wave must be perpendicular to the propagation of electromagnetic wave. Hence they lie in X – Y plane mutually perpendicular to each other.
Frequency of wave, v = 30MHz = 30 × 10⁶Hz.; Velocity of light, C = 3 × 10⁸m/s
Wavelength of the wave, λ = \(\frac{C}{V}=\frac{3 \times 10^8}{30 \times 10^6}\) = 10m

Question 2.
A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator ?
Answer:
According to Maxwell, a charged particle oscillating with a frequency produces electro-magnetic waves of same frequency. Hence frequency of EM waves produced is, 10⁹Hz.

Textual Examples

Question 1.
A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0, It is connected for charging in series with a resistor R = 1 M Q across a 2V battery (fig)- Calculate the magnetic field at a point P halfway between the centre and the periphery of the plates, after t = 10^-3 s. (The charge on the capacitor at time τ is q (t) = CV [1 – exp (-t/τ)], where the time constant τ is equal to CR.).
Solution:
The time constant of the CR circuit is τ = CR = 10^-3s. Then, we have
q(t) = CV [1 – exp (-t/τ) ]
= 2 × 10^-9 [1 – exp (-t /10^-3)
The electric field in between the plates at time t is
E = \(\frac{q(t)}{\varepsilon_0 A}=\frac{q}{\pi \varepsilon_0}\); A = π (1)² m² = area of the plates.

Consider now a circular loop of radius (1/2)m parallel to the plates passing through R The magnetic field B at all points on the loop is along the loop arid of the same value.
The flux Φ_E through this loop is
The flux Φ_E = E × area of the loop
= E × π × (\(\frac{1}{2}\))² = \(\frac{\pi \mathrm{E}}{4}=\frac{\mathrm{q}}{4 \varepsilon_0}\)
The displacement current
i_d = e₀ \(\frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\) = \(\frac{1}{4} \frac{\mathrm{dq}}{\mathrm{dt}}\) = 0.5 × 10^-6 exp (-1)
at t = 10^-3s. Now, applying Ampere-Maxwell law to the loop, we get
B × 2π × (\(\frac{1}{2}\)) = m₀.(i_c + i_d) = m₀(0 + i_d) = 0.5 × 10^-6 m₀ exp(-1)
or, B = 0.74 × 10^-13 T.

Question 2.
A plane electromagnetic wave of frequency 25 MHz travels in free space along the x – direction. At a particular point in space and time, E = \(6.3 \hat{\mathbf{j}}\) V/m. What is B at the point ?
Solution:
Using Eq. B₀ = [E₀/c] the magnitude of B is
B = \(\frac{\mathrm{E}}{\mathrm{c}}\)
= \(\frac{6.3 \mathrm{~V} / \mathrm{m}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}}\) = 2.1 × 10^-8 T
To find the direction, we note that E is along y-direction and the wave propagates along x- axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction. Since, \((+\hat{\mathrm{j}})\) × \((+\hat{\mathrm{k}})\) = i, B is along the z-direction. Thus. B = 2.1 × 10^-8 \(\hat{\mathrm{k}}\) T

Question 3.
The magnetic field in a plane electromagnetic wave is given by
B_y = 2 × 10^-7 sin (0.5 × 10³ × + 1.5 × 10¹¹ t)T.
a) What is the wavelength and frequency of the wave ?
b) Write an expression for the electric field.
Solution:
a) Comparing the given equation with
B_y = B₀ sin \(\left[2 \pi\left(\frac{\mathrm{x}}{\lambda}+\frac{\mathrm{t}}{\mathrm{T}}\right)\right]\)
We get, λ = \(\frac{2 \pi}{0.5 \times 10^3}\) m = 1.26 cm,
and \(\frac{1}{\mathrm{~T}}\) = v = (1.5 × 10¹¹)/2π = 23.9 GHz

b) E₀ = B₀C = 2 × 10^-7 T × 3 × 10⁸ m/s = 6 × 10¹ V/m
The electric field component is perpendicular to the direction of propagation and the di-rection of magnetic field. Therefore, the electric field component along the z-axis is obtained as E_z = 60 sin (0.5 × 10³x + 1.5 × 10¹¹t) V/m.

Question 4.
Light with an energy flux of 18 W/cm² falls on a non reflecting surface at normal incidence. If the surface has an area of 20 cm² find the average force exerted on the surface during a 30 minute time span.
Solution:
The total energy falling on the surface is
U = (18 W/cm²) × (20 cm²) × (30 × 60)
= 6.48 × 10⁵ J
Therefore, the total momentum delivered (for complete absorption) is
P = \(\frac{\mathrm{U}}{\mathrm{C}}=\frac{6.48 \times 10^5 \mathrm{I}}{3 \times 10^8 \mathrm{m} / \mathrm{s}}\) = 2.16 × 10^-3 kg m/s
The average force exerted on the surface is
F = \(\frac{\mathrm{p}}{\mathrm{t}}=\frac{2.16 \times 10^{-3}}{0.18 \times 10^4}\) = 1.2 × 10^-6 N

Question 5.
Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3m. Assume that the efficiency of the bulb is 2.5% and it is a point source.
Solution:
The bulb, as a point source, radiates light in all directions uniformly. At a distance of 3m, the surface area of the surrounding sphere is A = 4 πr² = 4π (3)² = 113m²
The intensity at this distance is
I = \(\frac{\text { Power }}{\text { Area }}=\frac{100 \mathrm{~W} \times 2.5 \%}{113 \mathrm{~m}^2}\) = 0.022 W.m²
Half of this intensity is provided by the electric field and half by the magnetic field.
\(\frac{1}{2} I=\frac{1}{2}\left(\varepsilon_0 \mathrm{E}_{\mathrm{rms}}^2 \mathrm{C}\right)\)
= \(\frac{1}{2}\) (0.022 W/m²)
E_rms = \(\sqrt{\frac{0.022}{\left(8.85 \times 10^{-12}\right)\left(3 \times 10^8\right)}}\) V/m = 2.9 V/m
The value of E found above is the root mean square value of the electric field. Since the electric field in a light beam is sinusoidal, the peak electric field, E₀ is
E₀ = \(\sqrt{2}\)E_rms = \(\sqrt{2}\) × 2.9 V/m
= 4.07 V/m
Thus, you see that the electric field strength of the light that you use for reading is fairly large. Compare it with electric field strength of TV or FM waves, which is of the order of a few microvolts per metre.
Now, let us calculate the strength of the magnetic field. It is
B_rms = \(\frac{E_{\mathrm{rms}}}{\mathrm{C}}=\frac{2.9 \mathrm{Vm}^{-1}}{3 \times 10^8 \mathrm{~ms}^{-1}}\) = 9.6 × 10^-9 T.
Again, since tbs field in the light beam is sinusoidal, the peak magnetic field is B₀ = \(\sqrt{2}\) B_rms = 1.4 × 10^-8 T. Note that although the energy in the magnetic field is equal to the energy in the electric field, the magnetic field strength is evidently very weak.

Students get through AP Inter 2nd Year Physics Important Questions 12th Lesson Dual Nature of Radiation and Matter which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 12th Lesson Dual Nature of Radiation and Matter

Very Short Answer Questions

Question 1.
What are “cathode rays” ? [A.P. Mar. 17]
Answer:
Cathode rays are streams of fast moving electrons or negatively charged particles.

Question 2.
What important fact did Millikan’s experiment establish?
Answer:
Millikan’s experiment established that electric charge is quantised. That means the charge on any body (oil drop) is always an integral multiple of charge of an electron, i.e., Q = ne.

Question 3.
What is “work function” ? [T.S. Mar. 17, 15]
Answer:
The minimum energy required to liberate an electron from photometal surface is called work function, Φ₀.

Question 4.
What is “photoelectric effect” ?
Answer:
When light of sufficierit energy is incident on the photometal surface, electrons are emitted. This phenomenon is called photoelectric effect.

Question 5.
Give examples of “photosensitive substances”. Why are they called so ?
Answer:
Examples of photosensitive substances are Li, Na, K, Rb and Cs etc.
The work function of alkali metals is very low. Even the ordinary visible light, alkali metals can produce photoelectric emission. Hence they are called photosensitive substances.

Question 6.
Write down Einstein’s photoelectric equation. [A.P. Mar. 15]
Answer:
Einstein’s photoelectric equation is given by K_max = \(\frac{1}{2}\) mv²_max = hυ – Φ₀

Question 7.
Write down de-Broglie’s relation and explain the terms therein. [T.S. & A.P. Mar. 16]
Answer:
The de-Broglie wave length (λ) associated with a moving particle is related to its momentum (p) is λ = \(\frac{h}{p}=\frac{h}{m v}\), where h is planck’s constant.

Question 8.
State Heisenberg’s Uncertainity Principle. [Mar. 14]
Answer:
Uncertainity principle states that “it is impossible to measure both position (∆x) and momentum of an electron (∆p) [or any other particle] at the same time exactly”, i.e., ∆x . ∆p ≈ h where ∆x is uncertainty in the specification of position and ∆p is uncertainty in the specification of momentum.

Question 9.
The photoelectric cut off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ? [Mar. 11]
Answer:
Cut off voltage, V₀ = 1.5 V; Maximum kinetic energy, (KE)_max = eV₀ = e × 1.5 = 1.5eV

Question 10.
An electron, an α-particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength ? [T.S. Mar. 15]
Answer:
For a particle,
de Broglie wavelength, λ = h/p
Kinetic energy, K = p²/2m
Then, λ = h/\(\sqrt{2 \mathrm{mK}}\)
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses.A proton is 1836 times massive than an electron and an a-particle \(\left({ }_2^4 \mathrm{He}\right)\) four times that of a proton.
Hence, α-particle has the shortest de Broglie wavelength.

Qeustion 11.
What is-the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts ? [A.P. Mar. 15]
Solution:
Accelerating potential V = 100 V The de Broglie wavelngth λ is
λ = h/p = \(\frac{1.227}{\sqrt{\mathrm{V}}}\) nm
λ = \(\frac{1.227}{\sqrt{100}}\) nm = 0.123 nm
The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.

Question 12.
If the wave length of electro magnetic radiation is doubled, what happens to energy of photon ? [IPE 2015 (TS)]
Answer:
E = \(\frac{\mathrm{hc}}{\lambda}\) ⇒ E ∝ \(\frac{1}{\lambda}\), since wave length of photon is doubled, its energy becomes halved.

Question 13.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10^-15 V s. Calculate the value of Planck’s constant.
Solution:
Given, slope of graph tan θ = 4.12 × 10^-15 V – s;

For slope of graph, tan θ = \(\frac{\mathrm{V}}{\mathrm{v}}\)
We know that hv = eV
\(\frac{\mathrm{V}}{\mathrm{v}}=\frac{\mathrm{h}}{\mathrm{e}} \Rightarrow \frac{\mathrm{h}}{\mathrm{e}}\) = 4.12 × 10^-15; h = 4.12 × 10^-15 × 1.6 × 10^-19 = 6.592 × 10^-34 J – s.

Question 14.
The energy flux of sunlight reaching the surface of the earth is 1.388 × 10³ W/m². How many photons (nearly) per square metre are incident on the Earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Solution:
Given, P = 1.388 × 10³ W/m²; λ = 550 nm = 550 × 10^-9 m ,
h = 6.63 × 10^-34 J-s; c = 3 × 10⁸ m/s
Energy of each photon E = \(\frac{\text { hc }}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}}\) = 3.616 × 10^-19 J
No. of photons incident on the earth’s surface, N = \(\frac{\mathrm{P}}{\mathrm{E}}=\frac{1.388 \times 10^3}{3.66 \times 10^{-19}}\)
∴ N = 3.838 × 10²¹ photons/m² – s.

Question 15.
Show that the wavelength of electromagnetic radiation is equal to the de-Brogile wavelength of its quantum (photon). [Mar. 14]
Answer:
Wave length of electromagnetic wave of frequency v and velocity C is given by,
λ = \(\frac{\mathrm{C}}{\mathrm{v}} \Rightarrow \lambda=\frac{\mathrm{C}}{\mathrm{v}} \times \frac{\mathrm{h}}{\mathrm{h}}=\frac{\mathrm{h}}{\left(\frac{\mathrm{hv}}{\mathrm{C}}\right)}=\frac{\mathrm{h}}{\mathrm{p}}\)    (∵ \(\frac{\mathrm{hv}}{\mathrm{C}}\) = p)
Hence, we can say wavelength of electromagnetic radiation is equal to the de-Brogile wavelength.

Sample Problem :

Question 1.
Calculate the (a) momentum and (b) dE-Brogile wavelength of the electrons accelerated through a potential difference of 56 V. [Mar. 14]
Answer:
a) Mass of the electron, m = 9 × 10^-31 kg;
Potential difference, V = 56V
Momentum of electron, mv = \(\sqrt{2 \mathrm{eVm}}=\sqrt{2 \times\left(1.6 \times 10^{-19}\right) \times 56 \times 9 \times 10^{-31}}\) = 4.02 × 10^-24kg ms^-1

b) de-Broglie wavelength, λ = \(\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.62 \times 10^{-34}}{4.04 \times 10^{-24}}\) = 1.64 × 10^-10m

Short Answer Questions

Question 1.
Describe an experiment to study the effect of frequency of incident radiation on ‘stopping potential’.
Answer:
Experimental study of the effect of frequency of incident radiation on stopping potential:

1. The experimental set up is shown in fig.
   
2. Monochromatic light of sufficient energy (E = hv) from source ‘s’ is incident on photosensitive plate ‘C’ (emitter), electrons are emitted by it.
3. The electrons are collected by the plate A (collector), by the electric field created by the battery.
4. The polarity of the plates C and A can be reversed by a commutator.
5. For a particular frequency of incident radiation, the minimum negative (retarding) potential V0 given to the plate A for which the photo current stops or becomes zero is called stopping potential.
6. The experiment is repeated with different frequencies, and their different stopping potential are measured with voltmeter.
7. From graph, we note that
   
   • The values of stopping potentials are different for different frequencies.
   • The value of stopping potential is more negative for radiation of highef incident frequency.
   • The value of saturation current depends on the intensity of incident radiation but it is independent of the frequency of the incident radiation.

Question 2.
What is the deBroglie wavelength of a ball of mass 0.12 Kg moving with a speed of 20 ms^-1? What can we infer from this result ?
Answer:
Given, m = 0.12 kg; υ = 20 m/s; h = 6.63 × 10^-34 J-s;
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.63 \times 10^{-34}}{0.12 \times 20}=\frac{6.63 \times 10^{-34}}{2.4}\)
∴ λ = 2.762 × 10^-34 m = 2762 × 10^-21 Å.
The wave length of ball is very very small. Hence, its motion can be observed.

Question 3.
What is the effect of (i) intensity of light (ii) potential on photoelectric current ?
Answer:
(i) Effect of intensity of light on photoelectric current:
1) When the intensity (I) of incident light, with frequency greater than the threshold frequency (υ > υ₀) is increased then the number of photoelectrons emitted increases i.e., the value of photoelectric current (i) increases, ie., i ∝ I.

ii) The effect of potential on photoelectric current:

1. On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
   
2. On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential.
3. Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.

Sample Problem :

Question 1.
The work function of caesium metal is 2.14 eV When light of frequency 6 × 10¹⁴ Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) stopping potential and (c) maximum speed of the emitted photoelectrons ? [A.P. Mar. 16]
Solution:
Given, Φ₀ = 2.14 eV; v = 6 × 10¹⁴ Hz
a) KE_max = hv – Φ₀ = \(\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}\) – 2.14
∴ KE_max = 0.35 eV

b) KE_max = eV₀ ⇒ 0.35 eV = eV₀
∴ V₀ = 0.35 V

Long Answer Questions

Question 1.
How did Einstein’s photoelectric equation explain the effect of intensity and potential on photoelectric current ? How did this equation account for the effect of frequency of ‘ incident light on stopping potential ?
Answer:

1. Einstein postulated that a beam of light consists of small energy packets called photons or quanta.
2. The energy of photon is E = hv. Where ‘h1 is Planck’s constant; v is frequency of incident light (or radiation).
3. If the absorbed energy of photon is greater than the work function (Φ₀ = hυ₀), the electron is emitted with maximum kinetic energy i.e., k_max = \(\frac{1}{2} \mathrm{mv}_{\max }^2\) = eV₀ = hv – Φ₀. This equation is known as Einstein’s photoelectric equation.
4. Effect of intensity of light on photoelectric current:
   When the intensity (I) of incident light, with frequency greater than the threshold frequency (υ > υ₀) is increased then the number of photoelectrons emitted decreases i.e., the value of photoelectric current (i) increases, ie., i ∝ I.
   
5. The effect of potential on photoelectric current:
   • On increasing the positive potential on collecting electrode, the photoelectric current increases. At a particular positive potential, the photocurrent becomes maximum which is known as saturated current.
     
   • On increasing the value of negative potential on collecting electrode, the photoelectric current gradually goes on decreasing. At a particular negative potential the value of photoelectric current becomes zero. This is known as stopping potential (v₀).
   • Stopping potential does not depend on the intensity of incident light. On increasing intensity, the value of saturated current increases, whereas the stopping potential remains unchanged.
6. The effect of frequency of incident radiation on stopping potential:
   On increasing the frequency of incident light, the value of stopping potential goes on increasing gradually as shown in fig. That means k_max increases eV₀ also increases.
   
7. From the graph, we note that
   • For a given photosensitive metal, the cut off potential (v₀) varies linearly with the frequency of the incident radiation.
   • For a given photosensitive metal, there is a certain minimum cut off frequency υ₀ (called threshold frequency) for which the stopping potential is zero.
     
8. From the graph we note that
   • The value of cut-off potential is different for radiation of different frequency.
   • The value of stopping potential is more negative for radiation of higher incident frequency.
9.  From above experiments, it is found that, if the incident radiation is of higher frequency than that of threshold frequency, the photoelectric emission is. possible.

Textual Examples

Question 1.
Monochromatic light of frequency 6.0 × 10¹⁴ Hz is produced by a laser. The power emitted is 2.0 × 10^-3 W. (a) What is the energy of a photon in the light beam ? (b) How many photons per second, on an average, are emitted by the source ?
Solution:
a) Each photon has an energy
E = hv = (6.63 × 10^-34 J s) (6.0 × 10¹⁴ Hz)
= 3.98 × 10^-19 J

b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = NE.
Then N = \(\frac{\mathrm{P}}{\mathrm{E}}=\frac{2.0 \times 10^{-3} \mathrm{~W}}{3.98 \times 10^{-19} \mathrm{~J}}\)
= 5.0 × 10¹⁵ photons per second.

Question 2.
The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. [A.P. Mar. 16]
Solution:
a) For the cut-off or threshold frequency, the energy hv₀ of the incident radiation must be equal to work function Φ₀, so that
v₀ = \(\frac{\phi_0}{\mathrm{~h}}=\frac{2.14 \mathrm{eV}}{6.63 \times 10^{-34} \mathrm{Js}}\)
= \(\frac{2.14 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}\) = 5.16 × 10¹⁴ Hz
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.

b) Photocurrent reduces to zero, when maximum kinetic energy of the emitted photoelectrons equals the potential energy e V₀ by the retarding potential V₀. Einstein’s Photoelectric equation is
eV₀ = hv – Φ₀ = \(\frac{\mathrm{hc}}{\lambda}\) – Φ₀
or λ = hc/(eV₀ + Φ₀)
= \(\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{(0.60 \mathrm{eV}+2.14 \mathrm{eV})}\)
= \(\frac{19.89 \times 10^{-26} \mathrm{~J} \mathrm{~m}}{(2.74 \mathrm{eV})}\)
λ = \(\frac{19.89 \times 10^{-26} \mathrm{~J} \mathrm{~m}}{2.74 \times 1.6 \times 10^{-19} \mathrm{~J}}\) = 454 nm

Question 3.
The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellow-green colour and about 760 nm for red colour.
(a) What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum ? (Take h = 6.63 × 10^-34 J s and 1 eV = 1.6 × 10^-19 J.)
(b) From which of the photosensitive materials with work functions listed in table and using the results of (i), (ii) and (iii) of (a) can you build a photoelectric device that operates with visible light ?

Solution:
a) Energy of the incident photon,
E = hv = hc/λ
E = (6.63 × 10^-34 J s) (3 × 10⁸ m/s)/λ.
= \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{\lambda}\)
i) For violet light,
λ₁ = 390 nm (lower wavelength end)
Incident photon energy,
E₁ = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{390 \times 10^{-9} \mathrm{~m}}\)
5.10 × 10^-19 J = \(\frac{5.10 \times 10^{-19} \mathrm{~J}}{1.6 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\)
= 3.19 eV

ii) For yellow-green light,
λ₂ = 550 nm (average wavelength)
Incident photon energy,
E₂ = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{550 \times 10^{-9} \mathrm{~m}}\)
= 3.62 × 10^-19 J = 2.26 eV.

iii) For red light,
λ₃ = 760 nm (higher wavelength end)
Incident photon energy,
E₃ = \(\frac{1.989 \times 10^{-25} \mathrm{~J} \mathrm{~m}}{760 \times 10^{-9} \mathrm{~m}}\)
= 2.62 × 10^-19 J = 1.64 eV

b) For a photoelectric device to operate, we require incident light energy E to be equal to or greater than the work function Φ₀ of the material. Thus, the photoelectric device will operate with violet light (with E = 3.19 eV) photosensitive material Na (with Φ₀ = 2.75 eV), K (with Φ₀ = 2.30 eV) and Cs (with Φ₀ = 2.14 eV). It will also operate with yellow-green light (with E = 2.26 eV) for Cs (with Φ₀ = 2.14 eV) only. However, it will not operate with red light (with E = 1.64 eV) for any of these photosensitive materials.

Question 4.
What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4 × 10⁶ m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s ?
Solution:
a) For the electron :
Mass m= 9.11 × 10^-31 kg, speed υ = 5.4 × 10⁶ m/s. Then, momentum
P = mυ = 9.11 × 10^-31 (kg) × 5.4 × 10⁶ (m/s)
P = 4.92 × 10^-24 kg m/s
de Broglie wavelength, λ = h/p
= \(\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{4.92 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}}\)
λ = 0.135 nm

b) For the ball:
Mass m’ = 0.150 kg, speed υ’ = 30.0 m/s.
Then momentum
p’ = m’υ’ = 0.150 (kg) × 30.0 (m/s)
p’ = 4.50 kg m/s
de Broglie wavelength λ’ = h/p’.
= \(\frac{6.63 \times 10^{\pm 34} \mathrm{Js}}{4.50 \times \mathrm{kg} \mathrm{m} / \mathrm{s}}\)
λ’ = 1.47 × 10^-34 m
The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10^-19 times the size of the proton, quite beyond experimental measurement.

Question 5.
An electron, an a-particle and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength ? [T.S. Mar. 15]
Solution:
For a particle,
de Broglie wavelength, λ = h/p
Kinetic energy, K = p²/2m
Then, λ = h /\(\sqrt{2 \mathrm{mK}}\)
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses. A proton \(\left({ }_1^1 \mathrm{He}\right)\) is 1836 times massive than an electron and an a-particle \(\left({ }_2^4 \mathrm{He}\right)\) four times that of a proton
Hence, α-particle has the shortest de Broglie wavelength.

Question 6.
A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10^-4. Calculate the particle’s mass and identify the particle.
Solution:
de Broglie wavelength of a moving particle, having mass m and velocity υ :
λ = \(\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}\)
Mass, m = h/A.
For an electron, mass m_e = h/λ_e υ_e
Now, we have υ/υ_e = 3 and
λ/λ_e = 1.813 × 10^-4
Then, mass of the particle,
m = m_e \(\left(\frac{\lambda_{\mathrm{e}}}{\lambda}\right)\left(\frac{v_{\mathrm{e}}}{v}\right)\)
m = (9.11 × 10^-31 kg) × (1/3) × (1/1.813 × 10^-4)
m = 1.675 × 10^-27 kg.
Thus, the particle, with this mass could be a proton or a neutron.

Question 7.
What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts? [A.P. Mar. 15]
Solution:
Accelerating potential V = 100 V The de Broglie wavelength λ is
λ = h/p = \(\frac{1.227}{\sqrt{\mathrm{V}}}\) nm
λ = \(\frac{1.227}{\sqrt{100}}\) nm = 0.123 nm
The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 1st Lesson Waves Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 1st Lesson Waves

Very Short Answer Questions

Question 1.
What does a wave represent ?
Answer:
A wave represent the transport of energy through a medium from one point to another without translation of the medium.

Question 2.
Distinguish between transverse and longitudinal waves.
Answer:
Transverse waves

1. The particles of the medium vibrate perpendicular to the direction of wave propagation.
2. Crests and troughs are formed alternatively.

Longitudinal waves

1. The particles of the medium vibrate parallel to the direction of wave propagation.
2. Compressions and rare fractions are formed alternatively.

Question 3.
What are the parameters used to describe a progressive harmonic wave ?
Answer:
Progressive wave equation is given y = a sin (ωt – kx)
Where ω = 2πv = \(\frac{2 \pi}{T}\)

Parameters :

1. a = Amplitude
2. λ = Wavelength
3. T = Time period
4. v = Frequency
5. k = Propagation constant
6. ω = Angular frequency.

Question 4.
Obtain an expression for the wave velocity in terms of these parameters.
Answer:
Let ‘v’ be the velocity of a wave, ‘v’ be frequency and ‘λ’ be the wavelength. If T is the time period, then v = \(\frac{1}{\mathrm{~T}}\)
The distance travelled by the wave in the time T = λ.
Distance travelled in one second = \(\frac{\lambda}{T}\)
which is equal to wave velocity v = \(\frac{\lambda}{T}\).
∴ v = vλ

Question 5.
Using dimensional analysis obtain an expression for the speed of transverse waves in a stretched string.
Answer:
Wave velocity v ∝ T^a µ^b ⇒ V = K T^a µ^b ——-> (1)
Dimensions of v = M⁰L¹ T^-1, Tension T = M¹L¹T^-2,
Linear mass µ = M¹L^-1, Constant K = M⁰L⁰T⁰
Now (1) becomes M⁰L¹T^-1 = [M¹L¹T^-2]^a [M¹L^-1]^b
M⁰L¹T¹ = M^{a + b}L^a-bT^-2a Comparing the powers of same physical quantity.
-1 = -2a ⇒ a = \(\frac{1}{2}\)
a + b = 0 ⇒ b = –\(\frac{1}{2}\)
⇒ v = (1)\(T^{\frac{1}{2}} \mu^{\frac{1}{2}}\) [∵ K = 1 Practically]
∴ v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\)

Question 6.
Using dimensional analysis obtain an expression for the speed of sound waves in a medium. .
Answer:
Speed of sound v ∝ B^a ρ^b ⇒ v = K B^a ρ^b ——–> (1)
Dimensions of v = M⁰L¹T^-1,
Elasticity of medium,
B = M¹L^-1T^-2, density ρ = M¹L^-3, constant K = M⁰L⁰T⁰.
Now (1) becomes M⁰L¹T^-1 = M⁰L⁰T⁰ [M¹L^-1T^-2]^a [M¹L^-3]^b
0 = a + b
1 = -a – 3b
-1 = -2a ⇒ a = \(\frac{1}{2}\)
b = –\(\frac{1}{2}\)
v = K \(B^{\frac{1}{2}} \rho^{\frac{1}{2}}\)
∴ v = \(\sqrt{\frac{B}{\rho}}\) [∵ K = 1, practically]

Question 7.
What is the principle of superposition of waves ?
Answer:
When two or more waves, are acting simultaneously on the particle of the medium, the resultant displacement is equal to the algebraic sum of individual displacements of all the waves. This is the principle of superposition of waves.

If y₁, y₂, …… y_n be the individual displacements of the particles,then resultant displacement y = y₁ + y₂ + ……. + y_n

Question 8.
Under what conditions will a wave be reflected ?
Answer:

1. When the medium ends abruptly at any point.
2. If the density and rigidity modulus of the medium changes at any point.

Question 9.
What is the phase difference between the incident and reflected waves when the wave is reflected by a rigid boundary.
Answer:
π Radian or 180°.

Question 10.
What is a stationary or standing wave ?
Answer:
When two identical progressive (Transverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are superimposed then the resultant wave is called stationary waves or standing wave.

Question 11.
What do you understand by the terms ‘node ‘and’ antinode’?
Answer:
Node : The points at which the amplitude is zero, are called nodes.
Antinodes : The points at which the amplitude is maximum, are called antinodes.

Question 12.
What is the distance between a node and an antinode in a stationary wave ?
Answer:
The distance between node and antinode is \(\frac{\lambda}{4}\)

Question 13.
What do you understand by ‘natural frequency’ or ‘normal mode of vibration’ ?
Answer:
When a body is set into vibration and then left to itself, the vibrations made by it are called natural or free vibrations. Its frequency is called natural frequency or normal mode of vibration.

Question 14.
What are harmonics ?
Answer:
The frequencies in which the standing waves can be formed are called harmonics.
(Or)
The integral multiple of fundamental frequencies are called harmonics.

Question 15.
A string is stretched between two rigid supports. What frequencies of vibration are possible in such a string ?
Answer:
The possible frequencies of vibrations in a stretched string between two rigid supports is given by
v_n = (n + \(\frac{1}{2}\))\(\frac{v}{21}\) where n = 0, 1, 2, 3, ……

Question 16.
The air column in a long tube, closed at one end, is set in vibration. What harmonics are possible in the vibrating air column ?
Answer:
The possible harmonics in the vibrating air column of a long closed tube is given by
v_n = [2n + 1]\(\frac{v}{4 l}\) where n = 0, 1, 2, 3, ……..

Question 17.
If the air column in a tube, open at both ends, is set in vibration; what harmonics are possible?
Answer:
The possible harmonics in vibrating air column of a long open tube is given by
V_n = \(\frac{n v}{21}\)
where n = 1, 2, 3, ……….

Question 18.
What are ‘beats’ ?
Answer:
Beats : When two sound notes of nearly frequency travelling in the same direction and interfere to produce waxing and waning of sound at regular intervals of time is called “Beats”.

Question 19.
Write down an expression beat frequency and explain the terms there in.
Answer:
Expression of beat frequency, Δv = v₁ ~ v₂
where v₁ and v₂ are the frequencies of two waves.

Question 20.
What is ‘Doppler effect’? Give an example.
Answer:
Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between source of sound and observer is called “Doppler effect”.

E.g.: When the whistling railway engine approaches the stationary observer on the platform, the frequency of sound appears to increase above the actual frequency. When it moves away from the observer, the apparent frequency decreases.

Question 21.
Write down an expression for the observed frequency when both source and observer are moving relative to each other in the same direction.
Answer:
Apparent frequency of sound heard by an observer,
v’ = \(\left[\frac{v-v_0}{v-v_s}\right] v\)

where v = frequency of sound
v = velocity of sound
v₀ = velocity of observer
v_S = velocity of source

Short Answer Questions

Question 1.
What are transverse waves ? Give illustrative examples of such waves.
Answer:
Transverse waves: In a wave motion, the vibration of the particles and the direction of the propagation of the waves are perpendicular to each other, the waves are said to be transverse waves.

Illustration:

1. Waves produced in the stretched strings are transverse.
2. When a stretched string is plucked, the waves travel along the string.
3. But the particles in the string vibrate in the direction perpendicular to the propagation of the wave.
4. They can propagate only in solids and on the surface of the liquids.
5. Ex : Light waves, surface water waves.

Question 2.
What are longitudinal waves ? Give illustrative example of such waves.
Answer:
Longitudinal waves : In a wave motion, the direction of the propagation of the wave and vibrations of particles are in the same direction, the waves are said to be longitudinal waves.
Illustration:

1. Longitudinal waves may be easily illustrated by releasing a compressed spring.
2. A series of compressions and rarefactions (expansions) propagate along the spring.
   
   C = Compression; R = Rarefaction.
3. They can travel in solids, liquids and gases.
4. Ex : Sound waves.

Question 3.
Write an expression for a progressive harmonic wave and explain the various parameters used in the expression.
Answer:
The expression of a progressive harmonic wave is written as y = a sin(ωt – \(\frac{2 \pi}{\lambda} x\))
or y = a sin(ωt – kx) where ω = 2πv, k = \(\frac{2 \pi}{\lambda}\)

Parameters:

1. Amplitude (a) : It is the maximum displacement of a vibrating particle from its mean position.
2. Frequence (v): It is the number of complete vibrations made by a vibrating body in one second.
3. Wave length (λ) : It is defined as the distance covered by a wave while it completes one vibration, (or) It is the distance between two consecutive points in the same phase.
4. Phase of vibration (ϕ) : The phase of vibration of a vibrating particle gives its state of displacement at a given instant. It is generally given by the phase angle.

Question 4.
Explain the modes of vibration of a stretched string with examples.
Answer:
Modes of vibrations of a stretched string :

1. In sitar or Guitar, a stretched string can vibrate, in different frequencies and form stationary waves. This mode of vibrations are known as harmonics.
2. If it vibrates in one segment, which is known as fundamental harmonic. The higher harmonics are called the overtones.
   
3. It vibrates in two segments then the second harmonic is called first overtone. Similarly the pattern of vibrations are shown fig.
4. If a stretched string vibrates with P ’Seg’ ments (loop) then frequency of vibration v = \(\frac{\mathrm{P}}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\) where T = tension in the string, µ = linear density = \(\frac{\text { mass }}{\text { length }}\)
5. In first mode of vibration, P = 1, then v = \(\frac{1}{2l} \sqrt{\frac{T}{\mu}}\) (1^st hamonic (or) fundamental frequency)
6. second mode of vibration, P = 2, then v₁ = \(\frac{2}{2l} \sqrt{\frac{T}{\mu}}\) = 2v (2^nd harmonic (or) 1^st overtone)
7. In third mode of vibration, P = 3, then v₂ = \(\frac{3}{2l} \sqrt{\frac{T}{\mu}}\) = 3v (3^rd harmonic (or) 2^nd overtone)
   The ratio of the frequency of Harmonics are, v : v₁ : v₂ = v : 2v : 3v = 1 : 2 : 3

Question 5.
Explain the modes of vibration of an air column in an open pipe.
Answer:
Modes of vibration of an air column in an open pipe :
1) For a open pipe both the ends are open. So antinodes will be formed at both the ends. But two antinodes cannot exist without a node between them.
2) The possible harmonics in vibrating air column of a open pipe is given by .
Where n = 1, 2, 3
(1^st harmonic or fundamental frequence)

3) In first normal Mode of vibrating air column in a open pipe v₁ = \(\frac{v}{2l}\) = v
(2nd harmonic 1^st overtone)

4) In second normal Mode of vibrating air column in a open pipe, v₂ = \(\frac{2 v}{2l}\) = 2v

5) In third, normal Mode of vibrating air column in a open pipe, v₃ = \(\frac{3 v}{21}\) = 3u
(3^rd harmonic 2^nd overtone)

6) In open pipe the ratio of frequencies of harmonics is
v₁ : v₂ : v₃ = v : 2v : 3v = 1 : 2 : 3

Question 6.
What do you understand by ‘resonance’ ? How would you use resonance to determine the velocity of sound in air ?
Answer:
Resonance: If the natural frequency of a vibrating body is equal to the frequency of external periodic force then the two bodies are said to be in resonance. At resonance the bodies will vibrate with increasing amplitude.

Determination of velocity of sound in air using resonance :

1) In resonance tube, an air column is made to vibrate by means of vibrating fork. At certain length of air column, the air column would have the same frequency as that of the fork. Then the air column vibrates with the maximum amplitude and the intense sound is produced.

2) The vibrating fork of known frequency (v) is placed above the open end of the tube.

3) The length of air column is gradually increased until the booming sound can be heard at two different lengths of air column.

4) In first resonance, l₁ be the length of air column, then \(\frac{\lambda}{4}\) = l₁ + C …….. (1)
Where λ is the wavelength of sound emitted by the fork and C is the end correction of the tube.

5) In second resonance, l₂ be the length of air column, then \(\frac{3 \lambda}{4}\) = l₂ + C …… (2)
(2) – (1) ⇒ \(\frac{\lambda}{2}\) = l₂ – l₁

λ = 2 (l₂ – l₁)
Speed of sound is given by
υ = vλ = v[2(l₂ – l₁)]
∴ υ = 2v (l₂ – l₁)

6) Hence by knowing v, l₁, l₂ the speed of sound is calculated.

Question 7.
What are standing waves ? Explain how standing waves may be formed in a stretched string.
Answer:
Standing wave or stationary: When two identical progressive (Transverse or longitudinal) waves travelling opposite directions in a medium along the same straight line, which are super-imposed then the resultant wave is called standing wave.

Formation of standing wave in a stretched string : –

1. If a string of length ‘l’ is stretched between two fixed points and set into vibration, a transverse progressive wave begins to travel along the string.
2. The wave is get reflected at the other fixed end.
3. The incident and reflected waves interfere and produce a stationary wave.
4. The stationary wave with nodes and antinodes is shown below.
   

Question 8.
Describe a procedure for measuring the velocity of sound in a stretched string.
Answer:
The velocity of a transverse wave travelling along a stretched string in fundamental mode is given by v = 2vl, where v = frequency, l = resonating length.

Measurement of velocity of sound in a stretched string using sonometer :

1. The wire is subjected to a fixed tension with suitable load.
2. A tuning fork of known frequency (v), is excited and the stem is held against the sono – meter box.
   
3. The distance between the two bridges is adjusted such that a small paper rider at the middle of B₁ B₂ vibrates vigorously and flies off due to resonance.
4. The resonating length ‘l’ can be measured between two bridges with scale.
5. By knowing v and l; we can find the velocity of a wave using υ = 2vl.

Question 9.
Explain, using suitable diagrams, the formation of standing waves in a closed pipe. How may this be used to determine the frequency of a source of sound ?
Answer:
Formation of standing waves in a closed pipe :

1. In closed pipe one end is closed and the other end is open. So antinode is formed at open end and antinode is formed at closed end.
2. The possible harmonics in vibrating air column in a closed pipe v_n = \(\frac{(2 n+1) v}{4 l}\) where v = 0, 1, 2, 3, …….
3. In first normal mode of vibrating air column in a closed pipe, v₁ = \(\frac{v}{41}\)
   [first harmonic (or) fundamental frequency]
   
4. In second normal mode of vibrating air column in a closed pipe,
   v₃ = \(\frac{3 \mathrm{v}}{4l}\) [Third harmonic (or) first overtone]
5. In third normal mode of vibrating air column in a closed pipe,
   v₅ = \(\frac{5 \mathrm{v}}{4 \mathrm{l}}\) [Fifth harmonic (or) second overtone]

Determination of frequency of a source of sound :

1. The vibrating fork of unknown frequency (v) is placed above the open end of the tube.
   
2. Reservoir is slowly lowered, until a large booming sound is heard. Measure 1st resonating, air column length l₁.
3. Further lower the reservoir, until second time a large booming sound is heard. Measure 2^nd resonating air column length l₂.
4. Velocity of a wave at 0°C is v = 331 m/s.
5. By knowing v, l₁ and l₂ we can find unknown frequency of a tuning fork using
   v = \(\frac{v}{2\left(l_2-l_1\right)}\)

Question 10.
What are ‘beats’ ? When do they occur ? Explain their use, if any.
Answer:
Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing (maximum) and waning (minimum) in the intensity of the resultant sound waves at regular intervals of time is called beats.

It two vibrating bodies have slightly difference in frequencies, beats can occur.
No. of beats can be heard Δυ = υ₁ – υ₂

Importance :

1. It can be used to tune musical Instruments.
2. Beats are used to detect dangerous gases

Explanation-for tuning musical instruments with beats :
Musicians use the beat phenomenon in tuning their musical instruments. If an instrument is sounded against a standard frequency and tuned until the beats disappear. Then the instrument is in tune with the standard frequency.

Question 11.
What is ‘Doppler effect’? Give illustrative examples.
Answer:
Doppler effect: The apparent change in the frequency heard by the observer due to relative motion between the observer and the source of sound is called doppler effect.

Examples:

1. The frequency of whistling engine heard by a person standing on the platform appears to increase, when the engine is approaching the platform and it appears to decrease when the engine is moving away from the platform.
2. Due to Doppler effect the frequency of sound emitted by the siren of an approaching ambulance appears to increase. Similarly the frequency of sound appears to drop when it is moving away.

Long Answer Questions

Question 1.
Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse wave in stretched strings. (A.P. Mar. ’19)
Answer:
A string is a metal wire whose length is large when compared to its thickness. A stretched string is fixed at both ends, when it is plucked at mid point, two reflected waves of same amplitude and frequency at the ends are travelling in opposite direction and overlap along the length. Then the resultant waves are known as the standing waves (or) stationary waves.

Let two transverse progressive waves of same amplitude a, wave length λ and frequency ‘v’, travelling in opposite direction be given by
y₁ = a sin (kx – ωt) and y₂ = a sin (kx + ωt)
where ω’ = 2πv and k = \(\frac{2 \pi}{\lambda}\)

The resultant wave is given by y = y₁ + y₂
y = a sin (kx – ωt) + a sin (kx + ωt)
y = (2a sin kx) cos ωt
2a sin kx = Amplitude of resultant wave.
It depends on ‘kx’. If x = 0, \(\frac{\lambda}{2}\), \(\frac{2 \lambda}{2}\), \(\frac{3 \lambda}{2}\),……. etc, the amplitude = zero
These positions are known as “Nodes”.
If x = \(\frac{\lambda}{4}\), \(\frac{2 \lambda}{2}\), \(\frac{3 \lambda}{2}\) ……… etc, the amplitude = zero

The positions are known as “Nodes”
If x = \(\frac{\lambda}{4}\), \(\frac{3 \lambda}{4}\), \(\frac{5 \lambda}{4}\) ……. etc, the amplitude = maximum (2a).
These positions are called “Antinodes”.

If the string vibrates in ‘P’ segments and ‘l’ is its length then length of each segment = \(\frac{l}{p}\)
Which is equal to \(\frac{\lambda}{2}\)
∴ \(\frac{l}{\mathrm{p}}\) = \(\frac{\lambda}{2}\) ⇒ λ = \(\frac{2 l}{\mathrm{P}}\)
Harmonic frequency v = \(\frac{v}{\lambda}=\frac{v \mathrm{P}}{2 l}\)
v = \(\frac{v P}{2 l}\) ——- (1)
If ‘ T’ is tension (stretching force) in the string and ‘μ’ is linear density then velocity of transverse wave (v) in the string is v = \(\sqrt{\frac{\mathrm{T}}{\mu}}\) —– (2)
From the Eqs (1) and (2)
Harmonic frequency v = \(\frac{p}{2 l} \sqrt{\frac{T}{\mu}}\)
If P = 1 then it is called fundamental frequency (or) first harmonic frequency
∴ Fundamental Frequency v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\) —— (3)

Laws of Transverse Waves Along Stretched String:

Fundamental frequency of the vibrating string v = \(\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mu}}\)

First Law : When the tension (T) and linear density (μ) are constant, the fundamental frequency (v) of a vibrating string is inversely proportional to its length.
∴ v ∝ \(\frac{1}{l}\) ⇒ vl = constant, when T and ‘μ’ are constant.

Second Law: When the length (I) and its, linear density (m) are constant the fundamental frequency of a vibrating string is directly proportional to the square root of the stretching force (T).
∴ v ∝ \(\sqrt{\mathrm{T}}\) ⇒ \(\frac{v}{\sqrt{T}}\) = constant, when ‘l’ and ‘m’ are constant.

Third Law: WHien the length (J) and the tension (T) are constant, the fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density (m).
∴ v ∝ \(\frac{1}{\sqrt{\mu}}\) ⇒ \(v \sqrt{\mu}\) = constant, when ‘l’ and ‘T’ are constant.

Question 2.
Explain the formation of stationary waves in an air column enclosed in open pipe. Derive the equations for the frequencies of the harmonics produced. (T.S. Mar. ’16, A.P. Mar. ’15)
Answer:
A pipe, which is opened at both ends is called open pipe. When a sound wave is sent through a open pipe, which gets reflected by the earth. Then incident and reflected waves are in same frequency, travelling in the opposite directions are super – imposed stationary waves are formed.

Harmonics in open pipe : To form the stationary wave in open pipe, which has two antinodes at two ends of the pipe with a node between them.
∴ The vibrating length (l) = half of the wavelength \(\left(\frac{\lambda_1}{2}\right)\)
l = \(\frac{\lambda_1}{2}\) ⇒ λ₁ = 2l
fundamental frequency v₁ = \(\frac{\mathrm{v}}{\lambda_1}\) where v is velocity of sound in air v₁ = \(\frac{v}{2 l}\) = v —— (1)
For second harmonic (first overtone) will have one more node and antinode than the fundamental.
If λ₂ is wavelength of second harmonic l = \(\frac{2 \lambda_2}{2}\) ⇒ λ₂ = \(\frac{2l}{2}\)
If ‘v₂‘ is frequency of second harmonic then v₂ = \(\frac{v}{\lambda_2}\) = \(\frac{v \times 2}{2 l}\) = 2v
v₂ = 2v —– (2)

Similarly for third harmonic (second overtone) will have three nodes and four antinodes as shown in above figure.
If λ₃ is wave length of third harmonic l = \(\frac{3 \lambda_3}{2}\)
λ₃ = \(\frac{2l}{3}\)
If ‘v₂’ is frequency of third harmonic then
v₃ = 3v —– (3)

Similarly we can find the remaining or higher harmonic frequencies i.e v₃, v₄ etc, can be determined in the same way.
Therefore the ratio of the harmonic frequencies in open pipe can be written as given below.
v : v₁ : v₂ = 1 : 2 : 3 ………

Question 3.
How are stationary waves formed in closed pipes ? Explain the various modes of vibrations and obtain relations for their frequencies. (A.P. & T.S. Mar. ’15)
Answer:
A pipe, which is closed at one end and the other is opened is called closed pipe. When a sound wave is sent through a closed pipe, which gets reflected at the closed end of the pipe. Then incident and reflected waves are in same frequency, travelling in the opposite directions are superimposed stationary waves are formed.

To form the stationary wave in closed pipe, which has atleast a node at closed end and antinode at open end of the pipe, it is known as first harmonic in closed pipe. Then length of the pipe (l) is equal to one fourth of the wave length.
∴ l = \(\frac{\lambda_1}{4}\) ⇒ λ₁ = 4l
If ‘v₁‘ is fundamental frequency then
v₁ = \(\frac{v}{\lambda_1}\) where ‘υ’ is velocity of sound in air
v₁ = \(\frac{v}{4l}\) = v ——- (1)

To form the next harmonic in closed pipe, two nodes and two antinodes should be formed. So that there is possible to form third harmonic in closed pipe. Since one more node and antinode should be included.
Then length of the pipe is equal to \(\frac{3}{4}\) of the wavelength.
∴ l = \(\frac{3 \lambda_3}{4}\) where ‘λ₃‘ is wave length of third harmonic.
λ₃ = \(\frac{4l}{3}\)
If ‘v₃‘ is third harmonic frequency (first overtone)
∴ v₃ = \(\frac{v}{\lambda_3}\) = \(\frac{3 v}{41}\)
v₃ = 3v ——- (2)

Similarly the next overtone in the close pipe is only fifth harmonic it will have three nodes and 3 antinodes between the closed end and open end.
Then length of the pipe is equal to \(\frac{5}{4}\) of wave length (λ₅)
∴ l = \(\frac{5 \lambda_5}{4}\) where ’λ₅‘ is wave length of fifth harmonic. .
λ₅ = \(\frac{4l}{5}\)
If ‘V₅‘ is frequency of fifth harmonic (second overtone)
V₅ = \(\frac{v}{\lambda_5}=\frac{5 v}{4 I}\)
v₅ = 5v —– (3)

∴ The frequencies of higher harmonics can be determined by using the same procedure. Therefore from the Eq (1), (2) and (3) only odd harmonics are formed.
Therefore the ratio of the frequencies of harmonics in closed pipe can be written as
v₁ : v₃ : v₅ = v : 3v : 5v
v₁ : v₃ : v₅ = 1 : 3 : 5

Question 4.
What are beats ? Obtain an expression for the beat frequency ? Where and how are beats made use of ?
Answer:
Beats : Two sound waves of nearly same frequency are travelling in the same direction and interfere to produce a regular waxing and waning in the intensity of the resultant sound waves at regular intervals of time are called Beats.
If v₁ and v₂ are the frequencies of two sound notes superimposed along the same direction, no of beats heard per second = Δv = v₁ – v₂.
Maximum no. of beats heard per sec is 10 due to persistence of hearing.

Expression for the beat frequency :

1. Consider the two wave trains of equal amplitude but of nearly equal frequencies.
2. Let the frequencies of the waves be v₁ and v₂. Say v₁ is slightly greater than v₂.
3. Let the beat period be T seconds.
4. No.of vibrations, made by the first wave train in T seconds – v₁T
   [∵ no.of oscillations in 1 sec = v]
   [∵ no.of oscillations in T sec = vt]
5. No.of vibrations, made by the second wave train in T seconds = v₂T
6. During the time interval T, the first wave train would have completed one vibration more than the second wave train.
7. Hence, v₁T – v₂T = 1 or v₁ – v₂ = \(\frac{1}{\mathrm{~T}}\)
8. Since, T is the beat period, no.of beats per seconds = \(\frac{1}{\mathrm{~T}}\)
9. Hence the beat frequency = \(\frac{1}{\mathrm{~T}}\) = v₁ – v₂ = Δv
10. That is the beat frequency is the difference between the frequencies of the two wave trains.

Practical applications of beats:

1. Determination of an unknown frequency: Out of two tuning forks, one is loaded with wax and the other is filed. The excited tuning forks are close together and no.of beats can be heard. Then after unknown frequencies of them will be found practically.
2. For tuning musical instruments : Musicians use the beat phenomenon in tuning their musical instruments.
3. For producing colourful effects in music: Sometimes, a rapid succession of beats is knowingly introduced in music. This produces an effect similar to that of human voice and is appreciated by the audience.
4. For detection of Marsh gas (dangerous gases) in mines.

Question 5.
What is Doppler effect? Obtain an expression for the apparent frequencý of sound heard when the source is in motion with respect to an observer at rest. (A.P. Mar. ’16, Mar. ’14)
Answer:
Doppler effect : The apparent change in the frequency heard by the observer due to the relative motion between the observer and the source of sound is called doppler effect.

When a whistling railway engine approaches an observer standing on the platform, the frequency of sound appears to increase. When it moves away the frequency appear to decrease.

Expression for apparent frequency when source is in motion and listener at rest:
Let S = Source of sound
O = listener

Let ‘S be the source, moving with a velocity ‘υ_s‘ towards the stationary listener.
The distance travelled by the source in time period T’ = υ_s. T
Therefore the successive compressions and rarefactions are drawn closer to listener.
∴ Apparent wavelength = λ’ = λ – υ_sT.

If ‘v’ “is apparent frequency heard by the listener
then v’ = \(\frac{v}{\lambda^{\prime}}\) where ‘υ’ is velocity of sound in air
v’ = \(\frac{v . V}{v-v_s}\)
Therefore the apparent frequency is greater than the actual frequency.
Similarly, if the source is away from the stationary listener then apparent frequency
v’ = \(\frac{v . V}{v+v_s}\), which is less than the actual frequency.

Limitation : Doppler effect is applicable when the velocities of the source and listener are much less than that of sound velocity

Question 6.
What is Doppler shift? Obtain an expression for the apparent frequency of sound heard when the observer Is In motion with respect to a source át rest.
Answer:
Doppler Shift: Due to the relative motion, when the source comes closer to listener, the apparent frequency is greater than actual frequency and source away from listener; the apparent frequency is less than actual frequency So the difference in apparent and actual frequencies is known as Doppler shift.
Expression for the apparent frequency heard by a moving observer:

Case (1) : When observer Is moving towards source:
Let ‘υ₀’ be velocity of listener ‘O’, moving towards the stationary source ‘s’ as shown in figure. So observer will receive more number of waves in each second.
The distance travelled by observer in one second = υ₀

The number of extra waves received by the observer = \(\frac{v_0}{\lambda}\)
We know v = vλ ⇒ λ = \(\frac{v}{v}\)

Where υ = Velocity of sound
v = Frequency of sound
If ‘v’ is apparent frequency heard by him then

Therefore the apparent frequency is greater than actual frequency.

Case (2) : When observer Is moving away from rest source:
If the observer is moving away from the stationary source then he loses the number of waves \(\frac{v_0}{\lambda}\)
∴ Apparent frequency v’ = v – \(\frac{v_0}{\lambda}\) = v – \(\frac{v_0 \cdot v}{v}\)
v’ = \(\left[\frac{v-v_0}{v}\right] \cdot v\)
Hence the apparent frequency is less than actual frequency.

Problems

Question 1.
A stretched wire of length 0.6m is observed to vibrate with a frequency of 30Hz in the fundamental mode. If the string has a linear mass of 0.05 kg/m find
(a) the velocity of propagation of transverse waves in the string
(b) the tension in the string.
Solution:
v = 30 Hz; l = 0.6 m ; μ = 0.05 kg m^-1 υ = ? ; T = ?
a) υ = 2vl = 2 × 30 × 0.6 = 36 m/s
b) T = υ²μ = 36 × 36 × 0.05 = 64.8 N

Question 2.
A steel cable of diameter 3 cm is kept under a tension of 10kN. The density of steel is 7.8 g/cm³. With what speed would transverse waves propagate along the cable ?
Solution:
T = 10 kN = 10⁴ N
D = 3 cm; r = \(\frac{\mathrm{D}}{2}\) = \(\frac{3}{2}\)cm
= \(\frac{3}{2}\) × 10^-2m;
A = πr² = \(\frac{22}{7} \times\left[\frac{3}{2} \times 10^{-2}\right]^2\)

Question 3.
Two progressive transverse waves given by y₁ = 0.07 sinπ(12x-500t) and y₂ = 0.07 sinπ(12x + 500t) travelling along a stretched string from nodes and antinodes. What is the displacement at the (a) nodes (b) antinodes ? What is the wavelength of the standing wave ?
Solution:
A₁ = 0.07; A₂ = 0.07; K = 12π
a) At nodes, displacement
y = A₁ – A₂ = 0.07 – 0.07 = 0
b) At antinodes, displacement
y = A₁ + A₂ = 0.07 + 0.07 = 0.14 m
c) Wavelength λ = \(\frac{2 \pi}{\mathrm{K}}=\frac{2 \pi}{12 \pi}\) = 0.16 m

Question 4.
A string has a length of 0.4m and a mass of 0.16g. If the tension in the string is 70N, what are the three lowest frequencies it produces when plucked ?
Solution:
I = 0.4 m; M = 0.16g = 0.16 × 10^-3 kg;

Question 5.
A metal bar when clamped at its centre resonantes in its fundamental frequency with longitudinal waves of frequency 4 kHz. If the clamp is moved to one end. What will be its fundamental resonance frequency ?
Solution:
When a metal bar of length l is clamped in the middle, it has one node in the middle and two antinodes at its free ends. In the fundamental mode. l = \(\frac{\lambda}{2}\) ⇒ λ = 21

In fundamental mode of frequency of bar
= frequency of wave = 4 kHz.
∴ v = \(\frac{\mathrm{v}}{\lambda}\) = \(\frac{\mathrm{v}}{2l}\) = 4kHz —– (1)
When clamp is moved to one end,
l = \(\frac{\lambda^{\prime}}{4}\) ⇒ λ’ = 4l
∴ V¹ = \(\frac{\mathrm{v}}{\lambda}\) = \(\frac{\mathrm{v}}{4 \mathrm{l}}\) = \(\frac{4 \mathrm{kHz}}{2}\) = 2kHz
[∵ from (1)]

Question 6.
A closed organ pipe 70 cm long is sounded. If the velocity of sound is 331 m/s, what is the fundamental frequency of vibration of the air column ?
Solution:
I = 70 cm = 70 × 10^-2m; v = 331 m/s ; v = ?
v = ?
v = \(\frac{v}{4 l}\) = \(\frac{331}{4 \times 70 \times 10^{-2}}\) = 118.2 Hz

Question 7.
A vertical tube is made to stand in water so that the water level can be adjusted. Sound waves of frequency 320 Hz are sent into the top of the tube. If standing waves are produced at two successive water levels of 20 cm and 73 cm, what is the speed of sound waves in the air in the tube ?
Solution:
v = 320 Hz; l₁ = 20cm = 20 × 10^-2
l₂ = 73 cm = 73 × 10^-2m; v = ?
v = 2v(l₂ – l₁)
= 2 × 320 (73 × 10^-2 – 20 × 10^-2)
∴ v = 339 m/s .

Question 8.
Two organ pipes of lengths 65cm and 70cm respectively, are sounded simultaneously. How many beats per second will be produced between the fundamental frequencies of the two pipes ? (Velocity of sound = 330 m/s).
Solution:
l₁ = 65 cm = 0.65 m
₂ = 70 cm = 0.7 m
v = 330 m/s
No. of beats per second ∆υ = υ₁ – υ₂
= \(\frac{v}{2 h}\) – \(\frac{\mathrm{v}}{2 l_2}\) = \(\frac{330}{2 \times 0.65}\) – \(\frac{330}{2 \times 0.7}\)
∴ ∆v = 253.8 – 235.8 = 18Hz

Question 9.
A train sounds its whistle as it approaches and crosses a level crossing. An observer at the crossing measures a frequency of 219 Hz as the train, approaches and a frequency of 184 Hz as it leaves. If the speed of sound is taken to be 340 m/s, find the speed of the train and the frequency of its whistle.
Solution:
When a whistling train approaches to rest observer,

v’ = \(\left[\frac{v}{v-v_s}\right] v\) ——– (1)
When a whistling train away from rest observer

v” = \(\left[\frac{v}{v+v_{\mathrm{S}}}\right] v\) —— (2)
Here v’ = 219 Hz; V” = 184 Hz;
v = 340 m/s
\(\frac{(1)}{(2)}\) ⇒ \(\frac{v^{\prime}}{v^{\prime \prime}}\) = \(\frac{\left(v+v_s\right)}{\left(v-v_s\right)}\)
\(\frac{219}{184}\) = \(\frac{340+v_{\mathrm{s}}}{340-v_{\mathrm{s}}}\)
219(340 – υ_s) = 184(340 + υ_s)
219 × 340 – 219 υ_s = 184 × 340 + 184 υ_s
403 υ_s = 35 × 340
∴ Velocity of train υ_s = 29.5 m/s
Frequency of whistle, v = v’ × \(\left[\frac{v-v_{\mathrm{S}}}{v}\right]\)
= 219 × \(\left[\frac{340-29.5}{340}\right]\)
= 199.98
∴ v = 200 Hz.

Question 10.
Two trucks heading in opposite directions with speeds of 60 kmph and 70 kmph respectively, approach each other. The driver of the first truck sounds his horn of frequency 400Hz. What frequency does the driver of the second truck hear ? (Velocity of sound = 330 m/ s). After the two trucks have passed each other, what frequency does the driver of the second truck hear ?
Solution:
v_s = 60 kmph = 60 × \(\frac{5}{18}\) m/s = \(\frac{300}{18}\) m/s
v₀ = 70 kmph = 70 × \(\frac{5}{18}\) m/s = \(\frac{350}{18}\) m/s
v = 400 Hz
When two trucks approach each other

When two trucks crossed each other,

Textual Exercises

Question 1.
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is caused at one end of the string, how long does the disturbance take to reach the other end ?
Answer:
Here M = 2.50 kg, T = 200 N, l = 20.0M
Mass per unit length; μ = \(\frac{\mathrm{M}}{l}\) = \(\frac{2.5}{20.0}\)
= 0.125 kg/m
Velocity V = \(\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{200}{0.125}}\) = 40 m/s
Time taken by disturbance to reach the other end
t = \(\frac{l}{\mathrm{~V}}\) = \(\frac{20}{40}\) = 0.5s.

Question 2.
A stone dropped from the top of a tower of height 300 m high splashes into the pond of water near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s^-1, (g = 9.8m s^-2)
Answer:
Here, h = 300m, g = 9.8 m/s², V = 340 m/s. If t₁ = time taken by stone to strike the surface of water in the pond, then from
S = ut + \(\frac{1}{2}\) at²
300 = 0 + \(\frac{1}{2}\) × 9.8 \(\mathrm{t}_1^2\)
t₁ = \(\sqrt{\frac{300}{4.9}}\) = 7.82s.
Time taken by sound to reach the top of tower t₂ = \(\frac{\mathrm{h}}{\mathrm{v}}\)
= \(\frac{300}{400}\) = 0.88s
Total time after which splash of sound is heard = t₁ + t₂ = 7.82 + 0.88 = 8.70s

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20° C = 343 m s^-1.
Answer:
Here, l = 12.0M, M = 2.10kg, T = ?
V = 343 m/s
Mass per unit length μ = \(\frac{\mathrm{M}}{l}\) = \(\frac{2.10}{12.0}\)
= 0.175 kg/m
As V = \(\sqrt{\frac{\mathrm{T}}{\mu}}\)
T = V² . μ = (343)² × 0.175 = 2.06 × 10⁴N.

Question 4.
Use the formula υ = \(\sqrt{\frac{\gamma P}{\rho}}\) to explain why the speed of sound in air
a) is independent of pressure,
b) increases with temperature,
c) increases with humidity.
Answer:
a) Effect of pressure:
The speed of sound in gases, υ = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\)
At constant temperature, PV = constant
P\(\frac{m}{\rho}\) = constant ⇒ \(\frac{\mathrm{P}}{\rho}\) = constant
If P increases, ρ also increases. Hence speed of sound in air is independent of pressure.

b) Effect of temperature:
As PV = nRT, P\(\frac{\mathrm{m}}{\rho}\) = \(\frac{m}{M} R T\)
⇒ \(\frac{P}{\rho}\) = \(\frac{\mathrm{RT}}{\mathrm{M}}\)
∴ υ = \(\sqrt{\frac{R T}{M}}\)
Since R, M are constants υ ∝ \(\sqrt{\mathrm{T}}\)
∴ Velocity of sound in air depends on temperature.

c) Effect of humidity:
As υ = \(\sqrt{\frac{\gamma \mathrm{P}}{\rho}}\) ∴ υ ∝ \(\frac{1}{\sqrt{\rho}}\)
As the density of water vapour is less than density of dry air at STP. So the presence of moisture in air decreases the
density of air. Since the speed of sound is inversely proportional to the square root of density. So sound travels faster in moist air than dry air. Hence velocity of sound
V ∝ humidity

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f(x, t) where x and t must appear in the combination x – υt or x + υt, i.e. y = f(x ± υ t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:
Answer:
No, the converse is not true. The basic requirement for a wave function to represent a travelling wave is that for all values of x & t, wave function must have a finite value.

Out of the given functions y, no one satisfies this condition therefore, none can represent a travelling wave.

Question 6.
A bat emits ultrasonic sound a frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of
(a) the reflected sound,
(b) the transmitted sound?
Speed of sound in air is 340 m s^-1 and in water 1486 m s^-1.
Answer:
Here V = 100 KHz = 10⁵Hz, V_a = 340m/s, V_w = 1486 m/s^-1
Wavelength of reflected sound, λ_a = \(\frac{\mathrm{V}^{\mathrm{a}}}{\mathrm{V}}\)
= \(\frac{340}{10^5}\) = 3.4 × 10^-3 m
Wavelength of transmitted sound,
λ_w = \(\frac{V_w}{V}\) = \(\frac{1486}{10^5}\) = 1.486 × 10^-2 m

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s^-1? The operating frequency of the scanner is 4.2 MHz.
Answer:
λ = ? υ = 1.7 Km/s = 1700 ms^-1
y = 4.2 MHz = 4.2 × 10⁶Hz
λ = \(\frac{v}{v}\) = \(\frac{1700}{4.2 \times 10^6} \mathrm{~m}\) = 0.405 × 10^-3 m
= 0.405 mm

Question 8.
A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right.
a) Is this a travelling wave or a stationary wave?
If it is travelling, what are the speed and direction of its propagation?
b) What are Its amplitude and frequency?
C) What is the initial phase at the origin?
d) What is the least distance between two successive crests in the wave?
Answer:
Compare the given equation with that of plane progressive wave of amplitude r, travelling with a velocity V from right to left.
y(x, t) = rsin\(\left[\frac{2 \pi}{\lambda}(v t+x)+\phi_0\right]\) ……… (1)
We find that
a) The given equation represents a transvërse harmonic wave travelling from right to left. It is ñot a stationary wave.

b) The given equation can be written as
Y(x, t) = 3.0sin[0.018(\(\frac{36}{0.018}\) + x) + \(\frac{\pi}{4}\)] ……… (2)
equating coefficient of t in the two
(1) & (2) we get. :
V = \(\frac{.36}{0.018}\) = 2000 cm/sec.
Obviously, r = 3.0 cm
Also, \(\frac{2 \pi}{\lambda}\) = 0.018
λ = \(\frac{2 \pi}{0.018} \mathrm{~cm}\)
Frequency, v = \(\frac{v}{\lambda}\) = \(\frac{2000}{2 \pi}\) × 0.018
= 5.7351.

c) Intial phase, φ₀ = \(\frac{\pi}{4}\) radian.

d) Least distance between two successive crests of the wave =
Wave length, λ = \(\frac{2 \pi}{0.018 \mathrm{~cm}}\) = 349 cm,

Question 9.
For the wave described in the last problem plot the displacement (y) versus (t) graphs for x = 0.2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
Answer:
The transverse harmonic wave is
y(x,t) = 3.0 sin[36t + 0.018x + \(\frac{\pi}{4}\)]
For x = 0
y(0, t) = 3.0 sin(36t + \(\frac{\pi}{4}\)) —— (i)
Here w = \(\frac{2 \pi}{T}\) = 36, T = \(\frac{2 \pi}{36}\) = \(\frac{\pi}{18}\)-sec.

For different values of t, we calculate y using eq(i). These values are tabulated below.

On plotting y versus t graph, we obtain a sinusoidal curve as shown in fig.

Similar graphs are obtained for x = 2cm & x = 4cm. The oscillary motion in travelling wave differs from one point to another only in terms of phase. Amplitude and frequency of oscillatory motion remain the same in all the three areas.

Question 10.
For the travelling harmonic wave
y(x, t) = 2.0 cos 2 π (10t – 0.0080 x + 0.35)
where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
a) 4 m,
b) 0.5 m,
c) λ/2,
d) 3λ/4,
Answer:
The given equation be written as
y = 2.0 cos[2π(10t – 0.0080x) + 2π × 0.35]
y = 2.0 cos[2π × (0.0080(\(\frac{10 \mathrm{t}}{0.0080}\) – x) + 0.7π]
Compare it with the standard equation of a travelling harmonic, we have
y = r.cos[\(\frac{2 \pi}{\lambda}(v t-x)+\phi_0\)
We get, \(\frac{2 \pi}{\lambda}\) = 2π × 0.0080
Further we know that phase diff. φ = \(\frac{2 \pi}{\lambda} \mathrm{x}\)
a) When x = 4m = 400 cm
φ = \(\frac{2 \pi}{\lambda}\) . x = 2π × 0.0080 × 400
= 6.4 π rad.

b) When x = 0.5 = 50 cm
φ = \(\frac{2 \pi}{\lambda}\) . x = 2π × 0.0080 × 50
= 0.8π rad.

c) When x = \(\frac{\lambda}{2}\)
φ = \(\frac{2 \pi}{\lambda} \times \frac{\lambda}{2}\) = π rad.

d) When x = \(\frac{3 \lambda}{4}\)
φ = \(\frac{2 \pi}{\lambda} \times \frac{3 \lambda}{4}\) = \(\frac{3 \lambda}{2} \mathrm{rad}\)

Question 11.
The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin\(\left(\frac{2 \pi}{3} x\right)\)cos (120 πt)
where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10^-2 kg
Answer the following:
a) Does the function represent a travelling wave or a stationary wave?
b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?
c) Determine the tension in the string.
Answer:
The given equation is
y(x, t) = 0.06 sin \(\frac{2 \pi}{3} \mathrm{x} \cos 120 \pi \mathrm{t}\)

a) As the equation involves harmonic functions of x and t seperately, it represents a stationary wave.

b) We know that when a wave pulse
y₁ = r sin \(\frac{2 \pi}{\lambda}(v t+x)\) —– (i)
travelling along + direction of x-axis is super imposed by the reflected wave
y = y₁ + y₂ = -2rsin\(\frac{2 \pi}{\lambda}\) xcos \(\frac{2 \pi}{\lambda}\) vt is formed. ——- (ii)
Comparing (i) & (ii) we find that
\(\frac{2 \pi}{\lambda}\) = \(\frac{2 \pi}{3}\) ⇒ λ = 3m.
Also \(\frac{2 \pi}{\lambda} v\) = 120π (Or)
V = 60λ = 60 × 3 = 180m/s.
frequency, v = \(\frac{v}{\lambda}\) = \(\frac{180}{3}\) = 60 hertz.
Note that both the waves have same wave length, same frequency and same speed.

c) Velocity of transverse waves is
υ = \(\sqrt{\frac{T}{\mu}}\) (or) υ² = T/μ
T = V² × μ where μ = \(\frac{3 \times 10^{-2}}{1.5}\)
= 2 × 10^-2 kg/m
T = (180)² × 2 × 10^-2 = 648 N.

Question 12.
i) For the wave on a string described in previous problem do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers.
(ii) What is the amplitude of a point 0.375 m away from one end?
Answer:
i) All the points on the string
a) have the same frequency except at the nodes (where frequency is cos θ),
b) have the same phase every where in one loop except at the nodes,
c) however, the amplitude of vibration at different points is different.

ii) From y(x, t) = 0.06 sin\(\left(\frac{2 \pi}{3} x\right)\) cos (120 πt)
The amplitude at x = 0.375 m is 0.06
sin \(\frac{2 \pi}{3} x \times 1\) = 0.06 × sin \(\frac{2 \pi}{3} \times 0.375\)
= 0.06sin\(\frac{\pi}{4}\) = \(\frac{0.06}{\sqrt{2}}\) = 0.042 M

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave State which of these represent
(i) a travelling wave,
(ii) a stationary wave or
(iii) none at all:
a) y = 2 cos (3x) sin (10t)
b) y = \(2 \sqrt{x-v t}\)
c) y = 3 sin(5x – 0.5t) + 4 cos(5x – 0.5t)
d) y = cos x sin t + cos 2x sin 2t
Answer:
a) It represents a stationary wave as harmonic functions of x & t are contained separetely in the equation.
b) It cannot represent any type of wave.
c) It represents a progressive / travelling harmonic wave.
d) This equation is sum of two functions each representing a stationary wave. Therefore it represents superposition of two stationary waves.

Question 14.
A wire stretched between two right supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10^-2 kg and its linear mass density is 4.0 × 10^-2 kg m^-1. What is (a) the speed of transverse wave on the string, and (b) the tension in the string?
Answer:
Here, v = 45Hz, μ = 3.5 × 10^-2 kg
Mass/length = μ = 4.0 × 10^-2 kg/m
l = \(\frac{\mu}{\mu}\) = \(\frac{3.5 \times 10^{-2}}{4.0 \times 10^{-2}}\) = \(\frac{7}{8}\)
As \(\frac{\lambda}{2}\) = l = \(\frac{7}{8}\) ∴ λ = \(\frac{7}{4}\)m = 1.75m
a) The speed of transverse wave
υ = vλ = 45 × 1.75 = 78.75 m/s.

b) As υ = \(\sqrt{\frac{\mathrm{T}}{\mu}}\)
∴ T = υ² × μ = (78.75)² × 4.0 × 10^-2
= 248.06 N.

Question 15.
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length Is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effect may be neglected.
Answer:
As there is a piston at one end of the tube, it behaves as a closed organ pipe, which produces odd harmonics only. Therefore the pipe is in resonance with the fundamental note at the third harmonic (79.3 cm is about 3 times 25.5 cm)
In the fundamental note = \(\frac{\lambda}{4}\) = l₁ = 25.5
λ = 4 × 25.5 = 102 cm = 1.02m
Speed of sound in air.
υ = vλ = 340 × 1.02
= 346.0 m/s

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel?
Answer:
Here, l = 100 cm = 1 m, y = 2.53 KHz
= 2.53 × 10³ Hz
When the rod is clamped at the middle, then in the fundamental mode of vibration of the rod, anode is formed at the middle and antinode is formed at each end.
Therefore, it is clear from fig

l = \(\frac{\lambda}{4}\) + \(\frac{\lambda}{4}\) + \(\frac{\lambda}{2}\)
λ = 2l = 2m
As v = λl
v = 2.53 × 10³ × 2
= 5.06 × 10³ ms^-1

Question 17.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s^-1).
Answer:
Here l = 20 cm = 0.2m, v_n = 430 Hz,
υ = 340 m/s
The frequency of n^th normal mode of vibration of closed pipe is
v_n = (2n – 1)\(\frac{v}{4l}\)
∴ 430 = (2n – 1)\(\frac{340}{4 \times 0.2}\)
2n – 1 = \(\frac{430 \times 4 \times 0.2}{340}\) = 1.02
2n = 2.02, n = 1.01
Hence it will be the 1^st normal mode of vibration. In a pipe, open at both ends we have
v_n = n × \(\frac{\mathrm{v}}{2l}\) = \(\frac{\mathrm{n} \times 340}{2 \times 0.2}\) = 430.
∴ n = \(\frac{430 \times 2 \times 0.2}{340}\) = 0.5
As n has to be an integer, therefore open organ pipe cannot be in resonance with the source.

Question 18.
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324Hz. What is the frequency of B?
Answer:
Let original frequency of sitar string A be n_a & original frequeny of sitar string B be n_b.
As number of beats / sec = 6
∴ n_b = n_a ± 6 = 330 (or) 318Hz.
When tension in A is reduced, its frequency reduces (∴ n ∝ \(\sqrt{T}\))
As number of beats/sec decreases to 3 therefore, frequency of B = 324 – 6
= 318Hz.

Question 19.
Explain why (or how):
a) in a sound wave, a displacement node is a pressure antinode and vice versa,
b) bats can ascertain distances, directions, nature and sizes of the obstacles without any ‘eyes”.
c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
d) Soils can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases and
e) The shape of a pulse gets distorted during propagation In a dispersive medium.
Answer:
a) Node (N) is a point where the amplitude of oscillation is 0. (and pressure is maximum)
Antinode (A) is a point where the amplitude of oscillation is maximum (and pressure is min).
These nodes & antinodes do not coincide with pressure nodes & antinodes.
Infact, N coincides with pressure antinode and A coincides with pressure node, as is clear from the definitions.

b) Bats emit ultrasonic wave of large frequencies, when these waveš are reflected from the obstacles in their path,
they give them the idea about the distance, direction, size & nature of the obstacle.

c) Though the violin note and sitar note have the same frequency, yet the over tones produced and their reactive strengths are different in the two flotes that is why we can distinguish between the two notes.

d) This is because solids have both, the elasticity of volume and elasticity of shape where as gases have only the volume elasticity.

e) A sound pulse is a combination of waves of different wavelengths. As waves of different wavelengths travel in a disperse medium with different velocities, therefore the shape of the pulse gets distorted.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.
i) What is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of 10ms^-1,
(b) recedes from the platform with a speed of 10 m s^-1 ?
ii) What is the speed of sound in each case ? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
i) Here, y = 400 Hz, υ = 340 m/s
a) Train approaches the platform
υ_s = 10m/s
v’ = \(\frac{v}{v-v_s}\) = \(\frac{340 \times 400}{340-10}\) = 412.12 Hz.

b) Train recedes from the platform
υ_s = 10m/s
v’ = \(\frac{v \times v}{v \times v_s}\) = \(\frac{340 \times 400}{340+10}\)
= 388.6Hz

ii) The speed of sound in each case is the same = 340 m/s

Question 21.
A train, stañdingin a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10m s^-1. What are the frequency, wavelength and speed of sound for an observer standing on the station’s platform ? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s^-1 ? The speed of sound in still air can be taken as 340 m^-1
Answer:
Here y = 400 Hz, υ_m = 10ms^-1, υ = 340m/s
As the wind is blowing in the direction of sound, therefore effective speed of sound
= υ + υ_m = 340 + 10 = 350m/s
As the source & Iistner both are at rest, therefore, frequency remains unchanged
i.e. v = 400 Hz.
Wavelength, λ = \(\frac{v+v_m}{v}\) = \(\frac{350}{400}\)
= 0.875 M.
When air is still, υ_m = 0
Speed of observer υ₁ = 10m/s υ_s = 0
As observer moves toward the source
υ’ = \(\frac{\left(v+v_l\right)}{v} \times v\) = \(\frac{(340+10)}{340} \times 400\)
= 411.76 Hz.
As source is at rest, wavelength does not change
i.e, λ’ = λ = 0.875M.
Also, speed of sound = υ + υ_m = 340 + 0
= 340 m/s
The situations in the two cases are entirely different.

Additional Exercises

Question 1.
A travelling harmonic wave on a string is described by
y(x, t) = 7.5 sin (0.0050x + 12t + π/4)
a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1cm
point at t = 2 s, 5 s and 11 s.
Answer:
a) The travelling harmonic wave is y(x, t)

From (1), y(1, 1) = 7.5 sin(732.55°)
= 7.5 sin (720 + 12.55°)
= 7.5 sin 12.55° = 7.5 × 0.2173 = 1.63 cm
velocity of oscillation, v = \(\frac{d y}{d t} y(1,1)\)
= \(\frac{\mathrm{d}}{\mathrm{dt}}\)[7.5 sin (0.005x + 12t + \(\frac{\pi}{4}\))
= 7.5 × 12 cos (0.005x + 12t + \(\frac{\pi}{4}\))
At x = 1 cm, t = 1 sec
= 7.5 × 12 cos (o.oo5 + 12 + \(\frac{\pi}{4}\))
= 90 cos (732.55°).
= 90 cos(720 + 12.55°)
= 90 cos (12.55°)
= 90 × 0.9765
= 87.89 cm/s.
Comparing the given equation with the standard form
y(x, t) = r sin\(\left[\frac{2 \pi}{\lambda}(v t+x)+\phi_0\right]\)
We get r = 7.5 cm, \(\frac{2 \pi v}{\lambda}\) = 12 (or) 2πV = 12
V = \(\frac{6}{\pi}\)
2\(\frac{\pi}{\lambda}\) = 0.005
∴ λ = \(\frac{2 \pi}{0.005}\) = \(\frac{2 \times 3.14}{0.005}\) = 1256 cm
= 12.56 m.
Velocity of wave propagation, υ = Vλ
= \(\frac{6}{\pi}\) × 12.56 .
= 24 m/s.
We find that velocity at x = 1 cm t = 1 sec is not equal to velocity of wave propagation.

b) Now, all points which are at a distance of ±λ, ± 2λ, ± 3λ from x = 1 cm will have same transverse displacement and velocity. As λ = 12.56 m, therefore, all points at distances ± 12.6m, ± 25.2 m displacement and velocity As λ = 12.56m, therefore all points at distances ± 12.6m, ± 25.2m, ± 37.8m from x = 11m will have same displacement & velocity at x = 1 cm point at t = 25.55 & 115s.

Question 2.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (I) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s (that is the whistle blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
Answer:
a) A short pip by a whistle has neither a definite wavelength nor a definite frequency. However its speed of propagation is fixed, being equal to speed of sound in air.

b) No, frequency of the note produced by whistle is not 1/20 = 0.05 Hz. Rather 0.05 Hz is the frequency of repetition of the short pipe of the whistle.

Question 3.
One end of a long string of linear mass density 8.0 × 10^-3 kg m^-1 is connected to an electrically driven tuniúg fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the Incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
Answer:
Here, μ = 8.0 × 10^-3 kg/m, y = 256 Hz, T = 90kg = 90 × 9.8 = 882N.
Amplitude of wave, r = 5.0 1m = 0.05m.
As the wave propagation along the string is a transverse travelling wave, the velocity of the wave is given by

As the wave is propagating along x direction, the equation of the wave is
y(x, t) = r sin (ωt – kx)
= 0.05 sin (1.61 × 10³t – 4.84x)
Here x, y are in mt & t in sec

Question 4.
A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine ? Take the speed of sound in water to be 1450 m s^-1.
Answer:
Here, frequency of SONAR,
v = 40 KHz = 40 × 10³ Hz.
Speed of observer / enemy’s submarine
υ₁ = 360km/h .
= 360 × \(\frac{5}{18}\)m/s = 100m/s.
Speed of sound wave in water; υ = 1450 m/s.
As the source is at rest & observer is moving towards the source, therefore, apparent frequency received by enemy submarine
v’ = \(\frac{\left(v+v_1\right) v}{v}\)
= \(\frac{(1450+100) 40 \times 10^3 \mathrm{~Hz}}{1450}\)
= 4.27 × 10⁴ Hz.
This frequency is reflected by enemy submarine (source) and is observed by SONAR. Therefore in this case,
υ_s = 360 km/s = 100 m/s, υ₁ = 0
∴ Apparent frequency, v¹¹ = \(\frac{v \times v}{v_i-v_s}\)
= \(\frac{1450 \times 4.276 \times 10^4}{1450-10}\)
= 4.59 × 10⁴ Hz = 45.9 Hz.

Question 5.
Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about 4.0 km s^-1 and that of P wave is 8.0 km s^-1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?
Answer:
Let υ₁, υ₂ be the velocity of S waves & P waves & t₁, t₂ be the time taken by these waves to travel to the position of seismograph. If l is the distance of occurrence of earth quake from the seismograph, then
l = υ₁t₁ = υ₂t₂ ——- (i)
now υ₁ = 4 km/s & υ₂ = 8 km/s .
∴ 4t₁ = 8t₂ (or) t₁ = 2t₂ ——- (ii)
Also t₁ – t₂ = 4min = 240s.
using (iii), 2t₂ – t₂ = 240s, t₂ = 240s
t₁ = 2t₂ = 2 × 240 = 480s.
Now from (i) l = υ₁t₁ = 4 × 480 = 1920 km.
Hence earthquake occurs 1920 km away from the seismograph.

Question 6.
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Here, the frequency of sound emitted by the bat, v = 40 kHz.
velocity of bat, υ_s = 0.03υ, where υ is velocity of sound.
Apparent frequency of sound striking the wall.
v’ = \(\frac{v \times v}{v-v_s}\) = \(\frac{v}{v-0.03 v}\) × 40 kHxz
= \(\frac{40}{0.97}\) kHZ.
This frequency is reflected by the wall & is received by the bat moving towards the wall, So υ_s = 0.
υ₁ = 0.03 υ
v’ = \(\frac{\left(v+v_1\right) v^{\prime}}{v}\) = \(\frac{(v+0.03 v)}{v}\left(\frac{40}{0.97}\right)\)
= \(\frac{1.03}{0.97} \times 40 \mathrm{kHz}\)
= 42.47 kHz

Students get through AP Inter 2nd Year Physics Important Questions 13th Lesson Atoms which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 13th Lesson Atoms

Very Short Answer Questions

Question 1.
What is the physical meaning of negative energy of an electron’ ?
Answer:
The ‘negative energy of an electron’ indicates that the electron is bound to the nucleus due to force of attraction.

Question 2.
Sharp lines are present in the spectrum of a gas. What does this indicate ?
Answer:
Sharp lines in the spectrum of gas, indicates bright lines against dark background.

Question 3.
Name a physical quantity whose dimensions are the same as those of angular momentum.
Answer:
Planck’s constant.

Question 4.
What is the difference between α – particle and helium atom ?
Answer:
Alpha particle

1. It is a + 2e charged Helium nucleus.
2. It contains 2 protons and 2 neutrons.

Helium atom

1. It has no charge.
2. It contains 2 protons, 2 electrons and 2 neutrons.

Question 5.
Among alpha, beta and gamma radiations, which get affected by the electric field ?
Answer:
Alpha and Beta radiations are get affected by the electric field.

Question 6.
What do you understand by the phrase ground state atom ?
Answer:
If the electron is present in the ground state, it is called ground state atom.

Question 7.
Why does the mass of the nucleus not have any significance in scattering in Rutherford’s experiment ?
Answer:
The size of the atom is 10^-10 m and size of the nucleus is 10^-15 m. Hence atom has large empty space. So the mass of nucleus has no significance in Rutherford’s scattering experiment.

Question 8.
The Lyman series of hydrogen spectrum lies in the ultraviolet region. Why ? [A.P. Mar. 15]
Answer:
The calculated values of wavelengths lie in the ultraviolet region of the spectrum will agree with the values of wavelengths observed experimentally by Lyman.

Question 9.
Write down a table giving longest and shortest wavelengths of different spectral series.
Answer:
Wavelength limits of some spectral series of hydrogen.

Question 10.
Give two drawbacks of Rutherford’s atomic model.
Answer:
Drawbacks of Rutherford’s atomic model:

1. As the revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. But matter .is stable, we cannot expect the atom collapse.
2. The atoms should emit continuous spectrum, but what we observe is only a line spectrum.

Question 11.
If the kinetic energy of revolving electron in an orbit is K, what is its potential energy and total energy ?
Answer:
For an electron revolving round the nucleus, total energy is always negative and it is numerically equal to kinetic energy.
∴ Total energy = -Kinetic energy = -K
Potential energy is always negative and PE = 2 × TE = -2K

Short Answer Questions

Question 1.
What is impact parameter and angle of scattering ? How are they related to each other ?
Answer:

1. Impact parameter (b) : Impact parameter is defined as the perpendicular distance of the initial velocity vector of the alpha particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.
2. Scattering angle (θ) : The scattering angle (θ) is the angle between the asymtotic direction of approach of the α – particle and the asymtotic direction in which it receeds.
3. The relation between b and θ is b = \(\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Ze}^2}{\mathrm{E}}\) cot \(\frac{\theta}{2}\) where E = K.E of α – particle = \(\frac{1}{2}\) mυ².

Question 2.
Explain the distance of closest approach and impact parameter.
Answer:
Distance of closest approach :

1. Suppose an α-particle with initial kinetic energy (K.E) is directed towards the centre of the nucleus of an atom.
2. On account of Coulomb’s repulsive force between nucleus and alpha particle, kinetic energy of alpha particle goes on decreasing and in turn, electric potential energy of the particle goes on increasing.
3. At certain distance d’ from the nucleus, K. E of α-particle reduces to zero. The particle stops and it cannot go closer to the nucleus. It is repelled by the nucleus and therefore it retraces its path, turning through 180°.
4. Therefore, the distance d is known as the distance of closest of approach.
   
   The closest distance of approach,
   d = \(\frac{1}{4 \pi \varepsilon_0} \times \frac{Z e^2}{\left(\frac{1}{2} m v^2\right)}\)
5. Impact parameter (b) : Impact parameter is defined as the ⊥^r distance of the initial velocity vector of the α – particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.
   

Question 3.
Describe Rutherford atom model. What are the draw backs of this model ?
Answer:
Rutherford atom model: The essential features of Rutherford’s nuclear model of the atom or planetary model of the atom are as follows :

1. Every atom consists of tiny central core, called the atomic nucleus, in which the entire positive charge and almost entire mass of the atom are concentrated.
2. The size of nucleus is of the order of 10^-15m, which is very small as compared to the size of the atom which is of the order of 10^-10 m.
3. The atomic nucleus is surrounded by certain number of electrons. As atom on the whole is electrically neutral, the total negative charge of electrons surrounding the nucleus is equal to total positive charge on the nucleus.
4. These electrons revolve around the nucleus in various circular orbits as do the planets around the sun. The centripetal force required by electron for revolution is provided by the electrostatic force of attraction between the electrons and the nucleus.

Drawbacks : According to classical E.M. theory,

1. the revolving electron loses energy continuously, it must spiral inwards and eventually fall into the nucleus. As matter is stable, we cannot expect the atoms to collapse.
2. since the frequency of radiation emitted is the same as the frequency of revolution, the atom should radiate a continuous spectrum, but what we observe is only a line spectrum.

Question 4.
What are the limitations of Bohr’s theory of hydrogen atom ? [A.P. Mar. 17; Mar. 14]
Answer:
Limitations of Bohr’s theory of Hydrogen atom :

1. This theory is applicable only to simplest atom like hydrogen, with z = 1. The theory fails in case of atoms of other elements for which z > 1.
2. The theory does not explain why orbits of electrons are taken as circular, while elliptical orbits are also possible.
3. Bohr’s theory does not say anything about the relative intensities of spectral lines.
4. Bohr’s theory does not take into account the wave properties of electrons.

Question 5.
Write a short note on Debroglie’s explanation of Bohr’s second postulate of quantization. [T.S. Mar. 17]
Answer:
Debroglie’s explanation of Bohr’s second postulate of quantization :

1. The second postulate of Bohr atom model says that angular momentum of electron orbiting around the nucleus is quantized i.e., mυr = \(\frac{\mathrm{nh}}{2 \pi}\) where n = 1, 2, 3,….
2. According to Debroglie, the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave.
3. When a string fixed at two ends is plucked, a large number of wavelengths are excited and standing wave is formed.
4. It means that in a string, standing waves form when total distance travelled by a wave down the string and back is an integral number of wavelengths.
5. According to Debroglie, a stationary orbit is that which contains an integral number of Debrogile waves associated with the revolving electron.
6. For an electron revolving in n^th circular orbit of radius r_n, total distance covered = circumference of the orbit = 2πr_n
   ∴ For permissible orbit, 2πr_n = nλ
7. According to Debrogile, λ = \(\frac{h}{m v_n}\)
   Where υ_n is speed of electron revolving in n^th orbit
   ∴ 2πr_n = \(\frac{\mathrm{nh}}{\mathrm{m} v_{\mathrm{n}}}\)
   mυ_nr_n = \(\frac{\mathrm{nh}}{2 \pi}=\mathrm{n}\left(\frac{\mathrm{h}}{2 \pi}\right)\)
   i.e., angular momentum of electron revolving in n^th orbit must be an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\), which is the quantum condition proposed by Bohr in second postulate.

Question 6.
Explain the different types of spectral series in hydrogen atom. [T.S. Mar. 16, A.P. Mar. 15]
Answer:
The atomic hydrogen emits a line spectrum consisting of five series.

1. Lyman series : v = Rc \(\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\) where n = 2, 3, 4
2. Balmer series : v = Rc \(\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\) where n = 3, 4, 5.
3. Paschen series : v = Rc \(\left(\frac{1}{3^2}-\frac{1}{n^2}\right)\) where n = 4, 5, 6
4. Brackett series: v = Rc \(\left(\frac{1}{4^2}-\frac{1}{n^2}\right)\) where n = 5, 6, 7
5. Pfund series: v = Rc \(\left(\frac{1}{5^2}-\frac{1}{n^2}\right)\) where n = 6, 7, 8

Long Answer Questions

Question 1.
State the basic postulates of Bohr’s theory of atomic spectra. Hence obtain an expression for the radius of orbit and the energy of orbital electron in a hydrogen atom. [A.P. Mar. 16]
Answer:
a) Basic postulates of Bohr’s theory are

1. The electron revolves round a nucleus in an atom in various orbits known as stationary orbits. The electrons cannot emit radiation when moving in their own stationary levels.
2. The electron can revolve round the nucleus only in allowed orbits whose angular momentum is the integral multiple of \(\frac{\mathrm{h}}{2 \pi}\)
   i.e., mυ_nr_n = \(\frac{\mathrm{nh}}{2 \pi}\) …………… (1)
   where n = 1, 2, 3 …………..
   If an electron jumps from higher energy (E₂) orbit to the lower energy (E₁) orbit, the difference of energy is radiated in the form of radiation.
   i.e., E = hv = E₂ – E₁ ⇒ v = \(\frac{E_2-E_1}{h}\) …………… (2)

b) Energy of emitted radiation: In hydrogen atom, a single electron of charge – e, revolves around the nucleus of charge e in a circular orbit of radius r_n.
1) K.E. of electron : For the electron to be in circular orbit, centripetal force = The electrostatic force of attraction between the electron and nucleus.

3) Radius of the oribit: Substituting the value of (6) in (2),
\(\frac{\mathrm{m}}{\mathrm{r}_{\mathrm{n}}}\left(\frac{\mathrm{n}^2 \mathrm{~h}^2}{4 \pi^2 \mathrm{r}_{\mathrm{n}}^2 \mathrm{~m}^2}\right)=\frac{\mathrm{Ke}^2}{\mathrm{r}_{\mathrm{n}}^2}\)
r_n = \(\frac{n^2 h^2}{4 \pi^2 m K e^2}\) ……………. (1)
∴ r_n = 0.53 n²

4) Total energy (E_n) : Revolving electron possess K.E. as well as P.E.

Problems

Question 1.
The radius of the first electron orbit of a hydrogen atom is 5.3 x HTum. What is the radius of the second orbit ?
Solution:
∴ r_n ∝ n²
\(\) ⇒ r₂ = 4r₁
∴ r₂ = 4 × 5.3 × 10^-11 = 2.12 × 10^-10m.

Question 2.
The total energy of an electron in the first excited state of the hydrogen atom is -3.4eV. What is the potential energy of the electron in this state?
Solution:
In 1^st orbit, E = -3.4 eV
Total energy E = \(\frac{\mathrm{KZe}^2}{2 \mathrm{r}}-\frac{\mathrm{KZe}^2}{\mathrm{r}}\)
\(\frac{\mathrm{KZe}^2}{\mathrm{r}}\) = U(say)
E = \(\frac{\mathrm{U}}{2}\) – u = \(\frac{\mathrm{-U}}{2}\)
U = -2E
∴ U = -2 × -3.4 = 6.8 eV.

Question 3.
The total energy of an electron in the first excited state of hydrogen atom is -3.4eV. What is the kinetic energy of the electron in this state ?
Solution:
In Hydrogen like atom, we know that
K = – Total energy E
Here E = – 3.4eV
∴ K = -(-3.4) = 3.4 eV . . ,

Question 4.
Prove that the ionisation energy of hydrogen atom is 13.6 eV.
Solution:
n = 1 corresponds to ground state.
E = \(\frac{-13.6}{\mathrm{n}^2}\) eV
E = \(\frac{-13.6}{\mathrm{1}^2}\) eV
E = -13.6 eV
The minimum energy required to free the electron from the ground state of hydrogen atom = 13.6 eV.
∴ Ionisation energy of hydrogen atom = 13.6 eV

Question 5.
Calculate the ionization energy for a lithium atom.
Solution:
For 3^Li7 atom, Z = 3, n = 2 [∵Li = 1s² 2s¹]
E_n = \(\frac{13.6 \mathrm{Z}^2}{\mathrm{n}^2}\) eV
= \(\frac{13.6 \times(3)^2}{4}\) = 30.6 eV
∴ Ionization energy of Lithium = 30.6eV

Question 6.
The wavelength of the first member of Lyman series is 1216 A. Calculate the wavelength of second member of Balmer series.
Solution:

Textual Examples

Question 1.
In the Rutherford’s nuclear model of the atom, the nucleus (radius about 10^-15m) is analogous to the sun about which the electron move in orbit (radius ≈ 10^-10m) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is ? The radius of earth’s orbit is about 1.5 × 10¹¹m. The radius of sun is taken as 7 × 10⁸ m.
Solution:
The ratio of the radius of electron’s orbit to the radius of nucleus is (10^-10 m)/(10^-15 m) = 105, that is, the radius of the electron’s orbit is 10⁵ times larger than the radius of nucleus. If the radius of the earth’s orbit around the sun were 10⁵ times larger than the radius of the sun, the radius of the earth’s orbit would be 10⁵ × 7 × 10⁸ m = 7 × 10¹³ m. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun.

It implies that an atom contains a much greater fraction of empty space than our solar system does.

Question 2.
In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV α-particle before it comes momentarily to rest and reverses its direction?
Solution:
The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an α-particle and a gold nucleus is conserved. The system’s initial mechanical energy is E_i; before the particle and nucleus interact, and it is equal to its mechanical energy E_f when the α-particle momentarily stops. The initial energy E_t is just the kinetic energy K of the incoming α-particle. The final energy E_f is just the electric potential energy U of the system. The potential energy U can be calculated from Equation.
F = \(\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{r}^2}\)
Let d be the centre-to-centre distance between the α-particle and the gold nucleus when the α-particle is at its stopping point. Then we can write the conservation of energy
E_i = E_f as
K = \(\frac{1}{4 \pi \varepsilon_0} \frac{(2 \mathrm{e})(\mathrm{Ze})}{\mathrm{d}}=\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{~d}}\)
Thus the distance of closest approach d is given by
d = \(\frac{2 \mathrm{Ze}^2}{4 \pi \varepsilon_0 \mathrm{~K}}\)
The maximum kinetic energy found in α-particles of natural origin is 7.7 MeV or 1.2 × 10^-12 J. Since 1/4πε₀ = 9.0 × 10⁹N m²/C². Therefore with e = 1.6 × 10^-19C, we have,
d = \(\frac{(2)\left(9.0 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^2 \mathrm{Z}}{1.2 \times 10^{-12} \mathrm{~J}}\)
= 3.84 × 10^-16 Zm
The atomic number of foil material gold is Z = 79, so that
d (Au) = 3.0 × 10^-14m = 30 fm. (1 fm (i.e. fermi) = 10^-15m.)
The radius of gold nucleus is, therefore, less than 3.0 × 10^-14 m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the a-particle. Thus, the α-particle reverses its motion without ever actually touching the gold nucleus.

Question 3.
It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.
Solution:
Total energy of the electron in hydrogen atom is – 13.6 eV = -13.6 × 1.6 × 10^-19J
= -2.2 × 10^-18 J.
Thus from equation, E = – \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) we have
– \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) = 2.2 10^-18 J
This gives the orbital radius
r = – \(\frac{\mathrm{e}^2}{8 \pi \varepsilon_0 \mathrm{r}}\) = \(\frac{\left(9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^2}{(2)\left(-2.2 \times 10^{-18} \mathrm{~J}\right)}\)
= 5.3 × 10^-11 m.
The velocity of the revolving electron can be computed from Equation r = –\(\frac{e^2}{4 \pi \varepsilon_0 r m v^2}\)
with
m = 9.1 × 10^-31kg,
υ = \(\frac{\mathrm{e}}{\sqrt{4 \pi \varepsilon_0 \mathrm{mr}}}\) = 2.2 × 10⁶ m/s

Question 4.
According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.
Solution:
From Examjple 3 we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10^-11 m is 2.2 × 10^-6 m/s. Thus, the frequency of the electron moving around the proton is
v = \(\frac{v}{2 \pi \mathrm{r}}=\frac{2.2 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}}{2 \pi\left(5.3 \times 10^{-11} \mathrm{~m}\right)}\)
≈ 6.6 × 10¹⁵ Hz.
According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 10¹⁵Hz.

Question 5.
A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite.
Solution:
From equation, we have
mυ_nr_n = nh/2π
Here m = 10 kg and r_n = 8 × 10⁶ m. We have the time period T of the circling satellite as 2h. That is T = 7200 s.
Thus the velocity υ_n = 2π r_n/T
The quantum number of the orbit of satellite
n = (2π r_n)² × m(T × h)
Substituting the values,
n = (2π × 8 × 10⁶m)² × 10/(7200 s × 6.64 × 10^-34 J s)
= 5.3 × 10⁴⁵
Note that the quantum number for the satellite motion is extremely large. In fact for such large quantum numbers the results of quantisation conditions tend to those of classical physics.

Question 6.
Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.
Solution:
The Rydberg formula is
hc/λ_if = \(\left(\frac{1}{n_{\mathrm{f}}^2}-\frac{1}{\mathrm{n}_{\mathrm{i}}^2}\right)\)
The wavelengths of the first four lines in the Lyman series correspond to transitions from n_i = 2, 3, 4, 5 to n_f = 1. We know that
\(\frac{\mathrm{me}^4}{8 \varepsilon_0^2 \mathrm{~h}^2}\) = 13.6 eV = 21.76 × 10^-19 J
Therefore,

Substituting, n_i = 2, 3, 4, 5 we get λ₂₁ = 1218 Å, λ₃₁ = 1028 Å, λ₄₁ = 974.3 Å, and λ₅₁ = 951.4 Å.

Students get through AP Inter 2nd Year Physics Important Questions 14th Lesson Nuclei which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 14th Lesson Nuclei

Very Short Answer Questions

Question 1.
The half life of ⁵⁸Co is 72 days. Calculate Its average life. [Board model Paper]
Answer:
T_half = 0.693 × T_Mean ⇒ T_Mean = \(\frac{T_{\text {half }}}{0.693}=\frac{72}{0.693}\) = 103.8 days.

Question 2.
Why do all electrons emitted during p-decay not have the same energy?
Answer:
When a neutron is converted into a proton, an electron and neutron are emitted along with it.
\({ }_1^1 \mathrm{n} \longrightarrow{ }_1^1 \mathrm{H}+{ }_{-1}^0 \mathrm{e}+\mathrm{v}\)
In β – decay proton remains in the nucleus, but electron and neutron are emitted with constant energy. The energy of neutron is not constant. So, ail electrons do not have same energy.

Question 3.
Neutrons are the best projectiles to produce nuclear reactions. Why ?
Answer:
Neutrons are uncharged particles. So they do not get deflected by the electric and magnetic fields. Hence Neutrons are considered as best projectiles in nuclear reaction.

Question 4.
Neutrons cannot produce ionization. Why ?
Answer:
Because neutrons are uncharged particles and cannot produce ionization.

Question 5.
What are delayed neutrons ?
Answer:
Neutrons are emitted in the fission products after, sometime are called delayed neutrons.

Question 6.
What are thermal neutrons ? What is their importance ?
Answer:
Neutrons having kinetic energies approximately 0.025 eV are called as slow neutrons or thermal neutrons. ²³⁵U undergoes fission only when bombarded with thermal neutrons.

Question 7.
What is the value of neutron multiplication factor in a controlled reaction and in an uncontrolled chain reaction ?
Answer:
In controlled chain reaction K = 1
In uncontrolled chain reaction K > 1

Question 8.
What is the role of controlling rods in a nuclear reactor ?
Answer:
In nuclear reactor, controlling rods are used to absorb the neutrons. Cadmium, boron materials are used in the form of rods in reactor. These control the fission rate.

Question 9.
Why are nuclear fusion reactions called thermo nuclear reactions ?
Answer:
Nuclear fusion occurs at very high temperatures. So it is called as thermo nuclear reaction.

Question 10.
Define Becquerel and Curie.
Answer:
Becquerel: 1 disintegration or decay per second is called Becquerel. It is SI unit of activity.
1 disintegration or decay
i.e., Becquerel = \(\frac{1 \text { disintegration or decay }}{\text { second }}\)
Curie : 3.7 × 10¹⁰ decays per second is called Curie.
1 Curie : 1 Ci = \(\frac{3.7 \times 10^{10} \text { decays }}{\text { second }}\) = 3.7 × 10¹⁰Bq.

Question 11.
What is a chain reaction ?
Answer:
Chain reaction : The neutrons produced in the fission of a nucleus can cause fission in other neighbouring nuclei producing more and more neutrons to continue the fission until the whole fissionable material is disintegrated. This is called chain reaction.

Question 12.
What is the function of moderator in a nuclear reactor ?
Answer:
They are used to slow down the fast moving neutrons produced during the fission process.
e.g.: Heavy water, Beryllium.

Question 13.
What is the energy released in the fusion of four protons to form a helium nucleus ?
Answer:
26.7 MeV energy is released.

Short Answer Questions

Question 1.
Write a short note on the discovery of neutron.
Answer:

1. Bothe and Becker found that when beryllium is bombarded with a – particles of energy 5 MeY which emitted a highly penetrating radiation.
2. The equation for above process can be written as
   \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{13} \mathrm{C}\) + γ – (radiation energy)
3. The radiations are not effected by electric and magnetic fields.
4. In 1932, James Chadwick, had subjected nitrogen and argon to the beryllium radiation. He interpreted the experimental results by assuming that the radiation is of a new kind of particles which has no charge and its mass is equal to proton. These neutral particles were named as neutrons. Thus the neutron was discovered.
5. The experimental results can be represented by the following equation.
   \({ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{12} \mathrm{C}+{ }_0^1 \mathrm{n}+\mathrm{Q}\)
   

Question 2.
What are nuclear forces ? Write their properties.
Answer:
The forces which hold the nucleons together in nucleus are called nuclear forces.
Properties of Nuclear forces:

1. Nuclear forces are attractive forces between proton and proton (P – N, proton and neutron (P – N) and neutron and neutron (N – N).
2. Nuclear forces are independent of charge. It was found that force between proton and proton is same as force between neutron and neutron.
3. These forces are short range forces i.e., these forces will act upto a small distance only. Generally the range of nuclear forces is upto few Fermi (10^-15 m).
4. These forces are non central forces, i.e., they do not act along the line joining the two nucleons.
5. These forces are exchange forces. The force between two nucleons is due to exchange of n-mesons.
6. These forces are spin dependent. These forces are strong when the spin of two nucleons are in same direction and they are weak when they are in opposite direction.
7. Nuclear forces are saturated forces i.e., the force between nucleons will extend upto the immediate neighbouring nucleons only.
8. These are the strongest forces in nature. They are nearly 10³⁸ times stronger than gravitational forces and nearly 100 times stronger than Coulombic forces.

Question 3.
Define half life period and decay constant for a radioactive substance. Deduce the relation between them.
Answer:
Half life period (T) : Time taken for the number of radio active nuclei to disintegrate to half of its original number of nuclei is called Half life period.

Decay constant (λ) : The ratio of the rate of radioactive decay to the number of nuclei present at that instant.
It is a proportional constant and is denoted by ‘λ’.
∴ λ = \(\frac{-\left(\frac{d N}{d t}\right)}{N}\)

Relation between half the period and decay constant:

1. The radioactive decay law N = N₀ e^-λt states that the number of radioactive nuclei in a radioactive sample decreases exponentially with time. Here A is called decay constant.
2. If N₀ is the number of nuclei at t = 0 and N is the radioactive nuclei at any instant of time’t’.
3. Substituting N = \(\frac{\mathrm{N}_0}{2}\) at t = T in N = N₀ e^-λt
4. \(\frac{\mathrm{N}_0}{2}\) = N₀ e^-λT
   e^λT = 2
   λT= ln 2
   T = \(\frac{\ln 2}{\lambda}=\frac{2.303 \log _{10}^2}{\lambda}\)
   ∴ T = \(\frac{0.693}{\lambda}\)

Question 4.
What is nuclear fission ? Give an example to illustrate it.
Answer:
Nuclear fission : The process of dividing a heavy nucleus into two or more smaller and stable nuclei due to nuclear reaction is called nuclear fission.
Ex: The fission reaction is \({ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_0^1 \mathrm{n}+\mathrm{Q}\)
Where Q is the energy released.
Q = (Total mass of reactants – Total mass of product) C²
= [(Mass of \({ }_{92}^{235} \mathrm{U}\) + Mass of \({ }_0^1 \mathrm{n}\)) – (Mass of \({ }_{56}^{141} \mathrm{Ba}\) + Mass of \({ }_{36}^{92} \mathrm{Kr}\) + Mass of three neutrons)] C²
= (235.043933 – 140.9177 – 91.895400 – 2 × 1.008665) amu × C².
= 0.2135 × 931.5 MeV = 198.9 MeV = 200 MeV

Question 5.
What is nuclear fusion ? Write the conditions for nuclear fusion to occur.
Answer:
Nuclear fusion : The process of combining lighter nuclei to produce a larger nucleus is known as nuclear fusion.
E.g : Hydrogen nuclei (₁H¹) are fused together to form heavy Helium (₂He⁴) along with 25.71 MeV energy released.
Conditions for nuclear fusion :

1. Nuclear fusion occurs at very high temperatures such as 10⁷ kelvin and very high pressures. These are obtained under the explosion of an atom bomb.
2. Higher density is also desirable so that collisions between light nuclei occur quite frequently.

Question 6.
Distinguish between nuclear fission and nuclear fusion.
Answer:
Nuclear fission

1. In this process heavy nucleus is divided into two fragments along with few neutrons.
2. These reactions will takes place even at room temperature.
3. To start fission atleast one thermal neutron from outside is compulsory.
4. Energy released per unit mass of participants is less.
5. In this process neutrons are liberated.
6. This reaction can be controlled.
   Ex: Nuclear reactor.
7. Atom bomb works on principle of fission reaction.
8. The energy released in fission can be used for peaceful purpose.
   Ex: Nuclear reactor and Atomic power stations.

Nuclear fusion

1. In this process lighter nuclei will join together to produce heavy nucleus.
2. These reactions will takes place at very high temperature such as 107 Kelwin.
3. No necessary of external neutrons.
4. Energy released per unit mass of participants is high. Nearly seven times more than fission reaction.
5. In this process positrons are liberated.
6. There is no control on fusion reaction.
7. Hydrogen bomb works on the principle of fusion reaction.
8. The energy released in fusion cannot be used for peaceful purpose.

Long Answer Questions

Question 1.
Define mass defect and binding energy. How does binding energy per nucleon vary with mass number ? What is its significance ?
Answer:
1. Mass defect (∆M) : The difference in mass of a nucleus and its constituents is called the mass defect. The nuclear mass M is always less than the total mass, Em, of its constituents.
Mass defect, (∆M) = [Zm_p + (A – Z)m_n – M]

2. Binding energy: The energy required to break the nucleus into its constituent nucleons is called the binding energy.
Binding Energy, (E_b) = ∆MC² = [Zm_p + (A – Z)m_n – M] 931.5 MeV
Nuclear binding energy is an indication of the stability of the nucleus.
Nuclear binding energy per nucleon E_bn = \(\frac{E_{\mathrm{b}}}{\mathrm{A}}\)

3. The following graph represents how the binding energy per nucleon varies with the mass number A.

4. From the graph that the binding energy is highest in the range 28 < A < 138. The binding energy of these nuclei is very close to 8.7 MeV

5. With the increase in the mass number the binding energy per nucleon decreases and consequently for the heavy nuclei like Uranium it is 7.6 MeV

6. In the region of smaller mass numbers, the binding energy per nucleon curve shows the characteristic minima and maxima.

7. Minima are associated with nuclei containing an odd number of protons and neutrons such as \({ }_3^6 \mathrm{Li},{ }_5^{10} \mathrm{~B},{ }_7^{14} \mathrm{~N}\) and the maxima are associated with nuclei having an even number of protons and neutrons such as \({ }_2^4 \mathrm{He},{ }_6^{12} \mathrm{C},{ }_8^{16} \mathrm{O}\).

Significance:
8. The curve explains the relationship between binding energy per nucleon and stability of the nuclei.

9. Uranium has a relatively low binding energy per nucleon as 7.6 MeV Hence to attain greater stability Uranium breaks up into intermediate mass nuclei resulting in a phenomenon called fission.

10. On the other hand light nuclei such as hydrogen combine to form heavy nucleus to form helium for greater stability, resulting in a phenomenon called fusion.

11. Iron is the most stable having binding energy per nucleon 8.7 MeV, and it neither undergoes fission per fusion.

Question 2.
What is radioactivity ? State the law of radioactive decay. Show that radioactive decay is exponential in nature. [T.S. Mar. 16]
Answer:
1. Radioactivity : The nuclei of certain elements disintegrate spontaneously by emitting alpha (α), beta (β) and gamma (δ) rays. This phenomenon is called Radioactivity or Natural radioactivity.

2. Law of radioactivity decay : “The rate of radioactive decay \(\left(\frac{d N}{d t}\right)\) (or) the number of
nuclei decaying per unit time at any instant, is directly proportional to the number of nuclei (N) present at that instant is called law of radioactivity decay”.

3. Radioactive decay is exponential in nature : Consider a radioactive substance. Let the number of nuclei present in the sample at t = 0, be N₀ and let N be the radioactive nuclei remain at an instant t.
\(\frac{\mathrm{dN}}{\mathrm{dt}}\) ∝ N ⇒ \(\frac{\mathrm{dN}}{\mathrm{dt}}\) = – λN
dN = – λ Ndt …………………….. (1)
The proportionality constant λ is called decay constant or disintegration constant. The negative sign indicates the decrease in the number of nuclei.

4. From eq. (1) \(\frac{\mathrm{dN}}{\mathrm{N}}\) = – λ dt ……………… (2)

5. Integrating on both sides
\(\int \frac{\mathrm{dN}}{\mathrm{N}}=-\lambda \int \mathrm{dt}\)
ln N = – λt + C …………….. (3)
Where C = Integration constant.

6. At t = O; N = N₀. Substituting in eq. (3), we get ln N₀ = C
∴ ln N = -λt + ln N₀
ln N – ln N₀ = – λt
ln (\(\frac{\mathrm{N}}{\mathrm{N}_0}\)) = – λt
∴ N = N₀ e^-λt
The above equation represents radioactive decay law.

7. It states that the number of radioactive nuclei in a radioactive sample decreases exponentially with time.

Sample Problems

Question 1.
The half life of radium is 1600 years. How much time does lg of radium take to reduce to 0.125g. [IPE 2016 (TS)]
Answer:
Half – life of radium = 1600 years;
\(\frac{N}{N_0}=\frac{1}{2^n} \Rightarrow \frac{0.125}{1}=\frac{1}{2^n} \Rightarrow \frac{125}{1000}=\frac{1}{2^n} \Rightarrow \frac{1}{8}=\frac{1}{2^n} \Rightarrow \frac{1}{2^3}=\frac{1}{2^n} \Rightarrow n=3\)
∴ Time taken = Half life × no. of Half lives = 1600 × 3 = 4800 years

Question 2.
Plutonium decays with a half-life of 24,000 years. If plutonium is stored for 72,000 years, what fraction of it remains ?
Answer:
Half life of plutonium, T = 24000 years; Stored time of plutonium, t = 72000 years
no. of half lives, n = \(\frac{t}{T}=\frac{72000}{24000}\) = 3; Fraction of plutonium remains = \(\frac{N}{N_0}=\frac{1}{2^n}=\frac{1}{2^3}=\frac{1}{8}\)

Question 3.
Explain the principle and working of a nuclear reactor with the help of a labelled diagram. [A.P. & T.S. Mar. 17, A.P. Mar. 16 15, Mar. 14]
Answer:
Principle : A nuclear reactor works on the principle of achieving controlled chain reaction in natural Uranium ²³⁸U enriched with ²³⁵U, consequently generating large amounts of heat.
A nuclear reactor consists of (1) Fuel (2) Moderator (3) Control rods (4) Radiation shielding (5) Coolant.

1. Fuel and clad : In reactor the nuclear fuel is fabricated in the form of thin and long cylindrical rods. These group of rods treated as a fuel assembly. These rods are surrounded by coolant, which is used to transfer of heat produced in them. A part of the nuclear reactor which is used to store the nuclear fuel is called the core of the reactor. Natural uranium, enriched uranium, plutonium and uranium – 233 are used as nuclear fuels.

2. Moderator : The average energy of neutrons released in fission process is 2 MeV. They are used to slow down the velocity of neutrons. Heavy water or graphite are used as moderating materials in reactor.

3. Control Rods : These are used to control the fission rate in reactor by absorbing the neutrons. Cadmium and boron are used as controlling the neutrons, in the form of rods.

4. Shielding : During fission reaction beta and gamma rays are emitted in addition to neutrons. Suitable shielding such as steel, lead, concrete etc., are provided around the reactor to absorb and reduce the intensity of radiations to such low levels that do not harm the operating personnel.

5. Coolant : The heat generated in fuel elements is removed by using a suitable coolant to flow around them. The coolants used are water at high pressures, molten sodium etc.

Working : Uranium fuel rods are placed in the aluminium cylinders. The graphite moderator is placed in between the fuel cylinders. To control the number of neutrons, a number of control rods of cadnium or beryllium or boron are placed in the holes of graphite block. When a few ²³⁵U nuclei undergo fission fast neutrons are liberated. These neutrons pass through the surrounding graphite moderator and loose their energy to become thermal neutrons. These thermal neutrons are captured by ²³⁵U. The heat generated here is used for heating suitable coolants which is turn heat water and produce steam. This steam is made to rotate steam turbine and there by drive a generator of production for electric power.

Question 4.
Explain the source of stellar energy. Explain the carbon – nitrogen cycle, proton – proton cycle occuring in stars.
Answer:
Scientists proposed two types of cyclic processes for the sources of energy in the sun and stars. The first is known as carbon-nitrogen cycle and the second is proton-proton cylce.

1. Carbon-Nitrogen Cycle : According to Bethe carbon-nitrogen cycle is mainly responsible for the production of solar energy. This cycle consists of a chain of nuclear reactions in which hydrogen is converted into Helium, with the help of Carbon and Nitrogen as catalysts. The nuclear reactions are as given below.

2. Proton – Proton Cycle: A star is formed by the condensation of a large amount of matter at a point in space. Its temperature rises to 2,00,000°C as the matter contracts under the influence of gravitational attraction. At this temperature the thermal energy of the protons is sufficient to form a deuteron and a positron. The deuteron then combines with another proton to form lighter nuclei of helium \({ }_2^3 \mathrm{He}\). Two such helium nuclei combine to form a helium nucleus latex]{ }_2^4 \mathrm{He}[/latex] and two protons releasing a total amount of energy 25.71 MeV The nuclear fusion reactions are given below.

Problems

Question 1.
Compare the radii of the nuclei of mass numbers 27 and 64.
Solution:
A₁ = 27; Asub>2 = 64
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left[\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right]^{1 / 3}\) [∵ R = R₀A^1/3]
\(\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left[\frac{27}{64}\right]^{\frac{1}{3}}=\frac{3}{4}\)
∴ R₁ : R₂ = 3 : 4

Question 2.
Find the energy required to split \({ }_8^{16} \mathrm{O}\) nucleus into four α-particles. The mass of an a-particle is 4.002603u and that of oxygen is 15.994915u.
Solution:
The energy required to split O = [Total mass of the products – Total mass of the reactants] c²
Mass of four \({ }_2^4 \mathrm{He}\) – Mass of \({ }_8^{16} \mathrm{O}\)] × c²
= [(4 X 4.002603) – 15.994915] u × c²
= [16.010412 – 15.994915] u × c²
= (0.015497) 931.5 MeV = 14.43 MeV

Question 3.
A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.
Solution:
Fraction of substance decays
= \(\frac{\text { Quantity remains }}{\text { Initial quantity }}\)
= \(\frac{1}{2^n}=\frac{1}{32}=\frac{1}{2^5}\)
∴ n = 5
Duration of time = 25 days
We know (n) = \(\frac{\text { Duration of time }}{\text { Half life time }}\)
∴ Half life time = \(\frac{\text { Duration of time }}{\mathrm{n}}\)
\(\frac{25}{5}\) = 5 days

Question 4.
One gram of radium is reduced by 2 milli- gram in 5 years by a-decay. Calculate the half¬life of radium.
Solution:
Initial (original) mass (N₀) = 1 gram
Reduced mass – 2 mg = 0.002 grams
Final mass (N)= 1 – 0.002 = 0.998 grams
t = 5 years
e^-λt = \(\frac{\mathrm{N}}{\mathrm{N}_0}\) ⇒ e^λt = \(\frac{\mathrm{N}_0}{\mathrm{~N}}\) ⇒ λt = log_e[latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex]
λt = 2.303 log [latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex]
λt = 2.303 log [latex]\frac{1}{0.998}[/latex]
= 2.303 log (1.002)
= 2.303 × 0.000868
= 0.001999
λ = \(\frac{0.001999}{5}\) = 0.0003998
T = \(\frac{0.693}{\lambda}=\frac{0.693}{0.0003998}\) = 1733.3 years

Question 5.
If one microgram of \({ }_92^{235} \mathrm{U}\) is completely destroyed in an atomhomb, how much energy will be released ?
Solution:
m = 1 μg = 1 × 10^-6 g = 1 × 10^-6 × 10^-3 kg
= 10^-9 kg
c = 3 × 10⁸ m/s
E = mc² = 1 × 10^-9 × 9 × 10⁶ = 9 × 10⁷ J

Question 6.
200 MeV energy is released when one nucleus of ²³⁵U undergoes fission. Find the number of fissions per second required for producing a power of 1 megawatt.
Solution:
E = 200 MeV
P = 1 × 10⁶ W
P = \(\frac{\mathrm{nE}}{\mathrm{t}} \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{P}}{\mathrm{E}}=\frac{10^6}{200 \times 10^6 \times 1.6 \times 10^{-19}}\)
= \(\frac{1}{32}\) × 10¹⁸
∴ P = 0.03125 × 10¹⁸
= 3.125 × 10⁶

Textual Examples

Question 1.
Given the mass of iron nucleus as 55.85u and A = 56, find the nuclear density ?
Solution:
m_Fe = 55.85, u = 9.27 × 10^-26 kg
Nuclear density = \(\frac{\text { mass }}{\text { volume }}\)
= \(\frac{9.27 \times 10^{-26}}{(4 \pi / 3)\left(1.2 \times 10^{-15}\right)^3} \times \frac{1}{56}\)
= 2.29 × 10¹⁷ kg m^-3
The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus.

Question 2.
Calculate the energy equivalent of 1 g of substance.
Solution:
Energy, E = 10^-3 × (3 × 10⁸)² J
E = 10^-3 × 9 × 10¹⁶ = 9 × 10¹³ J
Thus, if one gram of matter is converted to energy, there is a release of an enormous amount of energy.

Question 3.
Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of \({ }_8^{16} \mathrm{O}\) in MeV/c².
Solution:
1 u = 1.6605 × 10^-27 kg
To convert it into energy units, we multiply it by c² and find, that energy. equivalent
= 1.6605 × 10^-27 × 2.9979 × 10⁸ kg m²/s²
= 1.4924 × 10^-10 J
= \(\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}}\) eV
= 0.9315 × 10⁹ eV = 931.5 MeV
or, 1 u = 931.5 MeV/c².
For \({ }_8^{16} \mathrm{O}\), ∆M = 0.13691 u = 0.13691 × 931.5 MeV/c² = 127.5 MeV/c²
The energy needed to separate \({ }_8^{16} \mathrm{O}\) into hs constituents is thus 127.5 MeV/c².

Question 4.
The half-life of \({ }_{92}^{238} \mathrm{U}\) undergoing a – decay is 4.5 × 10⁹ years. What is the activity of 1 g sample of \({ }_{92}^{238} \mathrm{U}\) ?
Solution:
T_1/2 = 4.5 × 10⁹ y
= 4.5 × 10⁹ y × 3.16 × 10⁷ s/y.
= 1.42 × 10¹⁷ s
One k mol of any isotope contains Avogadro’s number of atoms, and so 1 g of \({ }_{92}^{238} \mathrm{U}\) contains
\(\frac{1}{238 \times 10^{-3}}\) k mol × 6.025 × 10²⁶ atoms/kmol
= 25.3 × 10²⁰ atoms.
The decay rate R is
R = λN
= \(\frac{0.693}{\mathrm{~T}_{1 / 2}}\) N = \(\frac{0.693 \times 25.3 \times 10^{20}}{1.42 \times 10^{17}}\) S^-1
= 1.23 × 10⁴ S^-1
= 1.23 × 10⁴ Bq

Question 5.
Tritium has a half-life of 12.5 y undergoing beta decay. What fraction of sample of pure tritium will remain undecayed after 25 y.
Solution:
By definition of half-life, half of the initial i sample will remain undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei would have decayed. Hence, one fourth of the sample of the initial pure s tritium will remain undecayed.

Question 6.
We are given the following atomic masses:
\({ }_{92}^{238} \mathrm{U}\) = 238.05079 u
\({ }_{2}^{4} \mathrm{He}\) = 4.00260 u
\({ }_{90}^{234} \mathrm{Th}\) = 234.04363 u
\({ }_{1}^{1} \mathrm{H}\) = 1.00783 u
\({ }_{91}^{237} \mathrm{Pa}\) = 237.05121 u
Here the symbol Pa is for the element protactinium (Z = 91).
a) Calculate the energy released during the alpha decay of \({ }_{92}^{238} \mathrm{U}\).
b) Show that \({ }_{92}^{238} \mathrm{U}\) cannot spontaneously emit a proton.
Solution:
a) The alpha decay of \({ }_{92}^{238} \mathrm{U}\) is given by

The energy released in this process is given by
Q = (M_U – M_Th – M_He) c²
Substituting the atomic masses as given in the data, we find
Q = (238.05079 – 234.04363 – 4.00260)u × c²
= (0.00456 u) c²
= (0.00456 u) (931.5 MeV/u)
= 4.25 MeV

b) If \({ }_{92}^{238} \mathrm{U}\) spontaneously emits a proton, the decay process would be
\({ }_{92}^{238} \mathrm{U} \rightarrow{ }_{91}^{237} \mathrm{~Pa}+{ }_1^1 \mathrm{H}\)
The Q for this process to happen is
= (M_U – M_Pa – M_H)c²
(238.05079 – 237.05121 – 1.00783) u × c²
=(- 0.00825 u) c²
= – (0.00825 u) (931.5 MeV/u)
= -7.68 MeV
Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply an energy of 7.68 MeV to a \({ }_{92}^{238} \mathrm{U}\) nucleus to make it emit a proton.

Students get through AP Inter 2nd Year Physics Important Questions 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits

Very Short Answer Questions

Question 1.
What is an n-type semiconductor ? What are the majority and minority charge carriers in it ?
Answer:

• If a pentavalent impurity is added to pure tetravalent semiconductor, it is called n-type semiconductor.
• In n-type semiconductor majority charge carriers are electrons and minority charge carriers are holes.

Question 2.
What are intrinsic and extrinsic semiconductors ? [A.P. Mar. 15]
Answer:

• Pure form of semiconductors are called intrinsic semiconductors.
• When impure atoms are added to increase their conductivity, they are called extrinsic semiconductors.

Question 3.
What is a p-type semiconductor ? What are the majority and minority charge carriers in it ? [A.P. & T.S. Mar. 17]
Answer:
If a trivalent impurity is added to a tetravalent semiconductor, it is called p-type semi-conductor.

In p-type semiconductor majority charge carriers are holes and minority charge carriers are electrons.

Question 4.
What is a p-n junction diode ? Define depletion’ layer.
Answer:
When an intrinsic semiconductor crystal is grown with one side doped with trivalent element and on the other side doped with pentavalent element, a junction is formed in the crystal. It is called p-n junction diode.

A thin narrow region is formed on either side of the p-n junction, which is free from charge carriers is called depletion layer.

Question 5.
How is a battery connected to a junction diode in i) forward and ii) reverse bias ?
Answer:
i) In p-n junction diode, if p-side is connected to positive terminal of a cell and n-side to negative terminal, it is called forward bias.

ii) In a p-n junction diode, p-side is connected to negative terminal of a cell and n-side to positive terminal, it is called reverse bias.

Question 6.
What is the maximum percentage of rectification in half wave and full wave rectifiers ?
Answer:

1. The percentage of rectification in half-wave rectifier is 40.6%.
2. The percentage of rectification in full-wave rectifiers is 81.2%.

Question 7.
What is Zener voltage (V_z) and how will a Zener diode be connected in circuits generally ?
Answer:

1. When Zener diode is in reverse biased, at a particular voltage current increases suddenly is called Zener (or) break down voltage.
2. Zener diode always connected in reverse bias.

Question 8.
Write the expressions for the efficiency of a full wave rectifier and a half wave rectifier.
Answer:

1. Efficiency of half wave rectifier (η) = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
2. Efficiency of full wave rectifier (η) = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)

Question 9.
What happens to the width of the depletion layer in a p-n junction diode when it is
i) forward-biased and ii) reverse biased ?
Answer:
When a p-n junction diode is forward bias, thickness of depletion layer decreases and in reverse bias, thickness of depletion layer increases.

Question 10.
Draw the circuit symbols for p-n-p and n-p-n transistors. [Mar. 14]
Answer:

Question 11.
Define amplifier and amplification factor.
Answer:

1. Rising the strength of a weak signal is known as amplification and the device is called amplifier.
2. Amplification factor is the ratio between output voltage to the input voltage.
   A = \(\frac{\mathrm{V}_0}{\mathrm{~V}_{\mathrm{i}}}\)

Question 12.
In which bias can a Zener diode be used as voltage regulator ?
Answer:
In reverse bias Zener diode can be used as voltage regulator.

Question 13.
Which gates are called universal gates ? [T.S. Mar. 15]
Answer:
NAND gate and NOR gates are called universal gates.

Question 14.
Write the truth table of NAND gate. How does it differ from AND gate ?
Answer:

Short Answer Questions

Question 1.
What is a rectifier ? Explain the working of half wave and full Wave rectifiers with diagrams. [A.P. Mar. 17]
Answer:
Rectifier: It is a circuit which converts ac into d.c. A p-n junction diode is used as a rectifier.
Half-wave rectifier:

1. A half wave rectifier can be constructed with a single diode. The ac input signal is fed to the primary coil of a transformer. The output signal is taken across the load resistance R_L.
2. During positive half cycle, the diode is forward biased and current flows through the diode.
3. During negative half cycle, diode is reverse biased and no current flows through the load resistance.
4. This means current flows through the diode only during positive half cycles and negative half cycles are blocked. Hence in the output we get only positive half cycles.
5. Rectifier efficiency is defined as the ratio of output dc power to the input ac power.
   η = \(\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}=\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
   Where r_f = Forward resistance of a diode; R_L = Load resistance
   The maximum efficiency of half wave rectifier is 40.6%.

Full wave rectifier : The process of converting an alternating current into a direct current is called rectification.
The device used for this purpose is called rectifier.

1. A full wave rectifier can be constructed with the help of two diodes D₁ and D₂.
2. The secondary transformer is centre tapped at C and its ends are connected to the p regions of two diodes D₁ & D₂.
3. The output voltage is measured across the load resistance R_L.
4. During positive half cycles of ac, the diode D₁ is forward biased and current flows through the load resistance R_L. At this time D₂ will be reverse biased and will be in switch off position.
5. During negative half cycles of ac, the diode D₂ is forward biased and the current flows through R_L. At this time D₁ will be reverse biased and will be in switch off position.
6. Hence positive output is obtained for all the input ac signals.
7. The efficiency of a rectifier is defined as the ratio between the output dc power to the input ac power.
   η = \(\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}=\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}\)
   The maximum efficiency of a full wave rectifier is 81.2%.

Question 2.
What is a junction diode ? Explain the formation of depletion region at the junction. Explain the variation of depletion region in forward and reverse-biased condition.
Answer:
p-n junction diode: When a p-type semiconductor is suitablyjoined to n-type semiconductor, a p-n junction diode is formed. ’
The circuit symbol of p-n junction diode is shown in figure.

Formation of depletion layer at the junction: When p-n junction is formed, the free electrons on n-side diffuse over to p-side, they combine with holes and become neutral. Similarly holes on p-side diffuse over to n-side and combihe with electrons become neutral.

This results in a narrow region formed on either side of the junction. This region is called depletion layer. Depletion layer is free from charge carriers.

The n-type material near the junction becomes positively charged due to immobile donor ions and p-type material becomes negatively charged due to immobile acceptor ions. This creates an electric field near the junction directed from n-region to p-region and cause a potential barrier.

The potential barrier stops further diffusion of holes and electrons across the* junction. The value of the potential barrier depends upon the nature of the crystal, its temperature and the amount of doping.

Forward bias :
“When a positive terminal of a battery is connected to p-side and negative terminal is connected to n-side; then p-n junction diode is said to be forward bias”.

The holes in the p-region are repelled by the positive polarity and move towards the junction. Similarly electrons in the n-region are repelled by the negative polarity and move towards the junction.

As a result, the width of the depletion layer decreases. The charge carriers cross the junction apd electric current flows in the circuit.
Hence in forward bias resistance of diode is low. This position is called switch on position.
Reverse bias:

“When the negative terminal of the battery is connected to p-side and positive terminal of the battery is connected to n-side of the p-n junction, then the diode is said to be reverse bias”.

The holes in the p-region are attracted towards negative polarity and move away from the junction. Similarly the electrons in the n-region are attracted towards positive polarity and move away from the junction.

So, width of the depletion layer and potential barrier increases. Hence resistance of p-n junction diode increases. Thus the reverse biased diode is called switch off position.

Question 3.
What is a Zener diode ? Explain how it is used as a voltage regulator.
Answer:
Zener diode : Zener diode is a heavily doped germanium (or) silicon p-n junction diode. It works on reverse bias break down region.
The circuit symbol of zener diode is shown in figure.

Zener diode can be used as a voltage regulator. In- general zener diode is connected in reverse bias in the circuits.

1. The zener diode is connected to a battery, through a resistance R. The battery reverse biases the zener diode.
2. The load resistance R_L is connected across the terminals of the zener diode.
3. The value of R is selected in such away that In the absence of load R_L maximum safe current flows in the diode.
4. Now consider that load is connected across the diode. The load draws a current.
5. The current through the diode falls by the same amount but the voltage drops across the load remains constant.
6. The series resistance ‘R’ absorbs the output voltage fluctuations, so as to maintain constant voltage across the load.
7. The voltage across the zener diode remains constant even if the load R_L varies. Thus, zener diode works as voltage regulator.
8. If I is the input current, I_Z and I_L are zener and load currents.
   I = I_Z + I_L; V_in = IR + V_Z
   But V_0ut = V_Z
   ∴ V_out = V_in – IR

Question 4.
Explain the working of LED and what are its advantages over conventional incandescent low power lamps.
Answer:
Light emitting diode (LED) : It is a photoelectronic device which converts electrical energy into light energy.

It is a heavily doped p-n junction diode which under forward bias emits spontaneous radiation. The diode is covered with a transparent cover so that the emitted light may not come out. Working : When p-n junction diode is forward biased, the movement of majority charge carriers takes place across the junction. The electrons move from n-side to p-side through the junction and holes move from p-side to n-side.

As a result of it, concentration of majority carriers increases rapidly at the junction.

When there is no bias across the junction, therefore there are excess minority carriers on either side of the junction, which recombine with majority carriers near the junction.

On recombination of electrons and hole, the energy is given out in the form of heat and light.
Advantages of LED’s over incandescent lamp :

1. LED is cheap and easy to handle.
2. LED has less power and low operational voltage.
3. LED has fast action and requires no warm up time.
4. LED can be used in burglar alarm system.

Question 5.
Define NAND and NOR gates. Give their truth tables. [T.S. Mar. 17]
Answer:
NAND gate : NAND gate is a combination of AND gate and NOT gate.

NAND gate can be obtained by connecting a NOT gate in the output of an AND gate. NAND gates are called universal gates.

1. If both inputs are low, output is high.
   A = 0, B = 0, X = 1
2. If any input is low, output is high.
   A = 0, B = 1, X = 1
   A = 1, B = 0, X =1
3. If both inputs are high, output is low.
   A = 1, B = 1, X = 0
   

NOR gate : NOR gate is a combination of OR gate and NOT gate when the output of OR gate is connected to NOT gate. It has two (or) more inputs and one output.

1. If both inputs are low, output is high.
   A = 0, B = 0, X = 1
2. If any input is high, the output is low.
   A = 0, B = 1, X = 0
   A = 1, B = 0, X = 0
3. If both inputs are high, the output is low.
   A = 1, B = 1, X = 0
   NOR gate is also a universal gate.
   

Question 6.
Explain the working of a solar cell and draw its I-V characteristics.
Answer:
Solar cell is a p-n junction device which converts solar energy into electric energy.
It consists of a silicon (or) gallium – arsenic p-n junction diode packed in a can with glass window on top. The upper layer is of p- type semiconductor. It is very thin so that the incident light photons may easily reach the p-n junction.

Working : When light (E = hv) falls at the junction, electron – hole pairs are generated near the junction. The electrons and holes produced move in opposite directions due to junction field. They will be collected at the two sides of the junction giving rise to a photo voltage between top and bottom metal electrodes. Top metal acts as positive electrode and bottom metal acts as a negative electrode. When an external load is connected across metal electrodes a photo current flows.

I-V characteristics : I – V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes. Because it does not draw current.

Uses : They are used in calculators, wrist watches, artificial satellites etc.

Question 7.
Explain the operation of a NOT gate and give its truth table. [IPE 15, T.S.]
Answer:
NOT gate: NOT gate is the basic gate. It has one input and one output. The NOT gate is also called an inverter. The circuit symbol of NOT gate is shown in figure.

1. If input is low, output is high.
   A = 0, X = \(\overline{0}\) = 1
2. If input is high, output is low.
   A = 1, X = \(\overline{1}\) = 0
   

Question 8.
Draw an OR gate using two diodes and explain its operation. Write the truth table and logic symbol of OR gate.
Answer:
OR Gate : It has two input terminals and one output terminal. If both inputs are low the output is low. If one of the inputs is high, or both the inputs are high then the output of the gate is high. The truth tables of OR gate.

In truth table logic function is written as A or B ‘OR’ logic function is represented by the symbol ‘plus’.
Q = A + B
Logic gate ‘OR’ is shown given below.

Implementation of OR gate using diodes :
Let D₁ and D₂ be two diodes.

A potential of 5V represents the logical value 1.
A potential of OV represents the logical value 0.

When A = 0, B = 0 both the diodes are reverse biased and there is no current through the resistance. So, the potential at Q is zero i.e., Q = 0. When A = 0 or B = 0 and the other equal to a potential of 5 V the diode whose anode is at a potential of 5 V is forward – biased and that diode behaves like a closed switch. The output potential then becomes 5 V i.e., Q = 1. When both A and B are 1, both the diodes are forward-biased and the potential at Q is same as that at A and B which is 5 V i.e., Q = 1. The output is same as that of the OR gate.

Question 9.
Sketch a basic AND circuit with two diodes and explain its operation. Explain how doping increases the conductivity in semiconductors ?
Answer:
AND gate : It has two input terminals and one output terminal.

• If both the inputs are low (or) one of the inputs is low.
  • The output is low in an AND gate.
• If both the inputs .are high
  • The output of the gate is high.
• Note : If A and B are the inputs of the gate and the output is ‘Q’ then ‘Q’ is a logical function of A and B.
  AND gate Truth Tables
  
  The logical function AND is represented by the symbol dot so that the output, Q = A.B and the circuit symbol used for the logic gate AND is shown in Fig.
  
  The logical function AND is similar to the multiplication.

Implementation of AND gate using diodes : Let D₁ and D₂ represents two diodes. A potential of 5 V represents the logical value 1 and a potential of 0 V represents the logical value zero (0). When A = 0, B = ,0 both the diodes D₁ and D₂ are forward-biased and they behave like closed switches. Hence, the output Q is same as that A or B (equal to zero.) When A = 0 or B = 0, D₁ or D₂ is forward – biased and Q is zero. When A = 1 and B = T both the diodes are reverse – biased and they behave like open switches. There is no current through the resistance R making the potential at Q equal to 5V i.e., Q = 1. The output is same as that of an AND gate.

Doping increases the conductivity in Semiconductors: If a pentavalent impurity (Arsenic) is added to a pure tetravalent semiconductor it is called n-type semiconductor. Arsenic has 5 valence electrons, but only 4 electrons are needed to form covalent bonds with the adjacent Germanium atoms. The fifth electron is very loosely bound and become a free electron. Therefore excess electrons are available for conduction and conductivity of semiconductor increases.

Similarly when a trivalent impurity Indium is added to pure Germanium it is called p-type semi-conductor. In this excess holes in addition to those formed due to thermal energy are available for conduction in the valence band and the conductivity of semiconductor increases.

Question 10.
Explain how transistor can be used as a switch ?
Answer:
To understand the operation of transistor as a switch.

1. As long as V_i is low and unable to forward bias the transistor, V₀ is high (at V_cc).
2. If V_i is high enough to drive the transistor into saturation then V₀ is low, very near to zero.
   
3. When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on.
4. We can say that a low input to the transistor gives a high output and high input gives a low output.
5. When the transistor is used in the cutoff (or) saturation state it acts as a switch.
   

Problems

Question 1.
In a half wave rectifier, a p-n junction diode with internal resistance 20 ohm is used. If the load resistance of 2 ohm is used in the circuit, then find the efficiency of this half wave rectifier.
Solution:
Internal resistance (r_f) = 20Ω
R_L = 2kΩ = 2000 Ω
η = \(\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.406 \times 2000}{20+2000} \times 100=\frac{812 \times 100}{2020}\)
η = 40.2%.

Question 2.
A full wave p-n junction diode rectifier uses a load resistance of 1300 ohm. The internal resistance of each diode is 9 ohm. Find the efficiency of this full wave rectifier.
Solution:
Given that R_L = 1300 Ω
r_f = 9Ω
η = \(\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.812 \times 1300}{9+1300} \times 100 ; \eta=\frac{8120 \times 13}{1309}\)
η = 80.64%.

Question 3.
Calculate the current amplification factor β(beta) when change in collector current is 1mA and change in base current is 20μA.
Solution:
Change in collector current (∆I_C) = 1mA = 10^-3 A
Change in base current (∆I_B) = 20 μA = 20 × 10^-6 A
β = \(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{10^{-3}}{20 \times 10^{-6}}\); β = \(\frac{1000}{20}\)
β = 50

Question 4.
For a transistor amplifier, the collector load resistance R_L = 2k ohm and the input resistance R_i = 1 k ohm. If the current gain is 50, calculate voltage gain of the amplifier.
Solution:
R_L = 2kΩ = 2 × 10³ Ω
R_i = 1kΩ = 1 × 10³ Ω
β = 50.
Voltage gain (A_V) = β × \(\frac{\mathrm{R}_{\mathrm{L}}}{\mathrm{R}_{\mathrm{i}}}=\frac{50 \times 2 \times 10^3}{1 \times 10^3}\)

A_V = 100.

Textual Examples

Question 1.
C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors ?
Solution:
The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out and electron from these atoms(i.e., ionisation energy E_g) will be least for Ge, following by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C.

Question 2.
Suppose a pure Si crystal has 5 × 10²⁸ atoms m^-3. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n. = 1.5 × 10¹⁶ m^-3.
Solution:
Note that thermally generated electrons (n_i ~ 10¹⁶ m^-3) are negligibly small as compared to those produced by doping.
Therefore, n_i ≈ N_D.
Since n_en_h = n_i², The number of holes, n_h = (1.5 × 10¹⁶)² / 5 × 10²⁸ × 16^-6
n_h = (2.25 × 10³²)/(5 × 10²²) ~ 4.5 × 10⁹m^-3

Question 3.
Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction ?
Solution:
No ! Any slab, howsoever flat, will have roughness much large than the inter-atomic crystal spacing(~2 to 3 A°) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.

Question 4.
The V-I characteristics of a silicon diode are shown in the Fig. Calculate the resistance of the diode at (a) I_D = 15 mA and (b) V_D = -10 V.

Solution:
Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law.
a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V
r_fb = ∆V/∆I = 0.1V/10mA = 10Ω
b) From the curve at V = -10 V, I = -1 µA.
Therefore r_rb = 10V/1µA = 1.0 × 10⁷ Ω

Question 5.
In a Zener regulated power supply a Zener diode’with V_z = 6.0 V is used for regulation. The load current is to be 4.0 mA and the unregulated input is 10,0 V. What should be the value of series resistor R_s?
Solution:
The value of Rs should be such that the current through the Zener diode is much larger than the load current. This is to have good load regulation. Choose Zener current as five times the load current, i.e., I_z = 20 mA. The total current through R_S is , therefore, 24 mA. The voltage drop across R_S is 10.0 – 6.0 = 4.0 V This gives R_S = 4.0V/(24 × 10^-3) A = 167Ω. The nearest value of carbon resistor is 150 Ω. So, a series resistor of 150 Ω is appropriate. Note that slight variation in the value of the resistor does not matter, what is important is that the current I_Z should be sufficiently larger than I_L.

Question 6.
The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~µA). What is the reason then to operate the photodiodes in reverse bias ?
Solution:
Consider the case of an n-type semiconductor. Obviously, the majority carrier density (n) is considerably larger than the minority hole density p (i.e., n> > p). On illumination, let the excess eletrons and holes generated be ∆n and ∆p, respectively,
n’ = n + ∆n
p’ = p + ∆p
Here n’ and p’ are the electron and hole concentrations at any particular illumination and n and p are carriers concentration when there is no illumination. Remember ∆n = ∆p and n > > p. Hence, the fractional change in the majority carriers (i.e., ∆n/n) would be much less than that in the minority carriers (i.e,, ∆p/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes-are preferably used in the reverse bias condition for measuring light intensity.

Question 7.
Why are Si and GaAs are preferred materials for solar cells ?
Solution:
The solar radiation spectrum received by us is shown in figure.

The maxima is near 1.5 eV. For photo-excitation, hv > E_g. Hence, semiconductor with band gap ~1.5 eV or lower is likely to give better solar conversion efficiency. Silicon has E_g ~ 1.1 eV while for GaAS it is ~ 1.53 eV. In fact, GaAs is better On spite of its higher band gap) than Si because of its relatively higher absorption coefficient. If we choose materials like CdS or CdSe(E_g ~ 2.4 eV), we can use only the high energy component of the solar energy for photo-conversion and a significant part of energy will be of no use.

The question arises: why we do not use material like pbS(E_g ~ 0.4 eV) which satisfy the condition hv > E_g for v maxima corresponding to the solar radiation spectra ? if we do so, most of the solar radiation will be absorbed on the top-layer of solar cell mid will not reach in or near the depletion region. For effective electron-hole separation, due to the junction field, we want the photo-generation to occur in the junction region only.

Question 8.
From the output charactristics shown in fig, calculate the values of β_ac and β_dc of the transistor when V_CE is 10 V and I_C = 4.0 mA.

Solution:
β_ac = \(\left(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right)_{\mathrm{V}_{\mathrm{CE}}}\) ; β_dc = \(\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}\)
For determining β_ac and β_dc at the stated values of V_CE and I_C one can proceed as follows. Consider any two characteristics for two values of I_B which lie above and below the given value of I_C. Here I_C = 4.0 mA. (Choose characteristics for I_B = 30 and 20μA.) At V _CE = 10V
we read the two values of Ic from the graph. Then
∆I_B = (30 – 20)μA = 10μA. ∆I_C
= (4.5 – 3.0) mA = 1.5 mA
Therefore, β_ac =1.5 mA/ 10μA = 150
For determining β_dc either estimate the value of I_B corresponding to I_C = 4.0 mA at V_CE = 10V or calculate the two values of β_dc for the two characteristics chosen and find their mean.
Therefore, for I_C = 4.5 mA and I_B = 30 μA
β_dc = 4.5 mA/30 μA = 150 and for I_C = 3.0 mA/ and I_B = 20 μA
β_dc = 3.0 mA / 20 μA= 150
Hence, β_dc = (150 + 150)/ 2 = 150

Question 9.
In Fig. the V_BB supply can be varied from OV to 5.0 V. The Si transistor has β_dc = 250 and R_B = 100 kΩ, R_C = 1 KΩ, V_CC = 5.0V. Assume that when the transistor is saturated, V_CE = 0V and V_BE = 0.8V. Calculate
(a) the minimum base current, for which the transistor will reach saturation. Hence,
(b) determine V₁ for when the transistor is ‘switched on’,
(c) find the ranges of V₁ for which the transistor is ‘switched of and ‘switched on’.

Solution:
Given at stauration
V_CE = OV, V_BE = 0.8V
V_CE = V_CC – I_CR_C
I_C = V_CC / R_C = 5.0V/1/0 kΩ = 5.0mA
Therefore, I_B = I_C/β
= 5.0 mA/250 = 20μA
The input voltage at which the transistor will go into saturation is given by
V_IH = V_BB = I_BR_B + V_BE
= 20μA × 100 kΩ + 0.8V = 2.8V
The value of input voltage below which the transistor remains cutoff is given by
V_IL = 0.6V, V_IH = 2.8V
Between 0.0V and 0.6V the transistor will be in the ‘switched off-state. Between 2.8V and 5.0V, it will be in ‘switched on’ state.
Note that the transistor is in active state when I_B varies from 0.0mA to 20mA. In this range, I_C = βI_B is valid. In the saturation range I_C ≤ βI_B.

Question 10.
For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2.0 kΩ is 2.0 V. Suppose the current amplification factor of the transistor is 100, what should be the value of R_B in series with V_BB supply of 2.0 V if the dc base current has to be 10 times the signal current. Also calculate the dc drop across the collector resistance. (Refer to Fig)

Solution:
The output ac voltage is 2.0 V. So, the ac collector current i_C= 2.0/2000 = 1.0 mA. The signal current through the base is, therefore given by i_B = i_C/β = 1.0 mA/100 = 0.010 mA. The dc base current has to be 10 × 0.010 = 0.10 mA
From V_BB = V_BE+ I_B R_B R_B = (V_BB – V_BE)/I_B. Assuming V_BE = 0.6V
R_B = (2.0 – 0.6)/0.10 = 14kΩ
The dc collector current I_C = 100 × 0.10 = 10 mA.

Question 11.
Justify the output waveform (Y) of the OR gate for the following inputs A and B given in fig.

Solution:
Note the following :

• At t ≤ t₁; A = 0, B = 0; Hence Y = 0
• For t₁ to t₂; A = 1, B = 0; Hence Y = 1
• For t₂ to t₃; A = 1, B = 1; Hence Y = 1
• For t₃ to t₄; A = 0, B = 1; Hence Y = 1
• For t₄ to t₅; A = 0, B = 0; Hence Y = 0
• For t₅ to t₆; A = 1, B = 0; Hence Y = 1
• For t > t₆; A = 0,B = 1; Hence Y = 1

Therefore the waveform Y will be as shown in the Fig.

Question 12.
Take A and B input waveforms similar to that in Example 11. Sketch the output waveform obtained from AND gate.
Solution:

• For t ≤ t₁; A = 0, B = 0; Hence Y = 0
• For t₁ to t₂; A = 1, B = 0; Hence Y = 0
• For t₂ to t₃; A = 1, B = 1; Hence Y = 1
• For t₃ to t₄; A = 0, B = 1; Hence Y = 0
• For t₄ to t₅; A = 0, B = 0; Hence Y = 0
• For t₅ to t₆; A = 1, B = 0; Hence Y = 0
• For t > t₆; A = 0,B = 1; Hence Y = 0

Based on the above, the output waveform for AND gate can be drawn as given below.

Question 13.
Sketch the output Y from a NAND gate having inputs A and B given below :
Solution:

• For t ≤ t₁; A = 1, B = 1; Hence Y = 0
• For t₁ to t₂; A = 0, B = 0; Hence Y = 1
• For t₂ to t₃; A = 0, B = 1; Hence Y = 1
• For t₃ to t₄; A = 1, B = 0; Hence Y = 1
• For t₄ to t₅; A = 1, B = 1; Hence Y = 0
• For t₅ to t₆; A = 0, B = 0; Hence Y = 1
• For t > t₆; A = 0,B = 1; Hence Y = 1
  

Students get through AP Inter 2nd Year Physics Important Questions 16th Lesson Communication Systems which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 16th Lesson Communication Systems

Very Short Answer Questions

Question 1.
What are the basic blocks of a communication system ?
Answer:
Basic blocks in communication system are

1. Transmitter
2. Receiver
3. Channel

Question 2.
What is ‘World Wide Web” (WWW) ?
Answer:
Tern Berners -Lee invented the World Wide Web.
It is an encyclopedia of knowledge accessible to every one round the clock through out the year.

Question 3.
Mention the frequency range of speech signals.
Answer:
Speech signals frequency range is 300 Hz to 3100 Hz.

Question 4.
What is sky wave propagation ?
Answer:
In the frequency range from a MHz upto about 30 MHz, long distance communication can be achieved by ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation.

Question 5.
Mention the various parts of the ionosphere ?
Answer:
Parts of ionosphere are

1. D Part of stratosphere (65-70 km day only),
2. E Part of stratosphere (100 km day only),
3. F₁ Part of mesosphere (170 km – 190 km),
4. F₂ Part of thermosphere [300 km at night 250 – 400 km during day time].

Question 6.
Define modulation. Why is it necessary ? [A.P. 17; A.P., T.S. Mar. 16, T.S. Mar. 15, Mar. 14]
Answer:
Modulation : The process of combining low frequency audio signal with high frequency carrier wave is called modulation.
Necessary: Low frequency signals cannot transmit directly. To reduce size of the antenna and to avoid mixing up of signal from different transmitters modulation is necessary.

Question 7.
Mention the basic methods of modulation. [T.S. Mar. 15, 17, A.P. Mar. 16]
Answer:
The basic methods of modulation are :

1. Amplitude modulation (AM)
2. Frequency modulation (FM)
3. Phase modulation (PM)

Question 8.
Which type of communication is employed in Mobile Phones ? [A.P. Mar. 15]
Answer:
Space wave mode of propagation is employed in mobile phones.

Textual Examples

Question 1.
A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna Is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given radius of earth 6.4 × 10⁶ m.
Solution:
d_m = \(\sqrt{2 \mathrm{Rh}_{\mathrm{T}}}+\sqrt{2 \mathrm{Rh}_{\mathrm{R}}}\)
d_m = \(\sqrt{2 \times 64 \times 10^5 \times 32}+\sqrt{2 \times 64 \times 10^5 \times 50 \mathrm{~m}}\)
=64 × 10² × \(\sqrt{10}\) +8 × 10³ × \(\sqrt{10}\)m = 144 × 10² × \(\sqrt{10}\)m
= 45.5 km.

Question 2.
A message signal of frequency 10 kHz and peak voltage of 10 volts Is used to modulate a carrier of frequency 1 MHz and peak voltage of 20 volts. Determine
(a) modulation index
(b) the side bands produced.
Solution:
a) Modulation index = \(\frac{10}{2}\) = 0.5
b) The side bands are at (1000 + 10 kHz) = 1010 kHz and (1000 – 10 kHz) = 990 kHz.