AP Inter 2nd Year Physics Study Material Chapter 12 Dual Nature of Radiation and Matter

Andhra Pradesh BIEAP AP Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter Textbook Questions and Answers.

AP Inter 2nd Year Physics Study Material 12th Lesson Dual Nature of Radiation and Matter

Very Short Answer Questions

Question 1.
What are “cathode rays” ?
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” ? (A.P. Mar. ’19 & T.S. Mar. ’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 sufficient 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 = \(\frac{1}{2} \mathrm{mv}_{\max }^2\) = hυ – ϕ₀.

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

Question 8.
State Heisenberg’s Uncertainly Principle. (A.P. Mar. ’19) (Mar. ’14)
Answer:
Uncertainity principle states that “it is impossible to measure both position (Δx) and momentum of an electron (Δp) [or any other particle] at the same time exactly”, i.e., Δx . Δp ≈ h where Δx is uncertainty in the specification of position and Δp is uncertainty in the specification of momentum.

Short Answer Questions

Question 1.
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.

Question 2.
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
   1. The values of stopping potentials are different for different frequencies.
   2. The value of stopping potential is more negative for radiation of higher incident frequency.
   3. The value of saturation current depends on the intensity of incident radiation but it is independent of the frequency of the incident radiation.
      

Question 3.
Summarise the photon picture of electromagnetic radiation.
Answer:
We can summarise the photon picture of electromagnetic radiation as follows.

1. In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons.
2. Each photon has energy E\(\left[\begin{array}{l}
   =\mathrm{hv} \\
   =\frac{\mathrm{hc}}{\lambda}
   \end{array}\right]\) and momentum P \(\left[\begin{array}{l}
   =\frac{h v}{c} \\
   =\frac{h}{\lambda}
   \end{array}\right]\) and speed c, the speed of light.
3. By increasing the intensity of light of given wave length, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation.
4. Photons are not deflected by electric and magnetic field. This shows that photons are electrically neutral.
5. In a photon-particle collision (such as photo-electron collision), the energy and momentum
   are conserved. However the number of photons may not be conserved in a collision. One photon may be absorbed or a new photon may be created.
6. The rest mass of photon is zero. According to theory of relativity, the mass of moving particle is given by m = \(\frac{\mathrm{m}_0}{\sqrt{1-\frac{v^2}{c^2}}}\) where v is velocity of particle and c is velocity of light.

Question 4.
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 = \(\frac{h}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{0.12 \times 20}\) = \(\frac{6.63 \times 10^{-34}}{2.4}\) ∴ λ = 2.762 × 10^-34 m = 2762 × 10^-21 A.

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 ? (T.S. Mar. ’19)
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 ‘h’ 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 = \(\frac{1}{2} m_{\max }^2\) = 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 thanthe threshold frequency (υ > υ₀) is increased then the number of photoelectrons emitted decreases i.e. the value of photoelectric current (i) increases, le. i ∝ I.
   
5. 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 (v₀).
      
   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.
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
   1. For a given photosensitive metal, the cut off potential (v₀) varies linearly with the frequency of the incident radiation.
   2. For a given photosensitive metal, there is a certain minimum cut off frequency v₀ (called threshold frequency) for which the stopping potential is zero.
      
8. From the graph we note that
   1. The value of cut-off potential is different for radiation of different frequency.
   2. 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.

Question 2.
Describe the Davisson and Germer experiment. What did this experiment conclusively prove?
Answer:
Davisson and Germer experiment:

1. The experimental arrangement is schematically shown in fig.
2. Electrons from a filament F are rendered into a fine beam by applying a positive potential to the cylinder A.
3. A fine narrow beam of electrons is incident on the nickel crystal. The electrons are scattered in all directions by the atoms of the crystal.
4. The intensity of the electron beam scattered in a given direction, is measured by the electron detector (collector). The detector can be moved on a circular scale and is connected to a sensitive galvanometer, which records the current.
5. The deflection of the galvanometer is proportional to the intensity of the electron beam entering collector.
6. The apparatus is enclosed in an evacuated chamber.
7. The experiment was performed by varying the accelerating voltage from 44 V to 68 V. It is found that the intensity is maximum at 50° for a critical energy of 54 V
   
8. For θ = 50°, the glancing angle, ϕ (angle between the scattered beam of electron with the plane of atoms of the crystal) for electron beam will be given by
   ϕ + θ + ϕ = 180°
   ϕ = \(\frac{1}{2}\left[180^{\circ}-50^{\circ}\right]\) = 65°
   
9. According to Bragg’s law for first order diffraction maxima (n = 1), we have 2 d sin ϕ = 1 × λ ⇒ λ = 2 × 0.91 × sin 65° = 1.65A = 0.165 nm. (experimentally).
   [∵ for Nickel crystal interatomic separation d = 0.91 A]
10. According to de-Broglie hypothesis, the wavelength of the wave associated with electron is given by λ =
    = 1.67A = 0.167 nm, (Theoritically).
11. The experimentally measured wavelength was found to be in confirmity with proving the existence of de-Broglie waves.

Textual Exercises

Question 1.
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Solution:
Given voltage V = 30 kV = 30 × 10³ V; e = 1.6 × 10^-19 C; h = 6.63 × 10^-34 j-s C = 3 × 10⁸ m/s
a) Maximum frequency, v = \(\frac{\mathrm{eV}}{\mathrm{h}}\) = \(\frac{1.6 \times 10^{-19} \times 30 \times 10^3}{6.63 \times 10^{-34}}\) = 7.24 × 10¹⁸ Hz

b) Minimum wavelength of X-ray, λ = \(\frac{\mathrm{C}}{\mathrm{v}}\) = \(\frac{3 \times 10^8}{7.24 \times 10^{18}}\) = 0.414 × 10^-10 Hz
∴ λ = 0.0414 × 10^-9m = 0.0414 nm.

Question 2.
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 ?
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
c) KE_max = \(\frac{1}{2} m v_{\max }^2\) ⇒ \(v_{\max }^2\) = \(\frac{2 K_{\max }}{m}\) = \(\frac{2 \times 0.35 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}\) (∴ e = 1.6 × 10^-19 C)
\(v_{\max }^2\) = 0.123 × 10¹² ⇒ υ_max = \(\sqrt{1230 \times 10^8}\) = 35.071 × 10⁴ m/s ∴ υ_max = 350.71 km/s.

Question 3.
The photoelectric cut-off voltage in certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ?
Solution:
Given, V₀ = 1.5 V; e = 1.6 × 10^-19 C, KE_max = eV₀ = 1.6 × 10^-19 × 1.5 = 2.4 × 10^-19 J.

Question 4.
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam ? (Assume the beam to have uniform cross-section which is less than the target area), and,
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon ?
Solution:
Given, λ = 632.8 nm = 632.8 × 10^-9m; p = 9.42 mW = 9.42 × 10^-3W
h = 6.63 × 10^-34 J-s; c = 3 × 10^-3 m/s

a) E = \(\frac{h c}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10}{632.8 \times 10^{-9}}\) = 3.14 × 10^-19 J.
Momentum of each photon, p = \(\frac{\mathrm{h}}{\lambda}\) = \(\frac{6.63 \times 10^{-34}}{632.8 \times 10^{-9}}\) = 1.05 × 10^-27kg \(\frac{\mathrm{m}}{\mathrm{s}}\)

b) No. of photons per second,

∴ N = 3 × 10¹⁶ photons/s
c) Since, P_Hydrogen = P_photon
⇒ mυ = p ⇒ υ = \(\frac{\mathrm{p}}{\mathrm{m}}\) = \(\frac{1.05 \times 10^{-27}}{1.66 \times 10^{-27}}\) [∴ m_H = 1.66 × 10^-27 kg] ∴ υ = 0.63 m/s.

Question 5.
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; e = 3 × 10⁸ m/s
Energy of each photon E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{550 \times 10^{-9}}\) = 3.616 × 10^-19 J
No. of photons incident on the earths 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 6.
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; .
e = 1.6 × 10^-19 c.

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

Question 7.
A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light ?
(b) At what rate are the photons delivered to the sphere ?
Solution:
Given, P = 100 W; λ = 589 nm = 589 × 10,sup>-9 m; h = 6.63 × 10^-34 J – S; c = 3 × 10⁸ m/s
a) E = \(\frac{\text { hc }}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{589 \times 10^{-9}}\) = 3.38 × 10^-19J = \(\frac{3.38 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV = 2.11 eV.
b) No. of photons delivered per second, N = \(\frac{\mathrm{P}}{\mathrm{E}}\) = \(\frac{100}{3.38 \times 10^{-19}}\) = 3 × 10²⁰ photons/s

Question 8.
The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Solution:
Given, v₀ = 3.3 × 10¹⁴ Hz; v = 8.2 × 10¹⁴ Hz; e = 1.6 × 10^-19 c; KE = eV₀ = hv – hv₀
V₀ = \(\frac{h\left(v-v_0\right)}{e}\) = \(\frac{6.63 \times 10^{-34} \times(8.2-3.3) \times 10^{14}}{1.6 \times 10^{-19}}\) = \(\frac{6.63 \times 10^{-34} \times 10^{14} \times 4.9}{1.6 \times 10^{-19}}\) ∴ V₀ = 2.03 V.

Question 9.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm ?
Solution:
Given, ϕ₀ = 4.2 eV = 4.2 × 1.6 × 10^-19 J = 6.72 × 10~1S J
λ = 330 nm = 330 × 10^-9 m; h = 6.63 × 10^-34 J – s ⇒ c = 3 × 10⁸ m/s
E = \(\frac{\text { hc }}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{330 \times 10^{-9}}\) ∴ E = 6.027 × 10^-19J
As E < ϕ₀, no photoelectric emission takes place.

Question 10.
Light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons ?
Solution:
Given, v = 7.21 × 10¹⁴ Hz; m = 9.1 × 10^-31 kg; υ_max = 6 × 10⁵ m/s
KE_max = \(\frac{1}{2} \mathrm{mv}_{\max }^2\) = hv – hv₀ = h(v – v₀)

Question 11.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made. .
Solution:
Given, λ = 488 nm = 488 × 10^-9 m; V₀ = 0.38 V; e = 1.6 × 10^-19 c; h = 6.63 × 10^-34 J – s
c = 3 × 10⁸ m/s ⇒ KE = eV₀ = \(\frac{\mathrm{hc}}{\lambda}\) – ϕ ⇒ 1.6 × 10^-19 × 0.38 = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}\) – ϕ₀
6.08 × 10^-20 = 40.75 × 10^-20 – ϕ₀ ⇒ (40.75 – 6.08) × 10^-20 = 34.67 × 10^-20 J
= \(\frac{34.67 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}\) ∴ ϕ₀ = 2.17 eV.

Question 12.
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V
Solution:
Given, V = 56 V; e = 1.6 × 10^-19 c; m = 9 × 10^-31 kg
a) As KE = \(\frac{\mathrm{P}^2}{2 \mathrm{~m}}\) ⇒ 2m (KE) = P² ⇒ P = \(\sqrt{2 \mathrm{~m}(\mathrm{KE})}\) = \(\sqrt{2 \mathrm{~m} \mathrm{eV}}\) [∵ KE = eV]
∴ P = \(\sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-31} \times 56}\) = 4.02 × 10^-24 kg – m/s
b) λ = \(\frac{12.27}{\sqrt{V}}\) A = \(\frac{12.27}{\sqrt{56}}\) A = 0.164 × 10^-9m ∴ λ = 0.164 nm.

Question 13.
What is the
(a) momentum,
(b) speed, and
(c) de Brogue wavelength of an electron with kinetic energy of 120 eV.
Solution:
Given, KE = 120 eV; m = 9.1 × 10^-3 kg; e = 1.6 × 10^-19 c
a) P = \(\sqrt{2 m(K E)}\) = \(\sqrt{2 \times 9.1 \times 10^{-31} \times\left(120 \times 1.6 \times 10^{-19}\right)}\) ∴ P = 5.91 × 10^-24 kg – m/s
b) υ = \(\frac{\mathrm{p}}{\mathrm{m}}\) = \(\frac{5.91 \times 10^{-24}}{9.1 \times 10^{-31}}\) = 6.5 × 10⁶ m/s .
c) λ = \(\frac{12.27}{\sqrt{\mathrm{V}}}\) A = \(\frac{12.27}{\sqrt{120}}\) A = 0.112 × 10^-9 m ∴ λ = 0.112 nm.

Question 14.
The wavelength of light from the spectral emission line of sodium is 589 nm Find the kinetic energy at which (a) an electron, and (b) a neutron, and would have the same de Brogue wavelength.
Solution:
Given, λ = 589 mn = 589 × 10^-9 m; m_e = 9.1 × 10^-31 kg.
m_n = 1.67 × 10^-27 kg; h = 6.62 × 10^-34 J – s.

Question 15.
What is the de Brogue wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 × 10^-9 kg drifting with a speed of 2.2 m/s?
Solution:
a) Given, for bullet m = 0.040 kg and o = 1000 m/s = 10³ m/s
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{0.040 \times 10^3}\) = 1.66 × 10^-35m
b)Given, for ball m = 0.060 kg and υ = 1 m/s ⇒ λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{0.060 \times 1}\) = 1.1 × 10^-32 m
c) Given, for a dust particle m = 1 × 10^-9 kg and υ = 2.2 m/s
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{1 \times 10^{-9} \times 2.2}\) = 3.0 × 10^-25 m.

Question 16.
An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Solution:
Given, λ = 1 mm = 10^-9m; h = 6.63 × 10^-34 J-S; c = 3 × 10⁸ m/S

Question 17.
(a) For what kinetic energy of a neutron will the associated de Brogue wavelength be 1.40 × 10^-10 m?
(b) Also find the de Brogue wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K.
Solution:
(a) Given, for neutron, λ = 1.40 × 10^-10 m and m = 1.675 × 10^-27 kg
KE = \(\frac{\mathrm{P}^2}{2 \mathrm{~m}}\) = \(\frac{h^2}{2 \mathrm{~m} \lambda^2}\) = \(\frac{\left(6.63 \times 10^{-34}\right)^2}{2 \times\left(1.40 \times 10^{-10}\right)^2 \times 1.675 \times 10^{-27}}\) ∴ KE = 6.686 × 10^-21J

b) Given, T = 300 k and K = 1.38 × 10^-23 J/K
KE = \(\frac{3}{2}\) KT = \(\frac{3}{2}\) × 1.38 × 10^-21 × 300 = 6.21 × 10^-21 J
λ = \(\frac{h}{\sqrt{2 m(K E)}}\) = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 6.21 \times 10^{-21}}}\) ∴ λ = 1.45 × 10^-10m = 1.45 A

Question 18.
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Solution:
The momentum of a photon of frequency v, wavelength λ is given by p = \(\frac{\mathrm{hv}}{\mathrm{c}}\) = \(\frac{\mathrm{h}}{\lambda}\)
λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) ⇒ de-Broglie wavelength of photon, λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{h}{p}\) = \(\frac{\frac{\mathrm{h}}{\mathrm{hv}}}{\mathrm{c}}\) = \(\frac{\mathrm{c}}{\mathrm{v}}\)
Thus, the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength.

Question 19.
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K ? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Solution:
Given, T = 300 k; K = 1.38 × 10^-23 J/k; m = 28.0152u = 28.0152 × 1.67 × 10^-27 kg;
h = 6.63 × 10^-34 Js; Mean KE of molecules \(\frac{1}{2}\) mυ² = \(\frac{3}{2}\) KT
υ = \(\sqrt{\frac{3 \mathrm{KT}}{\mathrm{m}}}\) = \(\sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{28.0152 \times 1.66 \times 10^{-27}}}\)
∴ υ = 516.78 m/s
de-Broglie wavelength, λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\frac{6.63 \times 10^{-34}}{28.0152 \times 1.66 \times 10^{-27} \times 516.78}\) = 2.75 × 10^-11 m
∴ λ = 0.0275 × 10^-19 m = 0.028 nm.

Additional Exercises

Question 1.
(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 10¹¹ C kg^-1.
(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified ?
Solution:
a) Given, V = 500 V, \(\frac{\mathrm{e}}{\mathrm{m}}\) = 1.76 × 10¹¹ C/kg; KE = \(\frac{1}{2} \mathrm{mv}^2\) = eV

b) V = 10 MV = 10⁷ V; υ = \(\sqrt{\frac{\mathrm{e}}{\mathrm{m}} \times 2 \mathrm{~V}}\) = \(\sqrt{1.76 \times 10^{11} \times 2 \times 10^7}\) ∴ υ = 1.8762 × 10⁹ m/s
This speed is greater than speed of light, which is not possible. As o approaches to c, then mass m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)

Question 2.
(a) A monoenergetic electron beam with electron speed of 5.20 × 10⁶ m s^-1 is subject to
a magnetic field of 1.30 × 10^-4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 10¹¹ C kg^-1.
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam ? If not, in what way is it modified ?
[Note : Exercises 20(b) and 21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]
Solution:
a) Given, υ = 5.20 × 10⁶ m/s; B = 1.30 × 10^-4 T; \(\frac{\mathrm{e}}{\mathrm{m}}\) = 1.76 × 10¹¹ C/kg
Since centripetal force is balanced by Force due to magnetic field, \(\frac{\mathrm{m} v^2}{\mathrm{r}}\) = Bυ
[∵ (\(\vec{v} \times \vec{B}\)) = e υ B sin 90° = Beυ]

b) Given, E = 20 MeV = 20 × 1.6 × 10^-13J; m_e = 9.1 × 10^-31 kg
E = \(\frac{1}{2} \mathrm{mv}^2\)
⇒ v = \(\sqrt{\frac{2 E}{m}}\) = \(\sqrt{\frac{2 \times 20 \times 1.6 \times 10^{-13}}{9.1 \times 10^{-32}}}\) ∴ v = 2.67 × 10⁹ m/s

As υ > C, the formula used in (a) r = \(\frac{\mathrm{mv}}{\mathrm{eB}}\) is not valid for calculating the radius of path of 20 MeV electron beam because electron with such a high energy has velocity in relatistic domain i.e., comparable with the velocity of light and the mass varies with the increase in velocity but we have taken it a constant.
∴ m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\) ⇒ Thus, the modified formula will be r = \(\frac{\mathrm{mv}}{\mathrm{eB}}\) = \(\left[\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\right] \frac{v}{e B}\)

Question 3.
An electron gun with its collector at a potential of 100V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~ 10^-2 mm of Hg). A magnetic field of 2.83 × 10^-4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons and emitting light by electron capture; this method is known as the fine beam tube’ method.) Determine e/m from the data.
Solution:
Given, V = 100 V; B = 2.83 × 10^-4 T; m = 9.1 × 10^-31 kg; e = 1.6 × 10^-19 C;
r = 12 cm = 0.12m; KE = \(\frac{1}{2} \mathrm{mv}^2\) = eV ⇒ \(\frac{1}{2}\) × 9.1 × 10^-31 × υ² = 1.6 × 10^-19 × 100
υ² = \(\frac{2 \times 1.6 \times 10^{-17}}{9.1 \times 10^{-3.1}}\) = 3.516 × 10¹³ ∴ υ = \(\sqrt{3.516 \times 10^{13}}\) = 5.93 × 10⁶ m/s
Specific charge of electron, \(\frac{\mathrm{e}}{\mathrm{m}}\) = \(\frac{v}{r B}\) [∵ \(\frac{\mathrm{mv}^2}{\mathrm{r}}\) = Beυ] = \(\frac{5.93 \times 10^6}{2.83 \times 10^{-4} \times 0.12}\)
∴ \(\frac{\mathrm{e}}{\mathrm{m}}\) = 1.74 × 10¹¹ C/kg.

Question 4.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 A. What is the maximum energy of a photon in the radiation ?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube ?
Solution:
a) Given, λ = 0.45 A = 0.45 × 10^-10 m; E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{0.45 \times 10^{-10} \times 1.6 \times 10^{-19}}\) eV
∴ E = 27.6 × 10³ eV = 27.6 KeV

b) In X-ray tube, accelerating voltage provides the energy to the electrons which produce X-rays. For getting X-rays, photons of 27.51 KeV is required that the incident electrons must posess kinetic energy of 27.61 KeV.
Energy = eV = E; eV = 27.6 KeV; V = 27.6 KV .
So, the order of accelerating voltage is 30 KV.

Question 5.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray ? (1 BeV = 10⁹ eV)
Solution:
Given, energy of 2 γ-rays, 2E = 10.2 BeV

⇒ 2\(\frac{\mathrm{hc}}{\lambda}\) = 10.2 BeV [∵ E = \(\frac{\mathrm{hc}}{\lambda}\)] ⇒ λ = \(\frac{2 \mathrm{hc}}{10.2 \mathrm{BeV}}\)
Here h = 6.63 × 10^-34 J-S; c = 3 × 10⁸m/s, 1 BeV = 10⁹ eV = 10⁹ × 1.6 × 10^-19J
⇒ λ = \(\frac{2 \times 6.63 \times 10^{-34} \times 3 \times 10^8}{10.2 \times 10^9 \times 1.6 \times 10^{-19}}\) ∴ λ = 2.436 × 10^-16 m

Question 6.
Estimating the following two numbers should bé interesting. The first number will tell you why radio engineers do not need to worry much about photons ! The second number tells you why our eye can never count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10^-10 W m^-2). Take the area of the pupil to be about 0.4 cm², and the average frequency of white light to be about 6 × 10¹⁴ Hz.
Solution:
a) Given, P = 10kW = 10 × 10³ W; λ = 500m; h = 6.63 × 10^-34 J – s; C = 3 × 10⁸
The no. of photons emitted per second, N = \(\frac{\mathrm{P}}{\mathrm{E}}\) = \(\frac{\mathrm{P}}{\frac{\mathrm{hc}}{\lambda}}\) = \(\frac{\mathrm{p} \lambda}{\mathrm{hc}}\) = \(\frac{10 \times 10^3 \times 500}{6.63 \times 10^{-34} \times 3 \times 10^8}\)
∴ N = 2.51 × 10³¹ photons/s

b) Given, v = 6 × 10^-4 Hz; I = \(\frac{E}{A-t}\) = 10^-10 W/m²; Area of pupil, A = 0.4 cm² = 0.4 × 10^-4 m².
Total energy falling on pupil in unit time, E’ = IA = 10^-10 × 0.4 × 10^-4 ∴ E’ = 4 × 10^-155 J/s
Energy of each photon, E” = hv = 6.63 × 10^-34 × 6 × 10¹⁴ = 3.978 × 10^-19 J
No. of photons per second, N = \(\frac{E^{\prime}}{E^{\prime \prime}}\) = \(\frac{4 \times 10^{-15}}{3.978 \times 10^{-19}}\) = 1.206 × 10⁴ photons/s
As this number is not so large a: in part (a), so it is large enough for us never to sense the individual photons by our eye.

Question 7.
Ultraviolet light of wavelength 2271 A from a 1oo W mercury source irradiates a photo cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (-10⁵ W m^-2) red light of wavelength 6328 A produced by a He-Ne laser?
Solution:
Given, for UV light, λ = 2271A = 2271 × 10^-10 m
V₀ = 1.3 V; P = 100W; h = 6.63 × 10^-34 J-s; c = 3 × 10⁸ m/s
From Einstein’s equation E = KE + ϕ₀, hυ = eV₀ + ϕ₀
ϕ₀ = \(\frac{\mathrm{hc}}{\lambda}\) – eV₀ = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2271 \times 10^{-10}}\) – 1.6 × 10^-19 × 1.3 = 8.758 × 10^-19 – 2.08 × 10^-19
ϕ₀ = \(\frac{6.678 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV = 4.17 eV ∴ ϕ₀ = 4.2 eV
Given, for red light, λ = 6328Å = 6328 × 10^-10m
E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6328 \times 10^{-10}}\) = \(\frac{3.143 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV ∴ E = 1.96 eV
Here, E < ϕ₀, So, the photocell will not respond to this red light. (It is independent of intensity).

Question 8.
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10^-9 m) from a neon, lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Solution:
Given, for Neon X = 640.2 nm = 640.2 × 10^-9 m ; V₀ = 0.54 V
h = 6.63 × 10^-34 J-s; c = 3 × 10⁸ m/s; e = 1.6 × 10^-19 C
ϕ = \(\frac{\mathrm{hc}}{\lambda}\) – eV₀ = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{640.2 \times 10^{-9}}\) – 1.6 × 10^-19 × 0.54
= 3.1 × 10^-19 – 0.864 × 10^-19 = 2.236 × 10^-19J = \(\frac{2.236 \times 10^{-19}}{1.6 \times 10^{-19}}\) eV ∴ ϕ = 1.4 eV
For iron; given ϕ₀ = 1.4eV; λ = 427.2 nm = 427.2 × 10^-9 m
Let V₀ be the new stopping potential, eV₀ = \(\frac{\mathrm{hc}}{\lambda}\) – ϕ₀
eV₀’ = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{427.2 \times 10^{-9} \times 1.6 \times 10^{-19}}\) – 1.4 = 1.51 eV. Required stopping potential V₀‘ = 1.51 V.

Question 9.
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:
λ₁ = 3650Å, λ₂ = 4047Å, λ₃ = 4358Å, λ₄ = 5461 Å, λ₅ = 6907Å,
The stopping voltages, respectively, were measured to be:
V₀₁ = 1.28 V, V₀₂ = 0.95 V, V₀₃ = 0.74 V, V₀₄ = 0.16 V, V₀₅ = 0V.
Determine the value of Plancks constant h, the threshold frequency and work function
for the material.
[Note : You will notice that, to get h from the data, you will need to know e(which you can take to be 1.6 × 10^-19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Solution:
Given λ₁ = 3650 A = 3650 × 10^-10 m
λ₂ = 4047 A = 4047 × 10^-10 m
λ₃ = 4358 A = 4358 × 10^-10 m
λ₄ = 5461 A = 5461 × 10^-10 m
λ₅ = 6907 A = 6907 × 10^-10 m
V₀₁ = 1.28V, V₀₂ = 0.95, V₀₃ = 0.74 V; V₀₅ = 0

a) v₁ = \(\frac{\mathrm{c}}{\lambda_1}\) = \(\frac{3 \times 10^8}{3650 \times 10^{-10}}\) = 8.219 × 10¹⁴ Hz
v₂ = \(\frac{\mathrm{c}}{\lambda_2}\) = \(\frac{3 \times 10^8}{4047 \times 10^{-10}}\) = 7.412 × 10¹⁴ Hz
v₃ = \(\frac{\mathrm{c}}{\lambda_3}\) = \(\frac{3 \times 10^8}{4358 \times 10^{-10}}\) = 6.884 × 10¹⁴ Hz
v₄ = \(\frac{\mathrm{c}}{\lambda_4}\) = \(\frac{3 \times 10^8}{5461 \times 10^{-10}}\) = 5.493 × 10¹⁴ Hz
v₅ = \(\frac{\mathrm{c}}{\lambda_5}\) = \(\frac{3 \times 10^8}{6907 \times 10^{-10}}\) = 4.343 × 10¹⁴ Hz
As the graph between V₀ and frequency v is a straight line.
The slope of this graph gives the values of \(\frac{\mathrm{h}}{\mathrm{e}}\)

∴ \(\frac{\mathrm{h}}{\mathrm{e}}\) = \(\frac{V_{01}-V_{04}}{v_1-v_4}\) = \(\frac{1.28-0.16}{(8.219-5.493) \times 10^{14}}\)
h = \(\frac{1.12 \times 1.6 \times 10^{-19}}{2.726 \times 10^{14}}\) = 6.674 × 10^-34 J . s

b) ϕ₀ = hv₀ = 6.574 × 10^-34 × 5 × 10¹⁴
= 32.870 × 10^-20 J = \(\frac{32.870 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}\)
= 2.05 eV

Question 10.
The work function for the following metals is given:
Na : 2.75 eV; K: 2.30 eV; Mo : 4.17 eV; Ni : 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 A from a He-Cd laser placed 1 m away from the photocell ? What happens if the laser is brought nearer and placed 50 cm away?
Solution:
Given λ = 3300 A = 3300 × 10^-10 m
Energy of incident photon, E = \(\frac{\mathrm{hc}}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3300 \times 10^{-10} \times 1.6 \times 10^{-19}}\) ∴ E = 3.75 eV
Here Na, K has lesser work function than 3.75 eV. So, they produce photoelectric effect. If the laser is brought nearer then only the intensity change or the number of photoelctrons change.

Question 11.
Light of intensity 10^-5 W m^-2 falls on a sodium photo-cell of surface area 2 cm². Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Solution:
Given, I = 10^-5 W/m²; A = 2 cm² = 2 × 10^-4 m²; ϕ₀ = 2eV
Let t be the time.
The effective atomic area of Na = 10^-20 m² and it contains one conduction electron per
atom.
No. of conduction electrons m five layers

We know that sodium has one free electron (or conduction electron) per atom.
Incident power on the surface area of photocell
= Incident intensity × Area on the surface area of photo cell
= 10^-5 × 2 × 10^-4 = 2 × 10^-9 W.
The electron present in all the 5 layers of sodium will share the incident energy equally.
Energy absorbed per second per electron, E =
= \(\frac{2 \times 10^{-9}}{10^{17}}\) = 2 × 10^-26 W.
Time required for emission by each electron,
which is about 0.5 yr.
The answer obtained implies that the time of emission of electron is very large and is not agreement with the observed time of emission. There is no time lag between the incidence of light and the emission of photoelectron.
Thus, it is implied that the wave theory cannot be applied in this experiment.

Question 12.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative. comparison. take the wavelength of the probe equal to 1 A, which is of the order of interatomic spacing in the lattice) (m_e = 9.11 × 10^-31 kg).
Solution:
Given λ = 1 A = 10^-10 m ; m_e = 9.11 × 10^-31 kg; h = 6.63 × 10^-34 J – s; c = 3 × 10⁸ m/s

Thus, for the same wavelength a X-ray photon has much KE than an electron.

Question 13.
(a) Obtain the de Brogue wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable ? Explain. (m_n = 1.675 × 10^-27 kg)
(b) Obtain the de Brogue wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffráction experiments.
Solution:
a) Given, KE = 150 eV; m = 1.675 × 10^-27 kg

The interatomic spacing is 10^-10 m, which is greater than this wavelength. So, neutron beam of 150 eV is not suitable for diffraction experiment.

b) T = t + 273 = 27 + 273 = 300 K; K = 1.38 × 10^-23 J/mol/K

This wavelength is order of interatomic spacing. So, the neutron beam first thermalised and then used for diffraction.

Question 14.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Solution:
Given, V = 50 KV s 50000 V
λ =

= 0.055 A ⇒ λ = 5.5 × 10^-12 m; For yellow light (λ) = 5.9 × 10^-7m
As resolving power (RP) ∝ \(\frac{1}{\lambda}\)
⇒

Question 15.
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10^-15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.5 11 MeV.)
Solution:
Given λ = 10^-15 m; E = 0.5 11 MeV; P = \(\frac{\mathrm{h}}{\lambda}\) = \(\frac{6.63 \times 10^{-34}}{10^{-15}}\) = 6.63 × 10^-19 kg^m/s
Rest mass energy; E₀ = m₀c² = 0.511 MeV = 0.511 × 1.6 × 10^-13 T.
From relativistic theory, E² = p²c² + \(m_0^2 c^4\)
= (3 × 10⁸ × 6.63 × 10^-19)² + (0.511 × 10^-13 × 1.6)² = 9 × (6.63)² × 10^-22.
As the rest mass energy is negligible ∴ Energy E = \(\sqrt{p^2 c^2}\) = pc = 6.63 × 10^-19 × 3 × 10⁸
= \(\frac{1.989 \times 10^{-10}}{1.6 \times 10^{-19}}\)eV = 1.24 × 10⁹ eV = 1.24 BeV
Thus, to energies the electron beam, the energy should be of the order of BeV.

Question 16.
Find the typical de Brogue wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare It with the mean separation between two atoms under these conditions.
Solution:
Given T = 27 + 273 = 300 K; K = 1.38 × 10^-23 J/mol/K; p = 1 atm = 1.01 × 10⁵ Pa

We can see that the wave length with mean separation r, it can be observed (r >> λ) that separation is larger than wave length.

Question 17.
Compute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10^-10 m.
[Note : Exercise 35 and 36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]
Solution:
Given, T = 27 + 273 = 300 K; r = 2 × 10^-10m
Momentum, P = \(\sqrt{3 \mathrm{mKT}}\) = \(\sqrt{3 \times 9.11 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}\) = 1.06 × 10^-25 kg-m/s
λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) = \(\frac{6.63 \times 10^{-34}}{1.06 \times 10^{-25}}\) = 62.6 × 10^-10m; Mean separation, r = 2 × 10^-10 m
\(\frac{\lambda}{r}\) = \(\frac{62.6 \times 10^{-10}}{2 \times 10^{-10}}\) = 31.3
We can see that de-Broglie wavelength is much greater than the electron separation.

Question 18.
Answer the following questions :
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e; (-1/3)e]. Why do they not show up in Millikan’s oil-drop experiment ?
Solution:
The quarks have fractional charges. These quarks are bound by forces. These forces become stronger when the quarks are tried to be pulled apart. That is why, the quarks always remain’ together. It is due to this reason that tough fractional charges exists in nature but the observable charges are always integral multiple of charge of electron.

(b) What is so special about the combination e/m ? Why do we not simply talk of e and m separately ?
Solution:
The motion.of electron in electric and magnetic fields are governed by these two equations.
\(\frac{1}{2} \mathrm{mv}^2\) = eV or Beυ = \(\frac{m v^2}{\mathrm{r}}\)
In these equations, e and m both are together i.e. there is no equation in which e or m are alone. So, we always take e/m.

(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures ?
Solution:
At ordinary pressure, only very few positive ions and electrons are produced by the ionisation of gas molecules. They are not able to reach the respective electrodes and becomes insulators. At low pressure, density decreases and the mean free path becomes large. So, at high voltage, they acquire sufficient amount of energy and they collide with molecules for further ionisation. Due to this, the number of ions in a gas increases and it becomes a conductor.

(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic ? Why is there an energy distribution of photoelectrons ?
Solution:
Because all the electrons in the metal do not belong to same level but they occupy a continuous band of levels, therefore for the given incident radiation, electrons come out from different levels with different energies.

(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations :
E = h v, p = \(\frac{\mathbf{h}}{\lambda}\)
But while the value of λ is physically significant, the value of v (and therefore, the value of the phase speed v λ) has no physical significance. Why ?
Solution:
As λ = \(\frac{\mathrm{h}}{\mathrm{p}}\) = p = \(\frac{h}{\lambda}\) ⇒ E = hv = \(\frac{\mathrm{hc}}{\lambda}\)
Energy of moving particle E’ = \(\frac{p^2}{2 m}\) = \(\frac{1}{2} \frac{\left(\frac{h}{\lambda}\right)^2}{m}\) = \(\frac{1}{2} \frac{h^2}{\lambda^2 \mathrm{~m}}\). For the relation of E and p, we note that there is a physical significance of λ but not for frequency v.