Structure of Atom Questions and Answers AP Inter 1st Year Chemistry Chapter 2

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 2nd Lesson Structure of Atom Class 11 Textbook Exercise Questions and Answers.

Structure of Atom Class 11 Questions and Answers AP Inter 1st Year Chemistry 2nd Lesson

I. Multiple Choice Questions (1 Mark)

Question 1.
Which of the following statements are correct about cathode rays ?
I. They move from cathode to anode.
II. these are visible rays.
III. Television picture tubes are cathode ray tubes.
IV. In theabsence of electrical or magnetic field, these rays travel in straight lines.
Choose the most appropriate answer from the options given below.
(1) I, II and III
(2) II, III and IV
(3) I, III and IV
(4) I, II, III and IV
Answer:
(3) I, III and IV

Question 2.
Increasing order for the values of e/m for electron (e), proton (p), neutron (n) and α-particles is
(1) e < p < n < α
(2) n < α < p < e
(3) n < p < e < a
(4) n < p < α < e
Answer:
(2) n < α < p < e

Question 3.
Rutherford’s experiment on the scattering of α – particles showed for the first time that the atom has
(1) Electrons
(2) Protons
(3) Nucleus
(4) Neutrons
Answer:
(3) Nucleus

Question 4.
Thomson’s model of the atom is also called
(1) Plum pudding model
(2) Raisin pudding model
(3) Watermelon model
(4) All of the above
Answer:
(4) All of the above

Question 5.
In photoelectric effect at which frequency electron will be ejected with certain kinetic energy (υ₀ = threshold frequency)
(1) υ = υ₀
(2) υ₀ > υ
(3) υ₀ ≥ υ
(4) υ > υ₀
Answer:
(4) υ > υ₀

Question 6.
A 600 Watt mercury lamp emits monochromatic radiation of wavelength 331.3 nm. How many photons are emitted from the lamp per second? (h = 6.626 * 10^-34) Js. Velocity of light: 3 x 10⁸ m/s)
(1) 1 × 10¹⁹
(2) 1 × 10²⁰
(3) 1 × 10²¹
(4) 1 × 10²³
Answer:
(3) 1 × 10²¹

Question 7.
Which of the following electronic transitions in the hydrogen atom will require highest energy?
(1) n = 4 to n = 5
(2) n = 1 to n = 2
(3) n = 3 to n = 5
(4) n = 2 to n = 3
Answer:
(2) n = 1 to n = 2

Question 8.
The ratio of radii of second orbit of hydrogen atom to 4^th orbit of He⁺ ion is
(1) 1 : 4
(2) 2 : 1
(3) 1 : 2
(4) 3 : 2
Answer:
(3) 1 : 2

Question 9.
The energies of an electron in first orbit of He⁺ and 3^rd of Li²⁺ in J are respectively.
(1) -8.72 × 10^-18, -2.18 × 10^-18 J
(2) -8.72 × 10^-18, -1.96 × 10^-17 J
(3) -1.96 × 10^-17, -2.18 × 10^-18 J
(4) -8.72 × 10^-17, -1.96 × 10^-17J
Answer:
(1) -8.72 × 10^-18, -2.18 × 10^-18 J

Question 10.
The value of Fydberg constant is
(1) 109777 cm^-1
(2) 109977 cm
(3) 109677 cm^-1
(4) 109277 cm^-1
Answer:

Question 11.
The velocity of a particle A is 0.1m/s and that of particle B is 6.05 m/s. If the mass of the particle B is. 5 times that of particle A. Then the ratio of de-Broglie wavelengths associated with the particles A and B is
(1) 2 : 5
(2) 3 : 4
(3) 6 : 4
(4) 5 : 2
Answer:
(4) 5 : 2

Question 12.
The number of radial nodes and angular nodes of a 4f orbital are respectively.
(1) 0, 3
(2) 1, 2
(3) 2, 1
(4) 2, 0
Answer:
(1) 0, 3

Question 13.
Maximum number of electrons possible with spin quantum number +1/2 principal quantum number n = 4 is
(1) 16
(2) 9
(3) 4
(4) 25
Answer:
(3) 4

Question 14.
The wavelength of second line in the Lyman series of hydrogen atomic spectrum is (Rydberg constant = R cm^-1)
(1) \(\frac{8 \mathrm{R}}{9}\)
(2) \(\frac{9}{8 R}\)
(3) \(\frac{4}{3 R}\)
(4) \(\frac{3 R}{4}\)
Answer:
(2) \(\frac{9}{8 R}\)

Question 15.
What is the correct orbital designation of an electron with the quantum numbers n = 4, l = 3, m = -2, s = +1/2 ?
(1) 3s
(2) 4f
(3) 5p
(4) 4d
Answer:
(2) 4f

II. Fill in the Blanks (1 Mark)

Question 1.
Splitting of spectral lines in magnetic field is called ____
Answer:
Zeman effect

Question 2.
Splitting of spectral lines in electric field is called ______
Answer:
Stark effect

Question 3.
The preference of three unpaired electrons in the nitrogen atom can explained by _____
Answer:
Hund’s Rule

Question 4.
The electronic configuration of copper is ____
Answer:
[Ar]4s¹3d¹⁰

Question 5.
The electronic configuration of chromium is ____
Answer:
[Ar]4s¹3d⁵

Question 6.
In potassium atom, the differentiating electrons enters into 4s orbital instead of 3d orbital. This is according to ____.
Answer:
Aufbau principle.

III. One Word Answer Questions (1 Mark)

Question 1.
Define isotopes.
Answer:
Isotopes are atoms of the same element that have the same number of protons but different number of neutrons. This means they have the same atomic number but different mass numbers.

Example: \({ }_6^{12} \mathrm{C}\) and \({ }_6^{14} \mathrm{C}\) are isotopes of carbon.
¹²C – Number of protons = 6
Mass Number = 12
¹⁴C – Number of protons = 6
Mass Number = 14

Question 2.
Write the difference between orbit and orbital ?
Answer:
Orbit: An orbit is a circular path around the nucleus of an atom where the elec-tron revolves.
Orbital: An orbital is the space around the nucleus where the probability of finding an electron in maximum.

Question 3.
How many nodal planes are present in d orbitals ?
Answer:
Two nodal planes are present in d-prbitals. The plane where the probability of finding an electron is zero.

Question 4.
How many ‘p’ electrons are present in Sulphur atom ?
Answer:
Electronic configuration of sulphur – 1s² 2s² 2p⁶ 3s² 3p⁴
Number of p – electrons = 6(2p⁶) + 4 (3p⁴)
= 10

Question 5.
What are degenerate orbitals?
Answer:
The orbitals having the same energy are called degenerate orbitals.

IV. Very Short Answer Questions (2 Marks)

Question 1.
Calculate the charge of one mole of electrons.
Answer:
Charge of Electron = -1.602 × 10^-19 coulombs
Charge of one mole of electrons
= 6.023 × 10²³ × 1.602 × 10^-19
= 9.648846 × 10⁴
= 96488.46 coulombs

Question 2.
How many neutrons and electrons are present in the nuclei of \({ }_6^{13} \mathrm{C},{ }_8^{18} \mathrm{P},{ }_{12}^{24} \mathrm{Mg},{ }_{26}^{56} \mathrm{Fe}\)
Answer:

Question 3.
What is black body ?
Answer:
The body which is perfect absorber and emmiter of all type of radiations incident on it is called a black body.

Question 4.
Draw the shape of \(d_{z^2}\) and \(d_{x^2-y^2}\) orbital.
Answer:

Question 5.
What is the frequency of radiation of wavelength 600 nm ?
Answer:
Given that wave length, λ = 600 nm
= 600 × 10^-9m
= 6 × 10^-7m
Velocity, c = 3 × 1o⁸ m/sec.
∴ Frequency, υ = \(\frac{c}{\lambda}\)
= \(\frac{3 \times 10^8}{6 \times 10^{-7}}\)
= \(\frac{1}{2}\) × 1o¹⁵
= 0.5 × 1o¹⁵ sec^-1 = 5 × 1o¹⁴ sec^-1

Question 6.
How many subshells are associated with n = 5 ?
Answer:
For n = 5, l = 0, 1, 2, 3, 4.
l = 0 → s-orbital
l = l → p – orbital
l = 2 → d – orbital
l = 3 → f – orbital
l = 4 → g – orbital
∴ Number of subshells are associated with n = 5 are 5. They are 5s, 5p, 5d, 5f, 5g

Question 7.
How many electrons in an atom may have if n = 4 and m_s = +1/2 ?
Answer:
For n = 4, l = 0, 1,2,3
If l = 0 → 4s, contains 1 electron with m_s = + \(\frac{1}{2}\)
l = 1 → 4p, contains 3 electron with m_s = + \(\frac{1}{2}\)
l = 2 → 4d, contains 5 electron with m_s = + \(\frac{1}{2}\)
l = 3 → 4f, contains 7 electron with m_s = + \(\frac{1}{2}\)
Each orbital can hold 2 electrons, but only 1 electron per orbital can have
m_s = + \(\frac{1}{2}\)
Since there are 16 orbitals [s = 1, p = 3, d = 5, f = 7] there are 16 electrons with m_s = + \(\frac{1}{2}\)

Question 8.
Explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle states that “No two electrons in an atom can have the same set of all the four quantum numbers”. The two electrons in an atom differ in atleast one qunatum number.

Question 9.
What is Hund’s rule ?
Answer:
Hund’s rule states that “pairing of electrons takes place in the orbitals after each orbital of the same subshell is filled with one electron.”

Question 10.
What is Aufbau principle?
Answer:
Aufbau principle states that “Electrons enter in to the orbitals in the increasing order of their energy”. This means that electrons occupy the lowest energy orbital first.
The energy of the orbital is given by (n + l) value.
where, n – principal quantum number
l – Azimuthal quantum number.
The new electron enters an empty orbital whose (n + l) value is minimum.

Example: 3s – n + l = 3 + 0 = 3
3p – n + l = 3 +1 = 4, 3s orbital is filled first.

V. Short Answer Questions (4 Marks)

Question 1.
Explain photoelectric effect.
Answer:
Based on plank’s quantum theory, Einstein explained photo electric effect in 1905.

“The phenomenon in which emission of electrons takes place from the surface of certain metals when light of suitable frequency falls on them is called photoelectric effect”. The electrons thus emitted are called photo electrons and the flow of these photo electrons is called photo electric current.

When a photon strikes the metal surface, it uses some part of its energy to eject the electrons from the metal atom. The remaining part of the total energy is given to the ejected electrons in the form of kinetic energy.

Hence we can write, hυ = W + KE
⇒ hυ = hυ₀ + \(\frac{1}{2}\)m_eV²
Where, hυ – Energy of photon
υ₀ – Threshold frequency
m_e – Mass of electron
W – Energy required to over come the attractive forces on the electron in the metal.
KE – Kinetic energy of ejected electron
v – Velocity of ejected electron

If a photon of sufficient energy struck the metal surface and could eject the electron. But if a photon has insufficient energy; it cannot eject the electron from the metal.

Question 2.
Which of the following sets of quantum numbers are not possible ?
(a) n = 0; l = 0; m_l = 0; m_s = +1/2
(b) n = 1; l = 0; m_l = 0, m_s = -1/2
(c) n = 1; l = 1; m_l = 0; m = +1/2
(d) n = 2; l = 1; m_l = 0; m_s = +1/2
Answer:
The following set of quantum numbers are not possible.
a) n = 0, l = 0, m_l = 0, m_s = + \(\frac{1}{2}\)
Reason: ‘n’ is principal quantum number, whose values are from 1 to n. The value of ‘n’ never equal to zero, but given n = 0.

c) n = 1, l = 1, m_l = 0, m_s = + \(\frac{1}{2}\)
Reason: The values of ‘l’ are from 0 to (n – 1). If n = 1, then the value of ‘l’ is zero but not equal to 1.

Question 3.
What is a Nodal plane ? How many nodal planes are possible for 2p, and 3d orbitals ?
Answer:
Nodal plane : The plane at which the probability of finding the electron is zero is called as Nodal plane.
For 2p orbitals, one nodal plane is possible for each p – orbital.
For 3d orbitals, two nodal planes are possible for each d – orbital.

Question 4.
Explain the difference between orbit and orbital.
Answer:

Orbit	Orbital
1) An orbit is a circular path around the nucleus of an atom where the electron revolves.	1) An orbital is the space around the nucleus where the probability of finding an electron is maximum.
2) Orbits are circular in shape.	2) Orbitals have different shapes.
3) It is a two dimensional representation.	3) It is a three dimensional representation.
4) Maximum number of electrons present in an orbit is 2n2	4) Maximum number of electrons present in an orbital is 2.
5) The distance of the orbit from the neucleus is fixed. So, the position of an electron is certainly known in an orbit.	5) It is impossible to know the exact position of an electron in an orbital of an atom.

Question 5.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:
According to Bohr’s atomic theory, only those orbits are stable in which the angular momentum is quantized.
∴ mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
Where, r – the radius of the orbit
n – principal quantum number
So, the angular momentum is an integral multiple of \(\frac{\mathrm{h}}{2 \pi}\).
∴ mvr = \(\frac{\mathrm{nh}}{2 \pi}\) ⇒ 2πr = \(\frac{\mathrm{nh}}{\mathrm{mv}}\) ⇒ 2πr = \(\mathrm{n} \cdot\left(\frac{\mathrm{~h}}{\mathrm{mv}}\right)\) …… (1)
According to de-Broglie’s hypothesis, a particle like electron has wavelength,
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\)
Substitute this value in equation (1)
∴ 2πr = nλ
Thus the circumference of the Bohr’s orbit (2πr) is an integral multiple of de-Broglie’s wave length.

VI. Long Answer Questions (8 Marks)

Question 1.
Explain briefly the Planck’s quantum theory.
Answer:
The postulater of planck’s quantum theory are:

• The emission of radiation is due to vibrations of charged particles (electrons) in the body.
• The emission is not continuous but in discrete packets of energy called “quanta”. The emitted radiation propagates in the form of waves.
• The energy (E) associated with each quantum for a particular radiation of frequency ‘υ’ is given by E = hυ. Here ‘h’ is planck’s constant.
• A body can emit or absorb either one quantum (hυ) of energy or some whole number multiples of it. Thus energy can be emitted or absorbed as hυ, 2hυ, 3hυ etc., but not fractional values. This is called quantisation of energy.
• The emitted radiant energy is propagated in the form of waves.
• The value of planck’s constant (h) in various units
  h = 6.6256 × 10^-27 erg. sec (or) gcm² s^-1
  = 6.6256 × 10^-34 J. sec (or) kg m²s^-1

Planck’s quantum theory successfully explains the black body radiation.
A black body is a perfect absorber and also a perfect radiator of radiations.

Question 2.
What are the postulates of Bohr’s model of hydrogen atom ? Discuss the importance of this model to explain various series of line spectra in hydrogen atom.
Answer:
Neils Bohr in 1913, postulated a theory based on Rutherford’ model and Plank’s quantum theory to explain the spectrum of atoms.

Postulates of Bohr’s theory:

• Electrons in an atom revolve around the nucleus with high velocity in certain, fixed circular paths called orbits.
• Each orbit is associated with a definite amount of energy. As long as the electrons revolve in an orbit, its energy is constant. Hence, these orbits are called stationary orbits or energy levels, or shells.

Each orbit is denoted by ‘n’, called Principal quantum number. The orbits are designated as 1,2,3,4,…. or K, L, M, N,…
If n = 1, the orbit is first orbit or K-Shell
n = 2, the orbit is second orbit or L-shell
n = 3, the orbit is third orbit or M-shell
n = 4, the orbit is forth orbit or N-shell

• As the value of ‘n’ increases; the size of the orbit as well as its energy also increases.

When an electron jumps from a higher energy level to a lower energy level, the difference in energy is emitted as radiation. When an electron jumps from a lower energy level to a higher energy level, the difference in energy is absorbed as radiation.
∴ ∆E = E₂ – E₁ = hυ
E₁ – Energy of first orbit
E₂ – Energy of second orbit
h – Plank’s constant = 6.625 × 10^-27erg.sec or 6.625 × 10^-34 J.sec
υ – Frequency of radiation

The electron revolves around the nucleus in certain fixed orbits whose angular momentum is equal to the integral multiple of \(\frac{\mathrm{h}}{2 \pi}\).
∴ mvr = n.\(\frac{\mathrm{h}}{2 \pi}\)
m – Mass of electron
v – Velocity of electron
r – Radius of the orbit
h – Plank’s constant
n – Principal quantum number

Thus the angular momentum of an electron is said to be quantized.

• The electrons keep revolving round the nucleus in the stationary states as the centrifugal force of the electron is balanced by the electrostatic attraction between nucleus and electron.
  Importance of Bohr’s model of hydrogen atom to explain various series of line
  spectra in hydrogen atom:
• The line spectrum of hydrogen is also known as the Emission spectrum of hydrogen. There are 5 series in the hydrogen spectrum, Lyman, Balmer, Paschen, Brackett, and Pfund series.
• The emission spectrum of hydrogen or line spectrum of hydrogen is produced when hydrogen gas is taken in the discharge tube and the light emitted on passing electric discharge at low pressure is examined with a spectroscope.
• It is found to consist of a large number of lines that are grouped into different series. The names of different series are the Lyman series, Balmer series, Paschen series, Brackett series, and Pfund series. The emission spectrum is similar except that in place of dark lines, there are colored lines with dark space in between.
• The Lyman series belongs to the ultraviolet region. The Balmer series belongs to the visible region. The Paschen series, Brackett series, and Pfund series belong to the infrared region.

Although a large number of lines are present in the hydrogen spectrum, Rydberg in 1890 gave a very simple theoretical equation for the calculation of the wavelength of these lines. The equation gives the calculation of the wave number \((\bar{v})\) of the lines by the formula:

Where R is a constant called Rydberg constant, and its value is equal to 109677 cm^-1, n₁ and n₂ are whole numbers and for a particular series, n₁ is constant and n₂ varies.

• For Lyman series, n₁ = 1, n₂ = 2, 3, 4 …………
• For Balmer series, n₁ = 2, n₂ = 3, 4, 5 ……….
• For Paschen series, n₁ = 3, n₂ = 4, 5, 6 ………
• For Brackett series, n₁ = 4, n₂ = 5, 6, 7 ………
• For Pfund series, n₁ = 5, n₂ = 6, 7, 8 ………
• For H-like particles, the formula is: \(\bar{v}\) = R\(\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)Z²

Where Z is the atomic number of the H-like particle.

Question 3.
How are quantum numbers n, 1 and m_l arrived at ? Explain the significance of these quantum numbers.
Answer:
Quantum numbers arise from the solutions to Schrodinger’s wave equation for the hydrogen atom. They describe the allowed energy states and spatial distribution of electrons in an atom. Quantum numbers are small integers by means of which any electron in an atom is completely defined. Every electron in an atom is characterized by a set of four quantum numbers. For each electron, the set of quantum numbers are unique.

1) PRINCIPAL QUANTUM NUMBER (n):

• This quantum number was proposed by Neils Bohr.
• It is denoted by the letter ‘n’. It has values from 1,2,3,4,…∞. The orbits with these numbers are also designated by K, L, M, N,…
• This quantum number indicates the orbit number.

Significance:

• It indicates the size and energy of the orbit. As the ‘n’ value increases, the size and energy of orbit also increases.
• With the increase in the value of ‘n’, the number of allowed orbitals increases and are given by ‘n²’.
• The maximum number of electrons in a given orbit number is given by ‘2n²’.

2) AZIMUTHAL QUANTUM NUMBER (l):

• This quantum number was proposed by Sommerfeld.
• It is denoted by the letter ‘ l ’.
• It is also called as orbital or angular momentum or subsidiary quantum number.
• ‘l’ has integer values from 0 to n-1 for each value of n i.e., 0, 1, 2, 3, .,..(n – 1).
• The value of ‘l ’ for a particular orbital is generally designated by the letters s, p, d, f,…. as follows.

Significance:

• It indicates the number of sub shells present in an orbit or shell.
• It indicates the angular momentum of electron.
• It indicates the shape of the orbital, s-orbital is spherical, p-orbital is dumb bell, d-orbital is double dumb bell and f-orbital is four fold dumb bell shape.

3) MAGNETIC QUANTUM NUMBER (m₁):

• It was proposed by Lande to explain Zeeman effect and Stark effect.
• It is denoted by ‘m_l’.
• ‘m_l’ value depends on V. ‘m_l’ can have all the integral values from -l to +l including zero, a total of (2l + 1) values.

Significance:

• It indicates the orientation of orbitals in space.
• It also indicates the number of orbitals present in a sub shell.

Question 4.
What are various ranges of electromagnetic radiation ? Explain the characteristics of electromagnetic radiation.
Answer:
When electrically charged particle mones under acceleration alternating electical and magnetic fields are transmitted in the form of waves called electro magnetic waves or electro magnetic radiations.

Ranges of electro magnetic radiations:
The electro magnetic spectrum is divided into different regions based on wavelength and frequency.

Characteristics of electro magnetic radiation:

1. Wavelength : The distance between two nearest crests or nearest troughs is called the wave length. It is denoted by λ and is measured in terms of centimetre (cm), Angstrom (A°), micro metre (μn), nano metre (nm) etc.
2. Frequency: It is defined as the number of waves which pass through a point in one second. It is denoted by the symbol ‘o’ and is measured in terms of cycles per second or Hertz (Hz).
   υ = \(\frac{c}{\lambda}\)
3. Velocity: It is defined as the distance travelled by a wave in one second. It is denoted by the letter ‘c’, and measured in cm/sec or m/sec. All electro magnetic waves travel with the same velocity. They can travel with the velocity of light i.e., 3 × 10⁸ m/sec or 3 × 1o¹⁰ cm/sec.
4. Wave number: It is defined as the number of wave lengths per unit length. It is the reciprocal of wavelength. It is denoted by \(\bar{v}\).
   ∴ \(\bar{v}\) = \(\frac{1}{\lambda}\)
   It is expressed in cm^-1 or m^-1.
5. Amplitude: It is defined as the height of the crest (or) depth of the trough of a wave. It is denoted by the letter A’. It determines the intensity or brightness of the wave.
6. Time period : It is defined as the time taken by the wave for one complete cycle. It is denoted by T.
7. Electro magnetic radiation propagate without a medium. They can travel through vacuum also.
8. Wave-Particle duality : Electro magnetic radiation behaves both as waves and as particles. It exhibits Interference and diffraction (wave behavior) and also photo electric effect (particle behaviour).

Question 5.
Define atomic orbital. Explain the shapes of s, p and orbitals with the help of diagram.
Answer:
ORBITAL : The region around the nucleus where the probability of finding an electron is maximum is called Orbital.

SHAPES OF ORBITALS : Four types of orbitals are recognized based on their geo-metrical shape and angular momentum of the electrons moving in them. These are s, p, d, f orbitals.

Shape of s-orbital: Only one orientation is possible in case of s-sub shell,
m = 2l + 1 = 2(0) + 1 = 1

s-orbital is spherical in shape i.e., the probability of finding the electron is same in all directions around the nucleus. However the size of the s-orbital increases with ‘n’ value.

Shape of p-orbital: Three orientations are possible in case of p-sub shell,
m = 2l + 1 = 2(1) + 1 = 3
These are represented as p_x, p_y, p_z. The shape of p-orbital is dumb bell. These p-orbitals consist of two lobes.

These are lying along X-axis, Y-axis and Z-axis respectively. The two lobes of a p-orbital are separated by a nodal plane. They have directional character. These are mutually perpendicular. All the p-orbitals of a particular shell have the same energy (degenerate orbitals).

Shape of d-orbital: Five orientations are possible in case of d-sub shell, m = 2l + 1 = 2(2) + 1 = 5
These are represented as d_xy, d_yz, d_zx, \(d_{x^2-y^2}\), \(d_{z^2}\). In d_xy, d_yz and d_zx orbitals, the probability is distributed on the axes, d_xy, d_yz, and d_zx orbitals contains four lobes in between the orbitals. \(d_{x^2-y^2}\) orbital also has four lobes on X and Y – axes. \(\mathrm{d}_{z^2}\) has two lobes on the Z-axis and a ring in the XY – plane. There are two nodal planes in each d-orbital, all these five d-orbitals are degenerate orbitals.

Question 6.
Explain the emission and absorption spectra. Discuss the general description of line spectra in hydrogen atom.
Answer:
Emission and absorption spectra are two fundamental tools in spectroscopy, used to study the composition and behavior of atoms and molecules by analyzing how they interact with light.

1. Emission Spectrum:
An emission spectrum is the spectrum of light emitted by atoms or molecules when they release energy (usually after being excited by heat, electricity, or radiation). When atoms absorb energy, their electrons jump to higher energy levels (excited states). When these electrons return to lower energy levels (ground state), they emit photons of specific energies. These photons appear as bright lines at characteristic wavelengths.

Types:

• Line Emission Spectrum: Produced by gases or atoms in low pressure; consists of discrete lines (e.g., hydrogen gas).
• Continuous Emission Spectrum: Produced by solids, liquids, or dense gases; contains all wavelengths (e.g., incandescent bulb).

Example:

• Hydrogen atom emits light at specific wavelengths like 656 nm (red), 486 nm (blue-green), etc.

2. Absorption Spectrum:
An absorption spectrum shows the wavelengths of light absorbed by a substance when light passes through it. A continuous spectrum of light (like sunlight) passes through a cooler gas or substance. Atoms or molecules absorb photons with energies corresponding to the difference between energy levels. This creates dark lines (absorption lines) in the spectrum where the light has been absorbed.

Types: –

• Always appears as a continuous background with dark lines corresponding to absorbed wavelengths.
  Example:
• The Sun’s spectrum shows dark lines (Fraunhofer lines) where elements in the solar atmosphere absorb specific wavelengths.

General description of line spectra in hydrogen atom:

• The line spectrum of hydrogen is also known as the Emission spectrum of hydrogen. There are 5 series in the hydrogen spectrum, Lyman, Balmer, Paschen, Brackett, and Pfund series.
• The emission spectrum of hydrogen or line spectrum of hydrogen is produced when hydrogen gas is taken in the discharge tube and the light emitted on passing electric discharge at low pressure is examined with a spectroscope.
• It is found to consist of a large number of lines that are grouped into different series. The names of different series are the Lyman series, Balmer series, Paschen series, Brackett series, and Pfund series. The emission spectrum is similar except that in place of dark lines, there are colored lines with dark space in between.
• The Lyman series belongs to the ultraviolet region. The Balmer series belongs to the visible region. The PasChen series, Brackett series, and Pfund series belong to the infrared region. Although a large number of lines are present in the hydrogen spectrum, Rydberg in 1890 gave a very simple theoretical equation for the calculation of the wavelength of these lines. The equation gives the calculation of the wave number \((\bar{v})\) of the lines by the formula:
• \(\bar{v}\) = \(R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)
  Where R is a constant called Rydberg constant, and its value is equal to 109677 cm^-1, n₁ and n₂ are whole numbers and for a particular series, n₁ is.
  constant and n₂ varies.
• For Lyman series, n₁ = 1, n₂ = 2, 3, 4 ……….
• For Balmer series, n₁ = 2, n₂ = 3, 4, 5 ……….
• For Paschen series, n₁ = 3, n₂ = 4, 5, 6 ……
• For Brackett series, n₁ = 4, n₂ = 5, 6, 7 ……..
• For Pfund series, n₁ = 5, n₂ = 6, 7, 8 ……
• For H-like particles, the formula is:
• \(\bar{v}\) = \(\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right) \mathrm{Z}^2\)
• Where Z is the atomic number of the H-like particle.
  

Question 7.
Write the postulates and limitations of Bohr’s model of an atom.
Answer:
Neils Bohr in 1913, postulated a theory based on Rutherford’ model and Plank’s
quantum theory to explain the spectrum of atoms.

Postulates of Bohr’s theory:

• Electrons in an atom revolve around the nucleus with high velocity in certain fixed circular paths called orbits.
• Each orbit is associated with a definite amount of energy. As long as the electrons revolve in an orbit, its energy is constant. Hehce, these orbits are called stationary orbits or energy levels, or shells.
• Each orbit is denoted by ‘n’, called Principal quantum number. The orbits are designated as 1, 2, 3, 4,…. or K, L, M, N,….
  If n = 1, the orbit is first orbit or K-shell
  n = 2, the orbit is second orbit or L-shell
  n = 3, the orbit is third orbit or M-shell
  n = 4, the orbit is forth orbit or N-shell
• As the value of ‘n’ increases, the size of the orbit as well as its energy also increases.
• When an electron jumps from a higher energy level to a lower energy level, the difference in energy is emitted as radiation. When an electron jumps from a lower energy level to a higher energy level, the difference in energy is absorbed as radiation.
  ∴ ∆E = E₂ – E₁ = hυ
  E₁ – Energy of first orbit
  E₂ – Energy of second orbit
  h – Plank’s constant = 6.625 × 10^-27 erg. sec or 6.625 × 10^-34 J.sec
  υ – Frequency of radiation
• The electron revolves around the nucleus in certain fixed orbits whose angular momentum is equal to the integral multiple of \(\frac{\mathrm{h}}{2 \pi}\).
  ∴ mvr = \(n \cdot \frac{\mathrm{~h}}{2 \pi}\)
  m – Mass of electron
  v – Velocity of electron
  r – Radius of the orbit
  h – Plank’sconstant
  n – Principal quantum number
  Thus the angular momentum of an electron is said to be quantized.
• The electrons keep revolving round the nucleus in the stationary states as the centrifugal force of the electron is balanced by the electrostatic attraction between nucleus and electron.

Limitations of Bohr’s atomic theory:

• Bohr’s theory fails to explain the spectra of atoms having more than one electron like He, Li, Be, B etc.
• Bohr’s theory fails to explain Zeeman and Stark effects. Splitting of spectral lines in the applied magnetic field is called Zeeman effect. Splitting of spectral lines in the applied electric field is called Stark effect.
• This theory could, not justify the quantization of angular momentum.
• Bohr’s theory assumes electron as a particle. This is against the wave nature of electron proposed by Debroglie.
• As per Bohr’s theory, the position and velocity of electron’can be determined accurately and simultaneously. This is against the Uncertainity principle proposed by Heisenberg.
• It could not explain the ability of atoms to form molecules by chemical bonds.

Question 8.
What are the consequences that lead to the development of quantum mechanical model of an atom ?
Answer:
The quantum mechanical model of the atom developed as a result of several failures of classical physics and the limitations of earlier atomic models like Rutherford’s model and Bohr’s model. The following are the key consequences that led to its developments.

1) Failure of classical physics to explain atomic stability:

• Rutherford’s model suggested that electrons in an atom revolve around the nucleus in orbite like planets revolve around the sun:
• According to classical electrodynamics, a moving charged particle i.e., electron should emit radiation and fell into the nucleus, making atoms unstable. But atoms are stable. Classical theory failed to explain why.

2) Inability to explain line spectra:

• Classical physics predicted a continuous spectrum from atoms. However atoms show discrete lime spectra.
• Bohr’s model explained Hydrogen’s lines, But failed for multi electron at-oms and could not explain fine structure in spectra.

3) Failure of Bohr’s model for complex atoms:

• Bohr’s model worked only for Hydrogen like atoms have one electron.
• It could not explain.
  • Spectra of multi electron atoms,
  • Zeeman effect
  • Stark effect

4) Dual nature of matter (de Broglie Hypothesis):

• DeBroglie proposed that electrons exhibit wave like behaviour.
• Classical model ignored wave behaviour, but electron diffraction experiment confirmed it.

5) Heisenberg’s Uncertainty principle:

• Werner Heisenberg showed that, it is impossible to know both the exact position and momentum of an electron simultaneously.
• This ruled out the idea of well defined electron orbits.

6) Schrodinger’s wave equation:

• Erwin schrodinger developed a mathematical equation treating electrons as wave functions, not particles in orbits.
• His model predicted probability distributions not exact paths.

Question 9.
Explain the following
a) De Broglie’s concept
b) Heisenberg’s uncertainty principle
Answer:
a) De Broglie’s concept: In 1924, de Broglie proposed that an electron like light behaves both as a particle and as a wave.
When electron is taken as wave, it must be associated with wavelength, fre-quency etc. If ‘υ’ is the frequency of this wave and ‘E’ is its energy then according to planck’s equation,
E = hυ ………. (1)
If electron is taken as particle, then its energy, E is given by Einstein equation.
E = mc² ….. (2)
Where, m – Mass of electron
c – velocity of light
From equations (1) & (2)
hu = mc²
⇒ mc = \(\frac{h \nu}{c}\)
But \(\frac{u}{c}\) = \(\frac{1}{\lambda}\)
∴ mc = \(\frac{\mathrm{h}}{\lambda}\)
⇒ p = \(\frac{\mathrm{h}}{\lambda}\)
where p – momentum
λ – wave length
h – planck’s constant
This is Debroglie’s equation.

b) Heisenberg’s uncertainty principle:
In 1927, Werner Heisenberg stated uncertainty principle. It states that “It is impossible to determine simultaneously, the exact position and exact momentum or velocity of an electron”.
Mathematically it can be given by h
∆x × ∆p ≥ \(\frac{h}{4 \pi}\)
Where ∆x is the uncertainty in position and ∆p is the uncertainty in momentum of the particle.
We know that momentum = Mass × velocity
∴ ∆x × m(∆v) ≥ \(\frac{\mathrm{h}}{4 \pi}\)
⇒ ∆x × ∆v ≥ \(\frac{\mathrm{h}}{4 \pi \mathrm{~m}}\)
If the position of the electron is known with high degree of accuracy i.e., ∆x is small, then the velocity of the electron will be uncertain i.e., ∆v is large.
If the velocity of the electron is known precisely i.e., ∆v is small, then the position of the electron will be uncertain i.e., ∆x is large.