Class 12 Physics Notes – Chapter 12: Atoms

1. Introduction & Early Models of Atom

The first model of the atom was proposed by J.J. Thomson in 1898. According to his “Plum Pudding Model”, the positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon.

2. Alpha-Particle Scattering and Rutherford’s Nuclear Model

In 1911, H. Geiger and E. Marsden (under the guidance of Ernst Rutherford) performed the alpha-particle scattering experiment. They directed a beam of 5.5 MeV α-particles at a thin gold foil.

Observations:

• Most of the α-particles passed straight through the gold foil without any deflection (implying most of the space inside the atom is empty).
• A small fraction of α-particles were deflected by small angles.
• A very small number (about 1 in 8000) bounced back (deflected by nearly 180°).

Rutherford’s Conclusions (Nuclear Model):

• The entire positive charge and most of the mass of the atom are concentrated in a tiny central core called the nucleus.
• The electrons revolve around the nucleus in circular orbits, similar to planets revolving around the sun.
• The electrostatic force of attraction provides the necessary centripetal force for the electrons.

Impact Parameter (b)

It is the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus. For a head-on collision, the impact parameter is minimum (zero), resulting in a rebound (scattering angle θ ≈ 180°).

3. Electron Orbits and Energy

In Rutherford’s model, a dynamically stable orbit in a hydrogen atom implies that the electrostatic force equals the centripetal force:

F_e = F_c ⇒ 1 / (4πε₀) · (e² / r²) = m v² / r

From this, the kinetic energy (K) and potential energy (U) of the electron are:

• Kinetic Energy (K): K = 1/2 m v² = e² / (8πε₀ r)
• Potential Energy (U): U = – e² / (4πε₀ r)

Total Energy (E) = K + U = – e² / (8πε₀ r)

The negative sign implies that the electron is bound to the nucleus.

4. Bohr Model of the Hydrogen Atom

Rutherford’s model could not explain the stability of the atom (an accelerating electron should radiate energy and spiral into the nucleus) and the discrete line spectra of atoms. Niels Bohr introduced three postulates to solve this:

1. Stationary Orbits: An electron revolves around the nucleus in certain stable circular orbits without emitting radiant energy.
2. Quantisation of Angular Momentum: The electron revolves only in those orbits for which its angular momentum (L) is an integral multiple of h / (2π).
   L = m v r = n h / (2π) (where n = 1, 2, 3…)
3. Energy Emission: An electron makes a transition from a higher energy state (E_i) to a lower energy state (E_f), emitting a photon of frequency ν.
   hν = E_i – E_f

5. Energy Levels of Hydrogen Atom

Using Bohr’s postulates, the radius of the n^th orbit is r_n ∝ n². The total energy of the electron in the n^th orbit is:

E_n = – 13.6 / n² eV

• Ground State: The lowest energy state (n = 1) has energy E₁ = -13.6 eV.
• Excited States: For n = 2, E₂ = -3.40 eV. For n = 3, E₃ = -1.51 eV.
• Ionisation Energy: The minimum energy required to free the electron from the ground state is +13.6 eV.

6. De Broglie’s Explanation of Bohr’s Quantisation

Louis de Broglie explained Bohr’s second postulate by proposing that the electron in its circular orbit must be seen as a standing matter wave. For a stable orbit, the circumference of the orbit must be an integral multiple of the de Broglie wavelength (λ).

2πr = n λ

Since λ = h / p = h / (mv), we get 2πr = n h / (mv), which rearranges to Bohr’s condition: mvr = nh / (2π).

Multiple Choice Questions (MCQs)

1. The first model of the atom, known as the plum pudding model, was proposed by:
a) Ernest Rutherford
b) Niels Bohr
c) J.J. Thomson
d) James Chadwick
Answer: (c) J.J. Thomson

2. In the Rutherford scattering experiment, what provides the force that deflects the alpha particles?
a) Gravitational force
b) Strong nuclear force
c) Magnetic force
d) Electrostatic force (Coulomb force)
Answer: (d) Electrostatic force (Coulomb force)

3. According to Bohr’s quantisation postulate, the angular momentum of an electron is an integral multiple of:
a) h / π
b) h / 2π
c) 2π / h
d) h
Answer: (b) h / 2π

4. The total energy of an electron in a hydrogen atom is:
a) Always positive
b) Zero
c) Always negative
d) Sometimes positive, sometimes negative
Answer: (c) Always negative

5. The energy of an electron in the ground state of a hydrogen atom is:
a) -1.51 eV
b) -3.4 eV
c) -13.6 eV
d) 0 eV
Answer: (c) -13.6 eV

6. The relation between kinetic energy (K) and total energy (E) of an electron in a Bohr orbit is:
a) K = E
b) K = -E
c) K = 2E
d) K = E/2
Answer: (b) K = -E

7. The impact parameter (b) for an alpha particle that rebounds (scatters by 180°) is:
a) Minimum (Zero)
b) Maximum
c) Infinity
d) Equal to the radius of the atom
Answer: (a) Minimum (Zero)

8. De Broglie explained Bohr’s second postulate by comparing the electron in orbit to:
a) A falling particle
b) A standing wave
c) A transverse light wave
d) A magnetic dipole
Answer: (b) A standing wave

9. In a hydrogen atom, the radius of the n^th Bohr orbit is proportional to:
a) n
b) n²
c) 1/n
d) 1/n²
Answer: (b) n²

10. The minimum energy required to completely remove an electron from the ground state of a hydrogen atom is called:
a) Kinetic energy
b) Potential energy
c) Excitation energy
d) Ionisation energy
Answer: (d) Ionisation energy

Textbook Exercise Solutions

Q 12.1: Choose the correct alternative from the clues given at the end of each statement:
(a) The size of the atom in Thomson’s model is ……… the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)
(b) In the ground state of ……… electrons are in stable equilibrium, while in ……… electrons always experience a net force. (Thomson’s model/ Rutherford’s model.)
(c) A classical atom based on ……… is doomed to collapse. (Thomson’s model/ Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a ……… but has a highly non-uniform mass distribution in ……… (Thomson’s model/ Rutherford’s model.)
(e) The positively charged part of the atom possesses most of the mass in ……… (Rutherford’s model/both the models.)

Solution:
(a) No different from
(b) Thomson’s model; Rutherford’s model
(c) Rutherford’s model
(d) Thomson’s model; Rutherford’s model
(e) Both the models

Q 12.2: Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?
Solution:
The nucleus of a hydrogen atom is a proton. The mass of an alpha particle (which is a helium nucleus) is about 4 times greater than the mass of a hydrogen nucleus (proton). Because the scattering particle is much more massive than the target nucleus, the alpha particle will not bounce back, even in a head-on collision. It is similar to a football colliding with a tennis ball at rest. Thus, there would be no large-angle scattering.

Q 12.3: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?
Solution:
Given: Energy difference ΔE = 2.3 eV.
Convert to Joules: ΔE = 2.3 × 1.6 × 10^-19 J = 3.68 × 10^-19 J.
Using Planck’s equation, ΔE = hν.
Frequency ν = ΔE / h = (3.68 × 10^-19) / (6.63 × 10^-34) = 5.6 × 10¹⁴ Hz.

Q 12.4: The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state?
Solution:
Given: Total Energy E = -13.6 eV.
In a hydrogen atom, Kinetic Energy K = -E, and Potential Energy U = 2E.
Kinetic energy K = -(-13.6) = +13.6 eV.
Potential energy U = 2 × (-13.6) = -27.2 eV.

Q 12.5: A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
Solution:
Ground state energy (n=1), E₁ = -13.6 eV.
Energy in n=4 level, E₄ = -13.6 / (4)² = -13.6 / 16 = -0.85 eV.
Energy of absorbed photon ΔE = E₄ – E₁ = -0.85 – (-13.6) = 12.75 eV.
Convert to Joules: ΔE = 12.75 × 1.6 × 10^-19 = 2.04 × 10^-18 J.
Frequency ν = ΔE / h = (2.04 × 10^-18) / (6.63 × 10^-34) = 3.1 × 10¹⁵ Hz.
Wavelength λ = c / ν = (3 × 10⁸) / (3.1 × 10¹⁵) = 0.97 × 10^-7 m = 97 nm (or 9.7 × 10^-8 m).

Q 12.6: (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.
Solution:
Speed of electron in n^th orbit is v_n = v₁ / n, where v₁ = 2.18 × 10⁶ m/s.
(a) For n = 1, v₁ = 2.18 × 10⁶ m/s.
For n = 2, v₂ = (2.18 × 10⁶) / 2 = 1.09 × 10⁶ m/s.
For n = 3, v₃ = (2.18 × 10⁶) / 3 = 7.27 × 10⁵ m/s.

(b) Orbital period T = 2πr_n / v_n.
Radius r_n = n² × 0.53 × 10^-10 m.
For n = 1: r₁ = 0.53 × 10^-10 m. T₁ = (2 × 3.14 × 0.53 × 10^-10) / (2.18 × 10⁶) = 1.52 × 10^-16 s.
For n = 2: r₂ = 4 × 0.53 × 10^-10 m = 2.12 × 10^-10 m. T₂ = (2 × 3.14 × 2.12 × 10^-10) / (1.09 × 10⁶) = 1.22 × 10^-15 s.
For n = 3: r₃ = 9 × 0.53 × 10^-10 m = 4.77 × 10^-10 m. T₃ = (2 × 3.14 × 4.77 × 10^-10) / (7.27 × 10⁵) = 4.11 × 10^-15 s.

Q 12.7: The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-11 m. What are the radii of the n = 2 and n = 3 orbits?
Solution:
The radius of the n^th orbit is given by r_n = n² r₁.
Given r₁ = 5.3 × 10^-11 m.
For n = 2: r₂ = (2)² × 5.3 × 10^-11 = 4 × 5.3 × 10^-11 = 2.12 × 10^-10 m.
For n = 3: r₃ = (3)² × 5.3 × 10^-11 = 9 × 5.3 × 10^-11 = 4.77 × 10^-10 m.

Q 12.8: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Solution:
At room temperature, hydrogen atoms are in the ground state (n = 1, E₁ = -13.6 eV).
Maximum energy of the atom after absorbing the electron’s energy = -13.6 eV + 12.5 eV = -1.1 eV.
Energy of n = 3 state is E₃ = -13.6 / 9 = -1.51 eV, and n = 4 state is E₄ = -13.6 / 16 = -0.85 eV.
Since -1.1 eV is between E₃ and E₄, the electron can only be excited up to the n = 3 state.
Possible transitions when it falls back:
1) n = 3 to n = 1 (Lyman series)
2) n = 2 to n = 1 (Lyman series)
3) n = 3 to n = 2 (Balmer series)
Calculating wavelengths (λ = hc / ΔE = 1240 eV nm / ΔE):
λ₃₁ = 1240 / (-1.51 – (-13.6)) = 1240 / 12.09 ≈ 103 nm (Lyman series).
λ₂₁ = 1240 / (-3.4 – (-13.6)) = 1240 / 10.2 ≈ 122 nm (Lyman series).
λ₃₂ = 1240 / (-1.51 – (-3.4)) = 1240 / 1.89 ≈ 656 nm (Balmer series).

Q 12.9: In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ m/s. (Mass of earth = 6.0 × 10²⁴ kg.)
Solution:
According to Bohr’s quantisation postulate: L = m v r = n h / (2π)
Therefore, quantum number n = (2π × m × v × r) / h
n = (2 × 3.14 × 6.0 × 10²⁴ × 3 × 10⁴ × 1.5 × 10¹¹) / (6.63 × 10^-34)
n = (169.56 × 10³⁹) / (6.63 × 10^-34)
n ≈ 2.6 × 10⁷⁴.
(This huge quantum number shows why macroscopic objects appear to have continuous energy levels and classical mechanics applies to them.)