(a) (i) In a diatomic molecule when one constituent atom is replaced by one of its heavier isotopes, what change takes place in the rotational spectrum ?

(ii) Calculate the change in rotational constant B when hydrogen is replaced by deuterium in th

Question

(a) (i) In a diatomic molecule when one constituent atom is replaced by one of its heavier isotopes, what change takes place in the rotational spectrum ?

(ii) Calculate the change in rotational constant B when hydrogen is replaced by deuterium in the hydrogen molecule.

(iii) Draw the spectra of rigid and non-rigid rotors by using the schematic representation of the rotational energy levels and comment on it. 20 marks

(b) (i) Briefly explain the effect of anharmonicity on the vibrational spectra of diatomic molecules.

(ii) Calculate the average period of rotation of HCl molecule if it is in the J = 3 state. The internuclear distance and the moment of inertia of HCl are 0·1274 nm and 0·0264×10⁻⁴⁵ kg.m² respectively. 15 marks

(c) Obtain the normalized eigenvectors of σₓ and σᵧ matrices. 15 marks

UPSC Answer Check · Accepted Answer

(a)(i) When one atom in a diatomic molecule is replaced by a heavier isotope, the reduced mass μ = m₁m₂/(m₁+m₂) increases. The bond length is almost unchanged, so the moment of inertia I = μr² increases. Since the rotational constant is B = h/(8π²Ic), B decreases. For a rigid rotor, rotational lines occur at 2B, 4B, 6B, … so the line spacing 2B decreases and the whole rotational spectrum shifts to lower frequency/wavenumber. If the original molecule is homonuclear and has no pure rotational dipole spectrum, isotopic substitution can break the symmetry and make a weak rotational spectrum appear.

(a)(ii) Taking the usual case of replacing one H atom by D in H₂ (i.e. H₂ → HD), let m_H = m and m_D ≈ 2m. Then μ(H₂) = m/2, μ(HD) = m(2m)/(m+2m) = 2m/3. Thus μ(HD)/μ(H₂) = (2m/3)/(m/2) = 4/3. Since B ∝ 1/μ, B(HD) = (3/4)B(H₂). Therefore ΔB = B(HD) − B(H₂) = −B(H₂)/4. Using B(H₂) ≈ 60.8 cm⁻¹, ΔB ≈ −15.2 cm⁻¹. If both H atoms are replaced (H₂ → D₂), then B(D₂) = B(H₂)/2 and ΔB ≈ −30.4 cm⁻¹.

(a)(iii) Rigid rotor: E_J = B J(J+1). Non-rigid rotor: E_J = B J(J+1) − D_c J²(J+1)², with D_c > 0 due to centrifugal stretching. Selection rule ΔJ = ±1.

Rigid levels:
J=3 ───── 12B
J=2 ───── 6B
J=1 ───── 2B
J=0 ───── 0

Rigid spectrum lines:
|--2B--|--2B--|--2B--| …
They are equally spaced.

Non-rigid levels:
J=3 ───── 12B − 144D_c
J=2 ───── 6B − 36D_c
J=1 ───── 2B − 4D_c
J=0 ───── 0

Non-rigid transition wavenumbers:
ν = 2B(J+1) − 4D_c(J+1)³.
So lines occur at
2B − 4D_c, 4B − 32D_c, 6B − 108D_c, …
The spacing decreases as J increases; the lines crowd together at higher frequency. Thus the rigid-rotor spectrum is a series of equally spaced lines, while the non-rigid-rotor spectrum shows unequal spacing due to centrifugal distortion.

(b)(i) Anharmonicity is described by a Morse-type potential rather than a pure harmonic potential. In the harmonic oscillator, levels are equally spaced: E_v = (v+1/2)ħω, and the selection rule is Δv = ±1, giving only one fundamental band. With anharmonicity,
E_v = ω_e(v+1/2) − ω_e x_e(v+1/2)² + …
The level spacing decreases as v increases, and the levels converge near the dissociation limit. The selection rule relaxes so that weak overtones Δv = ±2, ±3, … also appear. The fundamental band shifts slightly, overtones occur at frequencies lower than exact multiples of the fundamental, and hot bands arise from thermally populated excited vibrational levels. Thus the vibrational spectrum is no longer a single equally spaced set of lines.

(b)(ii) For a rigid rotor, the angular momentum is
L = √(J(J+1)) ħ = Iω.
Hence ω = √(J(J+1)) ħ / I, and the average period is
T = 2π/ω = 2πI/[ħ√(J(J+1))].
For J = 3, √(J(J+1)) = √12 = 2√3.
Given I = 0.0264×10⁻⁴⁵ kg.m² = 2.64×10⁻⁴⁷ kg.m² and ħ = 1.0546×10⁻³⁴ J.s,
T = 2π(2.64×10⁻⁴⁷)/[(1.0546×10⁻³⁴)(3.464)]
T = 4.54×10⁻¹³ s.
Thus the average rotational frequency is 1/T = 2.20×10¹² Hz.
T ≈ 4.54×10⁻¹³ s.

(c) For σₓ = [[0, 1], [1, 0]], the eigenvalue equation is det(σₓ − λI) = 0:
λ² − 1 = 0, so λ = ±1.
For λ = +1: b = a, choose a = 1, b = 1. Normalized eigenvector:
(1/√2)[[1], [1]].
For λ = −1: b = −a, choose a = 1, b = −1. Normalized eigenvector:
(1/√2)[[1], [−1]].

For σᵧ = [[0, −i], [i, 0]], again det(σᵧ − λI) = 0 gives λ² − 1 = 0, so λ = ±1.
For λ = +1: −i b = a and i a = b. Choose a = 1, b = i. Normalized eigenvector:
(1/√2)[[1], [i]].
For λ = −1: −i b = −a and i a = −b. Choose a = 1, b = −i. Normalized eigenvector:
(1/√2)[[1], [−i]].
These eigenvectors are mutually orthogonal and normalized up to an overall phase factor.

(a) (i) In a diatomic molecule when one constituent atom is replaced by one of its heavier isotopes, what change takes place in…

Model answer

What "Calculate" is asking you to do

Structure that answers it

Where marks are lost

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Practice this exact question

More from Physics 2022 Paper II