Q89 · CSIR-NET Chemistry, June 2016

Paper: CSIR-NET June 2016 · Subject: Physical Chemistry · Chapter: Quantum Chemistry · Topic: Quantum Chemistry – General · Marks: 2 · Difficulty: Medium

The transition moment integral for a rotational transition between $\mathrm{J}=1 ; \mathrm{M}_{\mathrm{J}}=0$ and $\mathrm{J}=2 ; \mathrm{M}_{\mathrm{J}}=0$ states for a diatomic molecule along the z-axis is proportional to
(a)$\int_{0}^{\pi} \cos ^{2} \theta\left(3 \cos ^{2} \theta-1\right) \mathrm{d} \theta$
(b)$\int_{0}^{\pi} \cos ^{2} \theta\left(3 \cos ^{2} \theta-1\right) \sin \theta \mathrm{d} \theta$
(c)$\int_{0}^{\pi} \cos \theta\left(3 \cos ^{2} \theta-1\right) \sin \theta \mathrm{d} \theta$
(d)$\int_{0}^{\pi} \cos ^{2} \theta\left(3 \cos ^{2} \theta-1\right) \sin ^{2} \mathrm{d} \theta$
Answer
Answer: B ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
Y₁₀ ∝ cos θ, Y₂₀ ∝ (3cos²θ − 1), μ_z ∝ cos θ, volume element sin θ dθ: the integral is ∫₀^π cos²θ(3cos²θ − 1) sin θ dθ.

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