The transition moment integral for a rotational transition between J=1; MJ=0 and J=2; MJ =0 states for a… · ABC26PH0170

Subject: Physical Chemistry · Chapter: Quantum Chemistry · Topic: Quantum Chemistry – General · Exam: CSIR NET 2018 · 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)$\int_{0}^{\pi} \cos ^{2} \theta\left(3 \cos ^{2} \theta-1\right) d \theta$
(b)$\int_{0}^{\pi} \cos ^{2} \theta\left(3 \cos ^{2} \theta-1\right) \sin \theta d \theta$
(c)$\int_{0}^{\pi} \cos \theta\left(3 \cos ^{2} \theta-1\right) \sin \theta d \theta$
(d)$\int_{0}^{\pi} \cos ^{2} \theta\left(3 \cos ^{2} \theta-1\right) \sin ^{2} d \theta$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
Y₁₀ ∝ cos θ, Y₂₀ ∝ (3cos²θ − 1), μ_z ∝ cos θ, and the volume element carries sin θ dθ. The z-transition moment is ∝ ∫₀^π cos²θ (3cos²θ − 1) sin θ dθ.

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