Q72 · CSIR-NET Chemistry, June 2017

Paper: CSIR-NET June 2017 · Subject: Physical Chemistry · Chapter: Statistical Thermodynamics · Topic: Statistical Thermodynamics – General · Marks: 2 · Difficulty: Medium

The first excited state $\left({ }^{2} \mathrm{P}_{1 / 2}\right)$ of fluorine lies at an energy of $400 \mathrm{cm}^{-1}$ above the ground state $\left({ }^{2} \mathrm{P}_{3 / 2}\right)$. The fraction of Fluorine atoms in the first excited state at $\mathrm{k}_{\mathrm{B}} \mathrm{T}=420 \mathrm{cm}^{-1}$ is close to
(a)$\frac{1}{1+\mathrm{e}}$
(b)$\frac{1}{2+\mathrm{e}}$
(c)$\frac{1}{1+4 \mathrm{e}}$
(d)$\frac{1}{1+2 \mathrm{e}}$
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
g(²P₃/₂) = 4, g(²P₁/₂) = 2 and e^{−400/420} ≈ e⁻¹: fraction = 2e⁻¹/(4 + 2e⁻¹) = 1/(1 + 2e).

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