ABC26PH1398 · Statistical Thermodynamics

Subject: Physical Chemistry · Chapter: Statistical Thermodynamics · Topic: Statistical Thermodynamics – General · Exam: NET JUNE 2017 · 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 (as printed): D
Explanation
No explanation was printed for this question.

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