Q33 · CSIR NET Chemistry, December 2012
Paper: CSIR NET December 2012 · Subject: Physical Chemistry · Chapter: Statistical Thermodynamics · Topic: Statistical Thermodynamics – General · Marks: 2 · Difficulty: Medium
The equilibrium population ratio $\left(n_{j} / n_{i}\right)$ of a doubly-degenerate energy level $\left(E_{j}\right)$ lying at energy 2 units higher than a lower non-degenerate energy level $\left(E_{j}\right)$, assuming $k_{B} T=1$ unit, will be
(a)$2e^{-2}$
(b)$2e^{2}$
(c)$\mathrm{e}^{2}$
(d)$\mathrm{e}^{-2}$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
n_j/n_i = (g_j/g_i)e^(−ΔE/kT) = 2e^(−2) (a).
Study loop for Statistical Thermodynamics