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

1. Practise the PYQsPrevious-year questions, with answers

· Browse this chapter in the app