The equilibrium population ratio (nj / ni ) of a doubly-degenerate energy level (Ej ) lying at energy 2 units higher than a lower non-degenerate energy level (Ej ), assuming kB T=1 unit, will be

ABC26PH0200 · Statistical Thermodynamics

Subject: Physical Chemistry · Chapter: Statistical Thermodynamics · Topic: Statistical Thermodynamics – General · Exam: CSIR-NET DEC 2012 · 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 (as printed): A

Explanation

$k_{B}T=1$, $\Delta\epsilon=2$. $\dfrac{n_{j}}{n_{i}}=\dfrac{g_{j}}{g_{i}}e^{-\Delta\epsilon/k_{B}T}=2e^{-2}$.

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