A Hint of Quantum Theory — Question 9

Question 9

Population ratios at two temperatures separate gap from degeneracy

An internal ground level and excited level have equilibrium population ratio $r(T)=P_1/P_0=g\,e^{-\Theta/T}$, with unknown positive degeneracy ratio $g$ and gap parameter $\Theta=\Delta/k_B$. Measurements give $r(200\,\mathrm K)=.125$ and $r(400\,\mathrm K)=.500$. The same two levels and gap apply in both runs.

1. Infer $\Theta$ and $g$ without assuming equal degeneracies.

2. Predict the ratio and excited fraction at 800 K. Explain what would go wrong if the first measurement alone were analyzed with $g=1$.

Question 9 - Solution

Population ratios at two temperatures separate gap from degeneracy

(a) Method and calculation

$$\ln\frac{r(400)}{r(200)}=\Theta\left(\frac1{200}-\frac1{400}\right)\ \Rightarrow\Theta=400\ln4\approx554.518\,\mathrm K.$$$$g=.125\,e^{\Theta/200}=.125e^{2\ln4}=2.00.$$

(b) Interpretation and further calculation

$$r(800)=2e^{-\ln4/2}=1,\quad P_1=\frac r{1+r}=\tfrac12.$$$$\Theta_{g=1}=-200\ln(.125)=200\ln8\approx415.888\,\mathrm K.$$One ratio at one temperature cannot separately identify both the gap and degeneracy; an assumed degeneracy biases the inferred gap.

Physical check

A plot of log population ratio against inverse temperature separates a slope from an intercept.