A Hint of Quantum Theory — Question 2

Question 2

Level degeneracy competes with the Boltzmann energy penalty

A ground level has one state and energy zero. An excited level at energy $\Delta$ has three distinct states of equal energy. Each state carries weight $e^{-E/(k_BT)}$. Let $\Theta=\Delta/k_B=600$ K.

1. Derive the total excited-level probability and find the temperature at which it equals .250.

2. Find the high-temperature limiting excited fraction and the probability of one particular excited state in that limit. Can the total excited level become more populated than the ground level?

Question 2 - Solution

Level degeneracy competes with the Boltzmann energy penalty

(a) Method and calculation

$$Z=1+3e^{-\Theta/T},\quad P_{\mathrm{ex}}=\frac{3e^{-\Theta/T}}{1+3e^{-\Theta/T}}.$$$$P_{\mathrm{ex}}=\tfrac14\ \Rightarrow\ 3e^{-\Theta/T}=\tfrac13\ \Rightarrow\ T=\frac{600}{\ln9}\approx273.072\,\mathrm K.$$

(b) Interpretation and further calculation

$$T\to\infty:\quad P_{\mathrm{ex}}\to\tfrac34,\quad P_{\mathrm{one\ excited\ state}}\to\tfrac14.$$$$P_{\mathrm{ex}}>P_0\ \Leftrightarrow\ 3e^{-\Theta/T}>1\ \Leftrightarrow\ T>\Theta/\ln3.$$Each excited state remains less probable than the unique ground state at finite positive temperature, but three such probabilities can have a larger sum.

Physical check

Boltzmann weights apply to states; a level probability includes the number of states at that energy.