Question

Each of the hydrogen atom states shown in Figure 6.2 is actually twofold degenerate, because the electron can be in two independent spin states, both with essentially the same energy. Repeat the calculation given in the text for the relative probability of being in a first excited state, taking spin degeneracy into account. Show that the results are unaffected.

Short answer

Therefore,

$\frac{P\left(E_2\right)}{P\left(E_1\right)}=4e^{-\left(E_2-E_1\right)/kT}$

Step-by-step solution

Given information

Each of the hydrogen atom states shown in Figure 6.2 is actually twofold degenerate, because the electron can be in two independent spin states, both with essentially the same energy.

Explanation

With n = 2, if E1 is the ground state energy and E2 is the first excited state energy, the probability is:

$P\left(s\right)=\frac{1}{Z}{e}^{-E/kT}$

There are four states in E2 and one state in E1, thus their probabilities are P(s2) multiplied by eight and P(s1) multiplied by two when we add the spin. Thus, the probability of E2 equals $P\left(s_2\right)$multiplied by eight and the probability of E1 equals $P\left(s_1\right)$multiplied by two when we include the spin.

$P\left(E_2\right)=\frac{8}{Z}{e}^{-E_2/kT}P\left(E_1\right)=\frac{2}{Z}{e}^{-E_1/kT}$

The ratio of these probabilities is:

role="math" localid="1647300792412" $$\begin{gathered} \frac{P\left(E_2\right)}{P\left(E_1\right)}=\frac{8e^{-E_2/kT}}{2e^{-E_1/kT}} \\ \frac{P\left(E_2\right)}{P\left(E_1\right)}=4e^{-\left(E_2-E_1\right)/kT} \end{gathered}$$