91Ó°ÊÓ

A hydrogen atom at room temperature is in one of its first excited states (relative to the probability of being in the ground state).

With n = 2, assuming E1 is ground state energy and E2 is first excited state energy, the chance of finding the atom in either of the first excited states is:

$P\left(s_2\right)=\frac{1}{Z}{e}^{-E_2/kT}$

The partition function is equal to the sum of the Boltzmann factors, but because the initial state has the greatest energy magnitude, we can approximate it as follows:

$Z=\underset{s}{∑}{e}^{-E\left(s\right)/kT}≈e^{-E_1/kT}$

Substitute in the above equation

$$\begin{gathered} P\left(s_2\right)=e^{E_1/kT}{e}^{-E_2/kT} \\ P\left(s_2\right)=e^{-\left(E_2-E_1\right)/kT} \end{gathered}$$

There are four such states, therefore the probability of E2 equals $P\left(s_2\right)$multiplied by 4, so:

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

Substitute the energies and Boltzmann constant in eV (k = 8.617 x 10^-5 eV/K) at room temperature T = 300 K for the energy of the ground state E1=-13.6 eV and the energy of the first excited state B2 =-3.4 eV.

$$\begin{gathered} P\left(E_2\right)=4e^{-(-3.4\mathrm{eV}-(-13.6\mathrm{eV}\left)\right)/\left(\left(8.617×{10}^{-5}\mathrm{eV}/\mathrm{K}\right)(300\mathrm{K})\right)} \\ P\left(E_2\right)=1.75×{10}^{-171} \end{gathered}$$

As a result, no Hydrogen atoms can be found in the first excited state.

We must now calculate this probability for the star $\mathrm{γ}$UMa, which has a surface temperature of 9500 K, as follows:

$$\begin{gathered} P\left(E_2\right)=4e^{-(-3.4\mathrm{eV}-(-13.6\mathrm{eV}\left)\right)/\left(\left(8.617×{10}^{-5}\mathrm{eV}/\mathrm{K}\right)(9500\mathrm{K})\right)} \\ P\left(E_2\right)=1.55×{10}^{-5} \end{gathered}$$

On this star, there is one atom in the first excited state out of 64500 hydrogen atoms.