Question

Estimate the probability that a hydrogen atom at room temperature is in one of its first excited states (relative to the probability of being in the ground state). Don't forget to take degeneracy into account. Then repeat the calculation for a hydrogen atom in the atmosphere of the star$\gamma$ UMa, whose surface temperature is approximately 9500 K.

Short answer

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

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

Step-by-step solution

Given information

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

Explanation

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:

$Z=\sum _s{e}^{-E\left(s\right)/kT}\approx e^{-E_1/kT}$

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.

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

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

Explanation

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

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

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