In the numerical example in the text, I calculated only the ratio of the probabilities of a hydrogen atom being in two different states. At such a low temperature the absolute probability of being in a first excited state is essentially the same as the relative probability compared to the ground state. Proving this rigorously, however, is a bit problematic, because a hydrogen atom has infinitely many states.

(a) Estimate the partition function for a hydrogen atom at 5800 K, by adding the Boltzmann factors for all the states shown explicitly in Figure 6.2. (For simplicity you may wish to take the ground state energy to be zero, and shift the other energies according!y.)

(b) Show that if all bound states are included in the sum, then the partition function of a hydrogen atom is infinite, at any nonzero temperature. (See Appendix A for the full energy level structure of a hydrogen atom.)

(c) When a hydrogen atom is in energy level n, the approximate radius of the electron wavefunction is $a_0{n}^2$, where a_o is the Bohr radius, about 5 x 10^-11 m. Going back to equation 6.3, argue that the PdV term is Tot negligible for the very high-n states, and therefore that the result of part (a), not that of part (b), gives the physically relevant partition function for this problem. Discuss.

In the numerical example in the text, only the ratio of the probabilities of a hydrogen atom being in two different states is calculated. At such a low temperature the absolute probability of being in a first excited state is essentially the same as the relative probability compared to the ground state. Proving this rigorously, however, is a bit problematic, because a hydrogen atom has infinitely many states.

(a)The partition function is equal to the total of all Boltzmann factors.

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

Consider a hydrogen atom; in figure 6.2, the ground state energy is zero, therefore the energies must be shifted by 15.6; the three energies are thus ${ϵ}_1=0\mathrm{eV},{ϵ}_2=-3.4+13.6=10.2\mathrm{eV}\text{and}{ϵ}_3=-1.5+13.6=12.1\mathrm{eV}$; the partition function is:

$Z=e^{-{ϵ}_1/kT}+4e^{-{ϵ}_2/kT}+9e^{-{ϵ}_3/kT}$

There are 9 degenerates in role="math" localid="1647326001026" ${ϵ}_3$and four degenerates in ${ϵ}_2$. Replace the energies and Boltzmann constant in eV (k = 8.617 x 10 eV/K) at T = 5800 K, as follows:

role="math" localid="1647326952362" $$\begin{gathered} Z=e^0+4e^{-12.1\mathrm{eV}/0.50\mathrm{eV}}+9e^{-10.2\mathrm{eV}/0.50\mathrm{eV}} \\ Z=1.000000013 \end{gathered}$$

(b) Because the $n^{th}$energy level has $n^2$states, the partition function is as follows:

$Z=\sum _n{n}^2{e}^{-E\left(s\right)/kT}$

$$\begin{gathered} Z>\sum _n{n}^2{e}^{-E_{\infty }/kT} \\ Z>e^{-E_{\infty }/kT}\sum _n{n}^2 \\ Z>e^{-E_{\infty }/kT}(\infty )=\infty \\ Z>\infty \end{gathered}$$

(c) If we keep the PdV term in equation 6.3, the Boltzmann factor will be:

$\text{Boltzmann Factor}=e^{-(E+PV)/kT}$

Where,

$P$is the pressure of reservoir

V is the volume

Consider the volume of a hydrogen atom in its ground state, $V={\left(1.0\times {10}^{-10}\mathrm{m}\right)}^3=1.0\times {10}^{-30}{\mathrm{m}}^3$at atmospheric pressure$P=1.0\times {10}^5\mathrm{Pa}$, the PV term is:

$PV=\left(1.0\times {10}^{-30}{\mathrm{m}}^3\right)\left(1.0\times {10}^5\mathrm{Pa}\right)=1.0\times {10}^{-25}\mathrm{J}$

$\text{using}1.0\mathrm{eV}=1.6\times {10}^{-19}\text{J, we get:}$

$PV=6.25\times {10}^{-7}\mathrm{eV}$

Now, because the radius is proportional to n, the volume will be 100³ greater than the volume of one atom for n = 10, hence the PV term is:

$$\begin{gathered} PV=\left(1.0\times {10}^6\right)\left(1.0\times {10}^{-30}{\mathrm{m}}^3\right)\left(1.0\times {10}^5\mathrm{Pa}\right)=1.0\times {10}^{-19}\mathrm{J} \\ \mathrm{PV}=0.625\mathrm{eV} \end{gathered}$$

At T = 5800 K, kT = 0.5 eV, the term PV reduces the Boltzmann factor by $e^{-PV/kT}=e^{-0.625\mathrm{eV}/0.50\mathrm{eV}}=0.2865$, which is a significant factor; as n increases, this component grows exponentially, forcing the Boltzmann factor to approach l.