91Ó°ÊÓ

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.

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

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 ${\mathrm{Ï\mu }}_1=0\mathrm{eV},{\mathrm{Ï\mu }}_2=-3.4+13.6=10.2\mathrm{eV}\text{and}{\mathrm{Ï\mu }}_3=-1.5+13.6=12.1\mathrm{eV}$; the partition function is:

$Z=e^{-{\mathrm{Ï\mu }}_1/kT}+4e^{-{\mathrm{Ï\mu }}_2/kT}+9e^{-{\mathrm{Ï\mu }}_3/kT}$

There are 9 degenerates in role="math" localid="1647326001026" ${\mathrm{Ï\mu }}_3$and four degenerates in ${\mathrm{Ï\mu }}_2$. 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:

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

$$\begin{gathered} Z>\underset{n}{∑}{n}^2{e}^{-E_{∞}/kT} \\ Z>e^{-E_{∞}/kT}\underset{n}{∑}{n}^2 \\ Z>e^{-E_{∞}/kT}(∞)=∞ \\ Z>∞ \end{gathered}$$

(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, $V={\left(1.0×{10}^{-10}\mathrm{m}\right)}^3=1.0×{10}^{-30}{\mathrm{m}}^3$at atmospheric pressure$P=1.0×{10}^5\mathrm{Pa}$, the PV term is:

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

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

$PV=6.25×{10}^{-7}\mathrm{eV}$

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:

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

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.