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## Q. 9.6

Assume that the Maxwell-Boltzmann distribution is valid in a gas of atomic hydrogen. What is the relative number of atoms in the ground state and first excited state at 293 K (room temperature), 5000 K (the temperature at the surface of a star), and $10^{6}$ K (a temperature in the interior of a star)?

Strategy The desired ratio is

$\frac{n\left(E_{2}\right)}{n\left(E_{1}\right)}=\frac{g\left(E_{2}\right)}{g\left(E_{1}\right)} \exp \left[\beta\left(E_{1}-E_{2}\right)\right]$

In the ground state (n = 1) of hydrogen there are two possible configurations for the electron, that is, $g\left(E_{1}\right)=2$. There are eight possible configurations in the first excited state (see Chapter 7), so $g\left(E_{2}\right)=8$. For atomic hydrogen $E_{1}-E_{2}=-10.2$ eV. Therefore

$\frac{n\left(E_{2}\right)}{n\left(E_{1}\right)}=4 \exp [\beta(-10.2 eV )]=4 \exp (-10.2 eV / k T)$

for a given temperature T. We need to insert numerical values for each of the temperatures.

## Verified Solution

The desired numerical results are

$\frac{n\left(E_{2}\right)}{n\left(E_{1}\right)}=4 \exp (-404) \approx 10^{-175} \quad \text { for } T=293 K$

$=4 \exp (-23.7) \approx 2 \times 10^{-10} \quad \text { for } 5000 K$

$=4 \exp (-0.118) \approx 3.55 \quad \text { for } 10^{6} K$

Notice that at very high temperatures $\left(T \gg 10^{6} K \right)$, the exponential factor approaches 1, so the ratio $n\left(E_{2}\right) / n\left(E_{1}\right)$ approaches 4, the ratio of the densities of states. In fact, atomic hydrogen cannot exist at such high temperatures ($10^{6}$ K or greater). The electrons dissociate from the nuclei to form a state of matter known as a plasma.