PX282 - C2b - an alternate form

the boltzmann equation can be written in an alternate form to get $\frac{N_{A}}{N}$
$N$ is the total number of electrons:

N = \sum_{m = 1}^{\infty} N_{m}

using the boltzmann equation:

\begin{matrix} \frac{N_{m}}{N_{1}} = \frac{g_{m}}{g_{1}} \exp (- \frac{E_{m} - E_{1}}{k_{B} T}) \\ ⟹ N = \frac{N_{1}}{g_{1}} \sum_{m = 1}^{\infty} g_{m} \exp (- \frac{E_{m} - E_{1}}{k_{B} T}) \end{matrix}

defining the partition function:

U (T) = \sum_{m = 1}^{\infty} g_{m} \exp (- \frac{E_{m} - E_{1}}{k_{B} T})

therefore, the fraction of electrons in state 1 is:

\begin{matrix} ⟹ \frac{N_{1}}{N} = \frac{g_{1}}{U (T)} \\ ∴ \frac{N_{m}}{N} = \frac{g_{m}}{U (T)} \exp (- \frac{E_{m} - E_{1}}{k_{B} T}) \end{matrix}

eg: what $T$ is needed to get $N_{2} = N_{1}$ ?
$⟹ \frac{N_{2}}{N_{1}} = 1$
- known:
  $g_{2} = 8$ , $g_{1} = 2$
  $E_{1} = - 13.6 e V$ , $E_{2} = - 3.4 e V$
- using the botzmann equation:
$- \frac{E_{1} - E_{2}}{k_{B} T} = l n (\frac{g_{1}}{g_{2}}) ⟹ T = 85 300 K$

example

calculating the temperature to get $N_{2} = N_{1}$
using the boltzmann equation:

\frac{N_{2}}{N_{1}} = \frac{g_{2}}{g_{1}} \exp (- \frac{E_{2} - E_{1}}{k_{B} T}) = 1

here, $g_{2} = 8$ and $g_{1} = 2$ , and $E_{2} = - 3.4 e V$ and $E_{1} = - 13.6 e V$
rearranging the equation:

T = \frac{E_{2} - E_{1}}{k_{B} \log (\frac{g_{2}}{g_{1}})} \approx 85300 K

balmer lines require lots of electrons in the $n = 2$ state, and they get weaker for $T >\sim 9000 K$ due to ionization
there exists a 'de-ionized' form of hydrogen, $H^{-}$ , with two electrons, which is useful in case of the sun
$H e I : 2 p 2 n 2 e$ ; $H e I I : 2 p 2 n 1 e$