PX282 - C2a - the boltzmann equation

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

derivation

from statistical mechanics, the probability that an electron is in state 1 , requiring an energy, $E_{1}$ , over state 2, with an energy, $E_{2}$ , for $E_{1} > E_{2}$ , is given by:

\frac{P_{1}}{P_{2}} = \frac{e^{- \frac{E_{1}}{k_{B} T}}}{e^{- \frac{E_{2}}{k_{B} T}}} = \exp (- \frac{E_{1} - E_{2}}{k_{B} T})

if the temperature is low, $k_{B} T \to 0 ⟹ \frac{P_{1}}{P_{2}} \to 0 :$ the electrons go to the lowest state
if the temperature is high, $k_{B} T \to \infty ⟹ \frac{P_{1}}{P_{2}} \to 1 :$ with infinite energy, all states are equally accessible
these are quantum states, defined by:
$n :$ energy
$l = n - 1 :$ angular momentum
$m_{l} \in [- l, + l], Z :$ alignment
$m_{s} = \pm \frac{1}{2} :$ spin
consider quantum numbers for $H :$
- at ground state:
  - $n = 1$
  - $l = n - 1 = 0$
  - $m_{l} = - l \to + l = 0$
  - $m_{s} = \pm \frac{1}{2}$
  - there are two energy states
- at $n = 2 :$
  - $l = 0, 1$
  - $m_{0} = 0$ , $m_{1} = - 1, 0, + 1$
  - $m_{s} = \pm \frac{1}{2}$
  - there are 8 energy states
states with the same $n$ have the same energy
all possible states need to be included
this gets done using 'statistical weights', $g$ , such that $g_{A}$ is the number of states with the energy, $E_{A}$
$\frac{P_{1}}{P_{2}}$ is for a given energy, not a given state, so accounting for states, the probability ratio becomes:

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

a stellar atmosphere has a large number of atoms, $N$
the probability ratio is equivalent to the number ratios of atoms in each state, therefore, the boltzmann equation is obtained:

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

considering the atoms of a single element in a specified state of ionization with number of electrons, $N_{x}$ , at an energy level, $E_{x}$