What fraction of hydrogen atoms is in the state with the principle quantum number `n=2` at a temperature `T=3000K`?

Explanation

By Boltzmann formula
`(N_(2))/(N_(1))=(g_(2))/(g_(1)) e^(DeltaE//kT)`.
Here `DeltaE=` energy difference between `n=1` and `n=2` states
`=13.6(1-(1)/(4))eV=10.22eV`
`g_(1)=2` and `g_(2)=8` (counting `2s` & `2P` states).Thus
`(N_(2))/(N_(1))=4e^(-10.22xx1.602xx10^(-19)//1.38xx10^(-23)xx3000)=2.7xx10^(-17)`
Explicity `eta=(N_(2))/(N_(1))=n^(2)e^(-DeltaE//kT),DeltaE_(n)= ħR(1-(1)/(n^(2)))`
for the nth excited state beacuse teh degeneracy of the state with principle quantum number `n is 2n^(2)`.