# What photon energy could be absorbed by a hydrogen atom that is in the n = 2 state?

May 17, 2017

Well, this is up in the air if you mean $n = 2$ as the starting state... I assume you mean that $n = 2$ is the final state...

The Rydberg equation describes the energy level transitions in the hydrogen atom only:

$\boldsymbol{\Delta E = - {R}_{H} \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)}$

where ${R}_{H}$ is the Rydberg constant, and is actually the magnitude of the ground-state energy of the hydrogen atom, $| - \text{13.61 eV"| = "13.61 eV}$.

${n}_{f}$ and ${n}_{i}$ are the final and initial values of the principal quantum number $n$, respectively.

Therefore, since going from $n = 1$ to $n = 2$ is the only possible transition upwards to $n = 2$:

$\textcolor{b l u e}{\Delta E} = - \text{13.61 eV} \left(\frac{1}{2} ^ 2 - \frac{1}{1} ^ 2\right)$

$=$ $\textcolor{b l u e}{\text{10.21 eV}}$

Or, in possibly more familiar units...

$\textcolor{b l u e}{\Delta E} = 10.21 \cancel{\text{eV" xx (1.602 xx 10^(-19) cancel"J")/(cancel"1 eV") xx "1 kJ"/(1000 cancel"J") xx (6.0221413 xx 10^(23) "things")/"1 mol H atom}}$

$=$ $\textcolor{b l u e}{\text{984.77 kJ/mol H atom}}$