Question

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

Answer 1

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:

`bb(DeltaE = -R_H(1/n_f^2 - 1/n_i^2))`

where `R_H` is the Rydberg constant, and is actually the magnitude of the ground-state energy of the hydrogen atom, `|-"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`:

`color(blue)(DeltaE) = -"13.61 eV"(1/(2)^2 - 1/(1)^2)`

`=` `color(blue)("10.21 eV")`

Or, in possibly more familiar units...

`color(blue)(DeltaE) = 10.21 cancel"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"`

`=` `color(blue)("984.77 kJ/mol H atom")`