Figuring out which energy shell the electron was in

[Axelneo]

I've gotten this problem of mine, which I cannot figure out myself, so I thought anyone here might be able to explain how it should be solved. The problem goes as follows:

"When a hydrogen atom deexcitates to its ground state it emits an photon with the energy of 12.75 eV, what energy state did the atom have before the deexcitation?"

 

I have tired and failed, the answer sould be that the atom was in its forth energy state, I would appreciate if someone explained how to approach such a problem and how to solve it, thanks.

[Enthalpy]

Hi Axelneo, welcome here!

 

- The photon carries the energy difference between two states. You know the final is the ground state.

- You know (or learn it) the energy of hydrogen's ground state. It's called one rydberg. It's also the ionization energy of an isolated hydrogen atom, not molecule.

- You're also supposed to know how energy levels go in the hydrogen atom. There is a simple law with integer numbers.

- Comparing the photon energy with one rydberg, you can tell what the initial energy was, in rydberg units. Then consider the law with integer numbers, it tells you the initial level.

[Axelneo]

Thanks for the help and for the welcoming, I did solve the problem and did it this way; I used Rydbergs formula, in which i put in energy as E=hc/lambda (multiplied both sides by hc) and after that it was pretty straight forward, just to plug in all of my values. Once again thanks.