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Initial and final state of hydrogen atom

by physgirl
Tags: atom, final, hydrogen, initial, state
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Jul14-07, 10:30 PM
P: 99
1. The problem statement, all variables and given/known data
given the wavelength of photon absorbed by H atom and wavelength of photon emitted by that H, find the final "n" state of H atom.

2. Relevant equations

3. The attempt at a solution
I tried... n1hf1=n2hf2
where n1=1
so: f1=n2f2

and I get an integer value for n. But how is that the right way? Because I thought n was the number of photons... so it doesn't make sense for me to be solving for n :( I tried using the Rydberg equation, first plugging in the first lambda given to figure out what level of energy H atom initially got excited to but I didnt get an integer...
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Jul14-07, 11:15 PM
HW Helper
G01's Avatar
P: 2,685

That equation is important, but I think you're a little confused. The "n" in that equation does stand for the number of photons, so solving for it isn't going to help. Also, we are only dealing with one photon at a time, so n will always be one in this problem. The relationship you are using is incorrect for another reason as well. The energy of the first and second states would not be equal. This relationship also does not mention the third and final state, whose "n" (energy level) we are trying to find.

Try starting this way:

Start with the ground state energy, E_0. The atom then gains gain some energy, and then looses some energy. So, the final energy of the atom will be of the form:

[tex]E_{Final}= E_0 + E_1 - E_2[/tex]

Now what would E_0, E_1 and E_2 be?

After you have found these values, do you know a relationship for the energy of a hydrogen atom involving "n" (here the energy level)?
Jul15-07, 08:11 AM
P: 99
So okay. What I tried was... since E_0 is ground state, it is equal to 0. E1 and E2 would be hc/lambda with corresponding lambda values given in our problem. Then I did E_final=E0+E1-E2=E1-E2. And then I converted that E_final value to eV by multiplying by 6.241E18 eV/J.... and then set that equal to E=-13.6eV/n^2... I solved for n and got 1.154.... I'm supposed to be expecting an integer though, right...? :(

And actually, I get a positive value for my E_final, which also doesn't make sense in terms of plugging in numbers to that E=-13.6eV/n^2 formula...

Doc Al
Jul15-07, 08:37 AM
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P: 41,436
Initial and final state of hydrogen atom

Quote Quote by physgirl View Post
So okay. What I tried was... since E_0 is ground state, it is equal to 0.
The H atom starts in the ground state, n = 1, which has an energy of -13.6 eV.

Redo your calculation for E_final and compare to the Bohr model to find the principal quantum number for the final state.
Jul15-07, 08:42 AM
P: 99
Ohh, so I would do:

E_final=E1-E2-13.6eV which is also equal to -13.6eV/n^2

Doc Al
Jul15-07, 08:45 AM
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P: 41,436
That's right.
Jul15-07, 08:46 AM
P: 99
Thank you so much!!!
Sep15-07, 05:08 PM
P: 1
would someone mind working out this problem a little more thoroughly? I have a problem that's pretty much just like it and I'm stuck too. Maybe if I saw it worked out a little further i would get "unstuck" Am I supposed to use the Rydberg formula to work out the intermediate energy states?

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