Photon Energies of Hydrogen in the n=6 State

[Jadehaan]

Homework Statement

A hydrogen atom is in the n=6 state.

a) Counting all possible paths, how many different photon energies can be emitted if the atom ends up in the ground state?

b) Suppose only$$\Delta$$n=1 transitions were allowed. How many different photon energies would be emitted?

c)How many different photon energies would occur in a Thomson-model hydrogen atom?

Homework Equations

The Attempt at a Solution

My attempts:

a) 11 different photon energies.

b) 5 different photon energies.

c) I do not understand what they are referring to.

 

[vorcil]

Remember that the absorbtion spectrum is all the wave lengths that can leave from the n=1 state to 6

where the emission spectrum is all the possible wave lengths from 6-1, 5-1, 4-1 ect..

use E=hf and v=f(lambda) to find the energies from the wave lengths

At least I think that's how you'd do it,

http://en.wikipedia.org/wiki/Emission_spectrum

 

[gabbagabbahey]

My attempts:

a) 11 different photon energies.

This doesn't look right, but I can't tell you where you've gone wrong if you don't show your work/reasoning. An answer alone does not qualify as a solution

b) 5 different photon energies.

Good.

c) I do not understand what they are referring to.

Surely your text/notes have some information on the Thomson-model Hydrogen atom?

 

[Jadehaan]

For a) I got the possible paths to be
1. from 6 to 1
2. from 6 to 2 to 1
3. from 6 to 3 to 1
4. from 6 to 3 to 2 to 1
5. from 6 to 4 to 1
6. from 6 to 4 to 3 to 1
7. from 6 to 4 to 3 to 2 to 1
8. from 6 to 4 to 2 to 1
9. from 6 to 5 to 1
10. from 6 to 5 to 2 to 1
11. from 6 to 5 to 3 to 1
12. from 6 to 5 to 3 to 2 to 1
13. from 6 to 5 to 4 to 1
14. from 6 to 5 to 4 to 3 to 2 to 1
15. from 6 to 5 to 4 to 3 to 1
16. from 6 to 5 to 4 to 2 to 1

Can I safely assume that these will produce all different photon energies for a total of 16 paths?

c) The Thomson model of a hydrogen atom would have an uniform positive charge. Do I use this to eliminate the Z from En=-(13.6 eV)(Z^2)/(n^2)?

 

[Jadehaan]

Ok I see now that there is only 1 photon energy from 6 to 1. Since the Thomson model did not take this into account does that mean there are 0 photon energies for that model?

 

[gabbagabbahey]

For a) I got the possible paths to be
1. from 6 to 1
2. from 6 to 2 to 1
3. from 6 to 3 to 1
4. from 6 to 3 to 2 to 1
5. from 6 to 4 to 1
6. from 6 to 4 to 3 to 1
7. from 6 to 4 to 3 to 2 to 1
8. from 6 to 4 to 2 to 1
9. from 6 to 5 to 1
10. from 6 to 5 to 2 to 1
11. from 6 to 5 to 3 to 1
12. from 6 to 5 to 3 to 2 to 1
13. from 6 to 5 to 4 to 1
14. from 6 to 5 to 4 to 3 to 2 to 1
15. from 6 to 5 to 4 to 3 to 1
16. from 6 to 5 to 4 to 2 to 1

Can I safely assume that these will produce all different photon energies for a total of 16 paths?

There are 16 possible paths (not counting the paths where the electron temporarily jumps back up a level or 2 before falling again). The question is, "how many unique transitions are there?". Each unique transition (eq. n=2 to n=1 or n=6 to n=3) gives rise to a photon of a different energy/wavelength, while each path includes several transitions, each of which will emit a photon of a certain energy.

c) The Thomson model of a hydrogen atom would have an uniform positive charge. Do I use this to eliminate the Z from En=-(13.6 eV)(Z^2)/(n^2)?

No, that equation is derived from the Bohr model.

In the Thomson model, the frequency of an emitted photon corresponds to the frequency of the electron's orbit. There is only one allowed orbit for Thomson model of Hydrogen, so only one spectral line occurs. In fact, there is no n=6 state at all, so I would say the transition is undefined and leave it at that.