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Initial energy level

  1. Sep 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Using the energies, determine from which initial energy level (n) the electrons decayed when it emitted photons at each of these wavelengths.

    2. Relevant equations
    [itex]E = - \frac{ me^4 }{ 8\epsilon_0^2h^2 }\frac{ 1 }{ n^2 }[/itex]

    [itex]Delta E = E_f-E_i= - \frac{ me^4 }{ 8\epsilon_0^2h^2 }\left( \frac{ 1 }{ n^2_f}-\frac{ 1 }{ n^2_i } \right)[/itex]

    3. The attempt at a solution
    I have the energy and wavelengths for this question but I'm not exactly sure what formula to use, this is for an hydrogen atom, so that's where my question lies but I have two equations (above)I just need guidance on which one to use.
     
  2. jcsd
  3. Sep 28, 2014 #2
    I was thinking to use the second equation but I don't have the the final energy I just have energy which is delta E.
     
  4. Sep 28, 2014 #3
    Or also, would rydberg formula work, where nf is 2 as it's the hydrogen atom?
     
  5. Sep 29, 2014 #4

    ehild

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    ni and nf are positive integers. Find an ni so as the given delta E-s correspond to positive integers less than ni.

    ehild
     
    Last edited: Sep 29, 2014
  6. Sep 29, 2014 #5
    Sorry, I don't quite understand what you mean? Could you interpret in different terms.

    What I think you're saying...

    [itex]E = - (13.6eV)\left( \frac{ 1 }{ n^2_f }-\frac{ 1 }{ n^2_i } \right)[/itex] so just solve for [itex]n_i[/itex]?
     
  7. Sep 29, 2014 #6

    ehild

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    What are the delta E-s? How many are given?

    You need to try different ni-s so as the delta E values can be fit to the equation. In case ni=2 there is only one possibility to decay, from the excited level 2--->1.
     
  8. Sep 29, 2014 #7
    Well for each wavelength I had, I had to calculate the energy of the photons in eV. So that's what I'm assuming are my delta E's. So for 1 I have E=2.1eV. So for the equation above, I would just put random n's for both [itex]n_f[/itex] and [itex]n_i[/itex] where n is 1,2,3..etc, and see if those correspond to the following E?

    Thanks for you help btw, really appreciate it.
     
  9. Sep 29, 2014 #8

    ehild

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    What are the other delta E-s and how many are given? Is 2.1 eV the smallest or the greatest?
     
  10. Sep 29, 2014 #9
    Three, and they're all similar, one is 2.4, the other 2.1, and 2.3. It's also the resulting wavelengths were all similar as well that's why the results are pretty weird.
     
  11. Sep 29, 2014 #10

    ehild

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    Can you cite the whole original problem text, please?
     
  12. Sep 29, 2014 #11
    It's part of a lab, so, just this was the part troubling me I did everything else.

    Hydrogen Spectrum

    Use the difference between these angles and the central maximum to determine the wavelengths of these colors. Use the grating line separation that you determined with the Sodium Spectrum.
    Which I used [itex]|right angle - central maximum|+|left angle-central maximum|[/itex]
    Using this value for theta I calculated the wavelength, [itex]lambda = \frac{ d sin theta }{ m }[/itex] m=1

    Then for each wavelengths, I had to calculate the energy of the photons in eV. Which I used
    [itex]\frac{ 1.24 times 10^-6 eV }{ lambda times 10^{-9} }[/itex]

    there are only three colours we used, so this the part which was sort of confusing for me.

    Now using these energies, determine from which initial energy level (n) the electron decayed when it emitted photons at each of these wavelengths.

    Energies: 2.1, 2.2, 2.3 eV
     
  13. Sep 30, 2014 #12

    ehild

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    You should have used the average deflection: half of the angle above. If you used the same formula when determining the grating parameter, it was wrong, too.

    As the angles were wrong, the energies are not correct.
    The hydrogen spectrum consists of 4 lines, 656.3 nm, 486.1 nm, 434.0 nm, 410.2 nm. a the last two lines are very close, the resolution of your set-up was not enough to show them as different lines. And the visible range corresponds to the Balmer series, so you know the final n.

    ehild
     
  14. Sep 30, 2014 #13
    ignore.
     
    Last edited: Sep 30, 2014
  15. Sep 30, 2014 #14
    Why is the formula wrong for the grating parameter, I was only given that formula for it unless you mean the angle I used to calculate d was wrong, but there was not much I could do for that as it was very close to the central maximum angle. And my first thought was to use the average of the angles I mentioned on the previous post but it didn't instruct me to do so, but I think I will do that anyways as I get a more clear result.

    I took the average of the angles and got different energies, being 3, 2.5, 2.

    And yes the final n is 2, so I just calculate the initial n now right? Using the formula I mentioned above.
     
  16. Sep 30, 2014 #15

    ehild

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    So the grating parameter was given? Then it must be correct. What was it?
    What were the angles? How did you measure the angles?
    The energies seem a bit inaccurate. What were the wavelengths ?
    Yes, the final level is n=2. Just calculate the initial ones.

    ehild
     
  17. Sep 30, 2014 #16
    The grating parameter wasn't given I had to calculate that myself, using the lambda = dsintheta/m but the central maximum was 80 degrees as was the average of the angles. Lambda was 589.3 nm for the grating line used to measure grating parameter. The measured wavelengths weren't very good haha, I know the experiment failed but I won't try to edit it as that would be bad science! But the wavelengths measured were 560nm, 600nm, 600nm.

    Once again, tyvm for the help.
     
    Last edited: Sep 30, 2014
  18. Sep 30, 2014 #17

    ehild

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    The energies you wrote are acceptable. What you get for the initial n? Round them to the nearest integer.

    ehild
     
  19. Sep 30, 2014 #18
    Alright so I got,
    (Rounded energy levels)
    3eV - 1
    2.5 eV - 2
    2 eV - 2
     
  20. Sep 30, 2014 #19

    ehild

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    NO! You have to get numbers greater than 2! The hydrogen atoms are excited in the lamp and they return to the level n=2 from higher levers when emitting a certain wavelength light.
    The formula is ##\Delta E = 13.6 (\frac{1}{2^2}-\frac{1}{n^2}) ##
     
  21. Sep 30, 2014 #20
    I just realized that haha, I found the mistake, I had -13.6 so that screwed up my whole calculation, thank you! So now, I have,

    3eV - 6
    2.5eV - 4
    2eV - 3

    Haha, thanks again ehild, you made this very fun for me xD.
     
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