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Homework Help: Determine energy levels of a electron in a hydrogen atom

  1. Dec 5, 2017 #1
    1. The problem statement, all variables and given/known data
    When an electron in a hydrogen atom makes a transiton between two levels with prinicipal quantum numbers n1 and n2, light is emitted with wavelength of 658.1 nm. If we assume that the energy levels of the atom are in agreement with the Bohr model, what are n1 and n2?

    2. Relevant equations

    3. The attempt at a solution
    I tried to calculate:
    E = hc/λ = 3.022 * 10-19
    which is the Energy for the wavelength.

    E = -hcR/n^2

    n^2 = -hcR/E
    n^2 = -R/λ
    which result in negative value(?)

    Then I tried with
    λ = (n1^2 - n2^2)/R

    but then I always get two unknowns, and cant really come any further than that. Maybe Im looking in the wrong way or missing some theory.
    Last edited by a moderator: Dec 5, 2017
  2. jcsd
  3. Dec 5, 2017 #2


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    First, get straight what the story is of how the photon is produced:

    1. The electron originally has energy [itex]E_1[/itex] (which can be written in terms of the quantum number [itex]n_1[/itex])
    2. It emits a photon of energy [itex]E[/itex] (which you calculate above).
    3. Afterward, the electron has energy [itex]E_2[/itex] (which can be written in terms of [itex]n_2[/itex])
    So by conservation of energy, what is the relationship between [itex]E_1, E_2[/itex] and [itex]E[/itex]?

    You're right, you will have two unknowns, [itex]n_1[/itex] and [itex]n_2[/itex]. But they are positive integers. So you can just try various values for [itex]n_1[/itex] and [itex]n_2[/itex] to see which combination works.
  4. Dec 5, 2017 #3


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    You use wrong formula. The wavelength is attributed to two levels, according to the Rydberg formula.
    where Z is the atomic number (1 for Hydrogen atom) and n , n' are the energy levels, n'<n. R is the Rydberg constant. (RH = 0.010972 nm-1)
  5. Dec 5, 2017 #4

    rude man

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    I tried adjacent orbits - no luck!
  6. Dec 5, 2017 #5


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    Don't expect the equation is exactly valid for any pair of n and n', but the emitted light is visible, what can be n'?
  7. Dec 5, 2017 #6

    rude man

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    Once n is picked, n' is easy to determine - but how to pick n other than by trial and error?
  8. Dec 6, 2017 #7


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    You know the wavelength ranges of the emission lines of Hydrogen atom. They are well separated according to the lower level. Emission in the visible range corresponds to the Balmer series, with n'=2 as lower level.
    Last edited: Dec 6, 2017
  9. Dec 6, 2017 #8


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    I think this solution is the simplest. If you are learning about electron transitions, then you must surely have learnt about the different series. Either you can check your textbook (or online) for the energy values of the different transitions from different "n" states.

    The value you calculate, E, is simply the difference between energy states. If you still don't understand what the value of E really is, take a look at this link:
  10. Apr 6, 2018 #9
    Re search: something like this perhaps. Spot the 'candidate' !

    ## \begin{array}{|c|c|c|c|c|c|}
    \hline & m & 1 & 2 & 3 & 4 & 5 \\
    \hline n \\
    \hline 1 & & & 0.750 & 0.889 & 0.938 & 0.960\\
    \hline 2 & & & & 0.139 & 0.188 & 0.210\\
    \hline 3 & & & & &0.049 & 0.071\\
    \hline 4 & & & & & & 0.023\\
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