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Electrons bombard gaseous hydrogen - photon energies emitted?

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data

    A beam of 13.0 eV electrons is used to bombard gaseous hydrogen in ground state.
    What photon energies will be emitted?


    2. Relevant equations

    ΔE = -13.6 ([itex]\frac{1}{n^{2}_{f}}[/itex] - [itex]\frac{1}{n^{2}_{i}}[/itex])

    However hydrogen is in the ground state therefore n_i= 1:

    ΔE = -13.6 ([itex]\frac{1}{n^{2}_{f}}[/itex] - 1)


    3. The attempt at a solution

    Rearranging the above equation I got n_f = 4.76.

    Am I correct in saying "the photon can be emitted from a maximum energy state of n=4"?
    I'm unsure how to word it.

    WRT the initial question, it asks what photon energies will be emitted so would I have to write them all down?

    ie

    4[itex]\rightarrow[/itex]3 = 0.66 ev
    4[itex]\rightarrow[/itex]2 = 2.55 ev
    4[itex]\rightarrow[/itex]1 = 12.75 ev

    3[itex]\rightarrow[/itex]2 = 1.89 ev
    3[itex]\rightarrow[/itex]1 = 12.09 ev

    2[itex]\rightarrow[/itex]1 = 10.2 ev
     
  2. jcsd
  3. Mar 25, 2014 #2

    maajdl

    User Avatar
    Gold Member

    13 eV is enough to reach the n=3 level which has an energy of 12.09 eV above ground state.
    13 eV is enough to reach the n=4 level which has an energy of 12.75 eV above ground state.
    13 eV is not enough to reach the n=5 level which has an energy of 13.056 eV above ground state.

    Yes.
    As you saw, there are not so many lines.
    You might present them in a level diagram, or in a table, if you like to present things ...
     
  4. Mar 25, 2014 #3
    Thanks maajdl
     
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