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Two times ionized Lithium

  1. Jan 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Two times ionized Lithium (Z=3) in excited states emits two photons, one with ##\lambda _1=72.91 nm## and another one with ##\lambda _2=13.5 nm##.
    In which excited state was originally the ion?


    2. Relevant equations



    3. The attempt at a solution

    Since two times ionized Lithium has only one electron I assume I can say that the energies of eigenstates are calculated as ##E_n=\frac{R_y}{n^2}##.

    I somehow imagined that I have to sum the energy of photons and equal that with the expression above.

    ##\frac{hc}{\lambda _1}+\frac{hc}{\lambda _2}+R_y=R_y(1-\frac{1}{n^2})##

    But this gives me some weird result...

    What do I have to do? :/
     
  2. jcsd
  3. Jan 28, 2014 #2

    DrClaude

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    Staff: Mentor

    That's not correct. The energy of one photon already corresponds to a transition between two levels: the difference between the initial excited state and the final state.

    Start again from the equation for the energy, but consider one transition at a time. After that you will have to combine both transitions to get the desired solution.
     
  4. Jan 28, 2014 #3
    Than first transition is:

    ##\frac{hc}{\lambda _1}=R_y(\frac{1}{(n')^2}-\frac{1}{(n)^2})##

    and second

    ##\frac{hc}{\lambda _2}=R_y(1-\frac{1}{(n')^2})##

    I hope that's exactly what you had in mind.

    However, to get rid of ##(n')^2## I woud simply sum the both equation.

    ##\frac{hc}{\lambda _2}+\frac{hc}{\lambda _1}=R_y(1-\frac{1}{n^2})##
     
  5. Jan 28, 2014 #4

    DrClaude

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    Staff: Mentor

    That looks ok. What do you get when you try to solve this for ##n##?

    (Note: the problem is not clearly stated, and I misread it the first time. This explains why I might have made you rederive the equation you had at the beginning, although now without the extra ##R_y##.)
     
  6. Jan 28, 2014 #5
    The next part confuses me the most...

    ##n^2=\left [1-\frac{hc}{R_y}(\frac{1}{\lambda _2}+\frac{1}{\lambda _1}) \right ]^{-1}##

    which gives me ##n=0.378##
     
  7. Jan 28, 2014 #6

    DrClaude

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    Staff: Mentor

    You're using the wrong Rydberg constant. It has to be scaled for the charge of the nucleus.
     
  8. Jan 28, 2014 #7
    ##R_y=13.6 eV##

    How do I do that?

    Lithum has 3 nucleus is than the right Rydberg constant multiplied by 3?
     
  9. Jan 28, 2014 #8

    DrClaude

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    Staff: Mentor

  10. Jan 28, 2014 #9
    Ok, that's good to now.

    So using ##Z^2R_y## gives me ##n=3##.

    Thank you!
     
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