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Energy of photons from Balmer series

  1. Mar 7, 2013 #1
    2ltn7r5.png

    I can see how I would do part b, but I am confused as to why I have got part a wrong. Apparently the answer is 12.09eV, but I cannot see how. It seems to me that using the formula as I did should give the answer. Thanks for any help.
     
  2. jcsd
  3. Mar 7, 2013 #2
    You've worked out the energy of the photon emitted going from n=3 to n=2.
    It doesn't ask for this.
    It wants the energy required to go from the ground level to to n=3 ( ie to the level where it can emit the Hα photon.
     
  4. Mar 7, 2013 #3
    But if I do [itex]\frac{1}{\lambda_\alpha} = R(\frac{1}{1^2} - \frac{1}{3^2})[/itex], I'm not using the Balmer series, am I? I thought it needed to be a [itex]2^2[/itex] for the Balmer series? If I use a 1, how can it be the [itex]H_\alpha[/itex] line?

    Unless you're saying the [itex]H_\alpha[/itex] is emitted during the transition from n=3 to n=2? Thus, it wants the energy to get to the n=3 in the first place, rather than the energy of the photon that would give the [itex]H_\alpha[/itex]?
     
  5. Mar 7, 2013 #4
    Yes (this agrees with the supplied answer.)
     
  6. Mar 7, 2013 #5
    Ahh, ok. This sorts out my confusion, then. Thank you for the help.
     
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