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Quick Question on the energy of a photon

  1. Apr 29, 2007 #1
    1. The problem statement, all variables and given/known data

    A free electron with negligible kinetic energy is captured by a stationary proton to form an excited state of the hydrogen atom. During this process a photon of energy Ea is emitted, followed shortly by another photon of energy 10.2 eV. No further photons are emitted. The ionization energy of hydrogen is 13.6 eV

    What is the energy Ea of the first photon?

    2. Relevant equations
    Ei - Ef = hf

    3. The attempt at a solution
    I am very... troubled by this step. I know that when the second photon is emitted, of energy 10.2 eV, the photon goes from the n = 1 state (ground state) to n=2 (first excited state) ... but how do I go about finding the energy Ea of the first photon? I believe it should have greater energy than the second one according to the equation above. I dont need an answer.. just something to ponder about because I'm rather stuck. The wavelength of the 10.2 eV photon is 121.875 nm if that helps.:uhh:
  2. jcsd
  3. Apr 29, 2007 #2


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    The first electron was unbound with negligible kinetic energy. It went from n=infinity to n=2, and the second went from n=2 to n=1. Why should the first emission have greater energy?
  4. Apr 29, 2007 #3
    I was maybe thinking the energy of an electron as it goes from one state to the next... but now I understand. Thank you so much ^_^
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