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Photoelectric effect and Rydberg's formula

  1. Sep 29, 2009 #1
    A peak in the energy spectrum is seen at 134.2 keV when using a germanium detector. In coincidence with this, an x-ray corresponding to the 2s to 1s electronic transition in germanium is detected. Find the energy (in keV) of the gamma ray that ejected the electron.

    Use Rydberg's formula for spectral lines of hydrogenic atoms:

    1/lambda = R Z2 ( ( 1/n12) - ( 1/n22) )

    where R, the Rydberg constant, is such that Rhc = 13.61 eV
    (c is the speed of light, h is Planck's constant).

    not really sure where to begin, do I solve for the wavelength and use that in the energy equation E=hc/lambda??
  2. jcsd
  3. Sep 29, 2009 #2


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    You might want to substitute the above equation into the equation for energy that you mentioned. Especiay since you are given R in terms of Rhc
  4. Sep 29, 2009 #3
    nevermind..i got it...thanks!
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