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Balmer Wavelength

  1. Jun 7, 2014 #1
    I have been having trouble with this Balmer wavelength problem and was hoping I can get a little guidance.

    The question: Find the balmer wavelength (n=3 --> n=2) emitted from a hydrogen-like Fe atom (z=26)

    The answer is supposed to be 0.971 nm

    My attempt:

    1/lambda = R(1/n'^2 - 1/n2)
    Since this is a hydrogen-like atom, it should be straight forward

    I get lambda = 36/5*R which comes out to be 654 nm

    And if I use one of Bohrs equations: E = -Z^2(R/n^2) and E3 = E2+ (hc/lambda) I get a value of 1200 nm
  2. jcsd
  3. Jun 7, 2014 #2
    I figured it out, my book did not give me this equation for hydrogen-like atoms:

    1/lambda = R*Z^2(1/n'^2 - 1/n^2) this gives me 0.97 nm

    Thanks wikipedia!
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