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Homework Help: Spectroscopy analysis

  1. May 16, 2004 #1
    In a lab, our class looked at a hydrogen lamp and calculated the wavelengths of the color red, blue and green. I needed to verify that only Balmer lines are visible using the equation:

    1/lambda=R(1/m^2)-(1/n^2)

    It says that R is Rydbergs constant and I need to find n. It says that m is the proper value for the Balmer series. What is m supposed to be?
     
  2. jcsd
  3. May 16, 2004 #2

    Doc Al

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    Staff: Mentor

    Balmer series: m = 2

    The emission spectrum for the hydrogen atom is given by:
    [tex]\frac{1}{\lambda} = R(\frac{1}{m^2} - \frac{1}{n^2})[/tex]
    This represents wavelengths emitted when the atom transitions from a higher state (n) to a lower state (m). (n > m) When the lower state is m = 2, the spectrum is called the Balmer series. It has several wavelengths in the visible range.

    Several other spectral series have names (after the experimentalists who discovered them). For example: The Lyman series (ultraviolet) come from transitions to the ground state (m = 1) and the Paschen series (infrared) has m = 3.
     
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