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Balmer Lines in Quantum Physics

  1. Jun 27, 2006 #1
    Hi again,
    Could someone explain a tiny bit about Balmer Lines to me?
    If I was asked what is the difference between the red Balmer Lines in Hydrogen and Deuterium, where would I begin?
    How do I know which energy level the electrons are in?
    If it's a red Balmer Line then I guess n = 2 for Hydrogen; but what does it equal for Deuterium?
    Also if [ En = -R / n squared ] then why is this true:

    1 / wavelength = exactly the same thing ???

    Thank you to anyone who can help,

  2. jcsd
  3. Jun 27, 2006 #2
    I forgot to say "red alpha Balmer Lines".
  4. Jun 27, 2006 #3


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    A given transition line is specified by providing *two* principal quantum numbers, [itex] n_i [/itex] and [itex] n_f [/itex]. It also depends if we are talking about an emission spectrum or an absorption spectrum. For an emission Balmer line, one has [itex] n_f =2[/itex] , [itex] n_i = 3,4,5...[/itex]. For the red line, [itex] n_i = 3 [/itex].

    For deuterium, the red Balmer line corresponds to the *same* quantum numbers! The only thing that changes is the reduced mass in the equation fo the energy.

    the equation for the inverse of the wavelength of the emitted photon is
    [itex] {1 \over \lambda} = R ( { 1 \over n_f^2 } - {1 \over n_i^2}) [/tex]

    but the equation for the *energy* is NOT R over n squared!!

  5. Jun 27, 2006 #4
    Thank you, I looked around some more and saw that when they say the 'Red Balmer Lines' that is like saying the 'H-alpha Balmer Lines'.
    The only thing I'm very unsure about is that in my notes it says: En = -R / n2
    This formula is right isn't it?
    This is confusing me a lot...
  6. Jun 27, 2006 #5


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    It's incorrect.

    [tex] R = {\mu \over 4 \pi c \hbar^3} ({e^2 \over 4 \pi \epsilon_0})^2 [/tex]
    where mu is the reduced mass (which is *slightly* different for deuterium and for hydrogen).

    whereas the energy levels of a hydrogen-like atom (one electron orbiting a central charge Ze) is

    [tex] - { \mu \over 2 \hbar^2} ({Z e^2 \over 4 \pi \epsilon_0})^2 {1 \over n^2} [/tex]

    With Z=1 and mu the reduced mass of hydrogen, this gives the usual -13.6 eV/n^2. This is NOT -R / n^2!!

    If you recall that [itex] E = hc/ \lambda = 2 \pi \hbar c / \lambda [/itex] then you will be able to get from the formula for the energy levels to the formula for the wavelength of the emitted photon (try it!). So if you want to write the equation for the enrgy levels in terms of R (for Z=1 let's say) you see that there is a factor of [itex] 2 \pi \hbar c [/itex] that will appear.

    Last edited: Jun 27, 2006
  7. Jun 27, 2006 #6
    Oh right. I think the R I'm using is R with a subscript infinity. Does that change anything?
  8. Jun 27, 2006 #7


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    No, I don't think so.

    I don't have my quantum mechanics books with me so I am not 100% sure but if I recall correctly, the infinity in [itex] R_\infty [/itex] refers to taking the nucleus mass to infinity. In that case the reduced mass becomes the mass of the electron. So the R value for a nucleus can be written as [tex] R = { \mu \over m_e} R_\infty[/tex].

    Again, this is from memory.

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