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Bohr Model Question

  1. Aug 16, 2006 #1
    Hi,

    I'm having trouble on the following question:
    A singly ionized helium atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The ion returns to the ground state by emitting two photons. What are the wavelengths of the lowest energy photon and highest energy photon?

    I think I have to use the conservation of energy approach where nf=1 and ni=3, and also account for the energy lost to the photon. But, I'm lost of how to actually do this, and how to differentiate between the lowest energy photon and highest energy photon (I figure it has something to do with the length of the wavelength - longer wavelength leading to less energy, and vice-versa).

    Thanks for your help!
     
  2. jcsd
  3. Aug 16, 2006 #2
    I think that problem suggests that the electron returns to the n = 1
    state via the n = 2 state. So when the electron goes from the n = 3
    state to the n = 2 state, it emits 1 photon then emits another photon
    going from the n = 2 state to the n = 1 state.
     
  4. Aug 16, 2006 #3
    what is the energy difference between level 3 and level 2? level 2 and level 1? wavelength=E/hc
     
  5. Aug 17, 2006 #4
    The wavelength will be very close to 1/4 of that of the corresponding
    transitions for hydrogen due to the Z^2 factor. For hydrogen the wavelengths in Angstroms are: N2 to N1 - 1215.66
    N3 to N2 - 6562.79

    The wavelength for He (He+) N2 to N1 - 303.777 Angstroms

    The general formula for hydrogen-like ions is wavenumber (1/wavelength in centimeters) =

    RZ^2 * (1/n2 ^2 - 1/n1^2) where n2 is the principal quantum
    number of the lower state and n1 the principal quantum
    number of the upper state.
    R is the Rydberg constant and for He+ = 109722.263 (1/cm)
    for H = 109677.581
    the difference being in the reduced mass of the nucleus-electron
    system.
     
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