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Calculate the wavelength of the n= 4 -> 3 transition in He

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the wavelength of the n = 4 → 3 transition in 4He+ to an accuracy of 4 significant figures. (R∞=109 737 cm-1.) (Fine structure effects can be neglected.)

    2. Relevant equations
    [itex] \frac{1}{λ} = \frac{m}{m_e} R_∞ (\frac{1}{n_1^2} - \frac{1}{n_2^2})[/itex]

    where λ is wavelength, m is the reduced mass, and [itex]R_∞[/itex] is Rydberg constant (1.1x105 cm-1)

    [itex] m = m_e \frac{m_N}{m_e + m_N} [/itex]

    where m is the reduced mass, [itex]m_e[/itex] is mass of an electron, and mN is the mass of the nucleus.

    3. The attempt at a solution
    In order to come out with a positive wavelength I set n1 to 3 and n2 to 4.

    Now the problem I have is with the reduced mass, m. To try and make sense of it I did everything in SI units:

    me = 9.11x10-31
    mN = 2 protons + 2 neutrons = 2x(1.673x10-27 kg) + 2x(1.675x10-27 kg) = 6.696x10-27 kg

    So that gives:

    m = me x [itex]\frac{6.696x10^-27}{9.11x10-31 + 6.696x10^-27}[/itex] = 0.9999me

    Plugging all the numbers in gives:

    [itex]\frac{1}{λ}[/itex] = 5347 cm-1

    ∴ λ = 1.870x10-4 cm = 1.870x10-6 m = 1870 nm


    However the answer is 468.7 nm which means the reduced mass m should equal 3.99 me.

    It's only a 1 mark question so I am obviously understanding it all wrong somewhere.

    Any help would be very much appreciated.
     
  2. jcsd
  3. Jan 10, 2015 #2
    You are dealing with helium and not hydrogen. Helium has a nuclear charge of 2. You need to take that into account in modifying your Rydberg formula.
     
  4. Jan 10, 2015 #3
    I had no idea the equation would differ. Just looked up 'Rydberg formula' and a Z2 appears. Big thanks, I was beginning to pull my hair out.
     
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