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Special Relativity, AP French 6.10: Atom Emits Photon

  1. Dec 12, 2013 #1
    1. The problem statement, all variables and given/known data

    I'll copy down the exact phrasing, so there's no question that this is what we're being asked:

    "An atom in an excited state of energy Qo (as measured in its rest frame) above the ground state moves towards a scintillation counter with speed v. The atom decays to its ground state by emitting a photon of energy Q, as recorded by the counter, coming completely to rest as it does so. If the rest mass of the atom is M, show that Q = Qo[1 + Qo/2Mc2 ]

    Now, I wasn't sure if the relative motion between the emitter (atom) and observer (scintillation counter) might have some impact on the energy measured by the counter, or - if we are meant to take account of that - how exactly to do so.


    2. Relevant equations

    E2 = (pc)2 + (m0c2)2 -- can be applied to the atom before and after

    Q = pc
    for the photon

    E = γm0c2 --- relativistic energy for a particle of rest mass m0


    3. The attempt at a solution

    My first thought was to use conservation of energy and momentum in the rest frame of the atom.

    This gives us energy before = m0c2 + Q0 as the atom is at an energy Q0 above the ground state as measured in its rest frame.

    Energy after = γm0c2 + Q - that is the relativistic energy of the atom, which recoils at a velocity v in its rest frame (because it is at rest in the lab frame) added to the energy of the photon it emits.

    We can also use conservation of momentum; this is a zero-momentum frame, so the total momentum has to be zero before and after the photon emission.

    Therefore Q/c (photon momentum) + p (atom momentum) = 0

    I tried to combine these to give the following expression:

    m0c2 + Q0 = Q + [(pc)2 + (m0c2)2]0.5

    But this didn't lead to the required expression, as I ended up with some unwanted cross terms QQ0.

    This is one of about five or six attempts along the same lines, and another unsuccessful attempt in the Lab Frame, and I'm beginning to think I must be missing something: every time, I can get an expression fairly close to what we're being required to show, but with some additional, unwanted cross terms.

    I'd appreciate any guidance! Thank you :)
     
  2. jcsd
  3. Dec 12, 2013 #2

    tiny-tim

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    hi raving_lunatic! welcome to pf! :smile:
    no, Q (of the photon) is measured in the rest-frame of the atom, so you'll need to adjust it :wink:
     
  4. Dec 12, 2013 #3
    I'm a little confused. Are you saying that Q is measured in the lab frame by the counter, and so when we're dealing with it in the atom's rest frame, we need to transform it using the Lorentz transformation?

    I.e

    Q' = γ(1-v/c)Q?

    So we'd then have:

    m0c2 + Q0 = γm0c2 + γ(1-v/c)Q
     
  5. Dec 12, 2013 #4

    tiny-tim

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    hi raving_lunatic! :smile:
    yes (what is worrying you about that? :confused:)

    the question gives you measurements in two different frames, so you have to convert all to the same frame

    (and similarly for the momentum equation!)
     
  6. Dec 12, 2013 #5
    I think I have it now! Thank you for your help
     
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