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Special relativity and quantum mech

  1. Nov 19, 2013 #1
    An atom can exist in two states , ground state of M and excited state of mass M+Δ , it goes to excited state by absorption of an photon, what is the photon frequency in the in the lab frame of reference ( where the atom is initially at rest )?

    A simple energy equation gives the answer:
    E=( M+Δ -M)[itex]c^{2}[/itex]=hγ
    γ=Δ[itex]c^{2}[/itex]/h
    But my intuition says that I am overlooking something and can't be that simple..
     
  2. jcsd
  3. Nov 20, 2013 #2

    PAllen

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    It is that simple.
     
  4. Nov 20, 2013 #3

    Meir Achuz

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    You should include the recoil of the excited atom in the final state.
    Use Energy^2=(M+K)^2=(M+Delta)^2+K^2, and solve for K (the photon energy). (with c=1)
     
  5. Nov 20, 2013 #4

    PAllen

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    Yes, I ignored recoil because its contribution is small for common cases. I, of course, agree with the above formula. Note that even for as extreme a case as a 10 Mev delta for a hydrogen atom, including recoil only changes the required photon energy to about 10.05 Mev. For Sodium, to about 10.002 Mev. For delta of 1 Mev and hydrogen, including recoil would require a 1.0005 Mev photon.
     
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