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I Hydrogen electron transition intensity 2p - 1s

  1. Nov 27, 2016 #1
    Let's say we have a transition from state 2p to 1s of an hydrogen atom's electron. The intensity radiated by the electron is given by I = 4/3 ωfi4 e2/c3 |< f | r | i >|2.

    If we take the the | 2 1 1 > → | 1 0 0 > transition for example, we must compute the following integral :

    < f | r | i > = C1 ∫∫∫ e-r/a r sin(θ) ei*Φ * e-r/2a sin(theta) r2 sin(θ)dr dθ dΦ

    , r = [ 0→ ∞], θ = [0 → pi], Φ = [0 → 2π]

    Which can be separated in three integrals (depending on r, θ and Φ respectively). The integral in Φ is

    IΦ = ∫ ei*Φ dΦ = 0

    , Φ = [0 → 2π]

    Which means that < f | r | i > = 0 and that the transition is not allowed.


    However in the literature (and logically), this transition is allowed.

    I'm not quite sure if it's a comprehension issue or a simple computation error.

    Thank you !
     
  2. jcsd
  3. Nov 27, 2016 #2

    DrClaude

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    Staff: Mentor

    You forgot the polarization of the photon. The term in the middle of the integral is not simply r, but ##\mathbf{r} \cdot \mathbf{\epsilon}##. That will add an angular compnent, which can be expressed in terms of ##Y_{1,0}##, ##Y_{1,+1}##, or ##Y_{1,-1}##. The angular part is then the integral of three spherical harmonics.
     
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