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Deuteron Wavefunction Question

  1. Jan 7, 2009 #1
    1. The problem statement, all variables and given/known data
    This is the first question from a past exam paper I'm doing at the moment, and I'm not sure if it's a case that I'm doing something stupid, or if there is a problem with the question.

    Q: The wavefunction of a deuteron can be approximated by:
    [tex] \psi (r) = \frac{C}{r} e^{-\alpha r} [/tex]
    Where [itex] \alpha = 0.23 fm^{-1}[/itex]Calculate the Normalisation Constant, C.

    What is the probability that the separation of the two nucleons in the deuteron exceeds 2 fm, and what is their average separation.

    3. The attempt at a solution
    To work out C, I did:
    [tex] \psi^{*}(r) = \frac{C}{r}e^{-\alpha r} [/tex]
    [tex] C^2 \int^{\infty}_{0} r^{-2} e^{-2 \alpha r} = 1 [/tex]

    But if you work out this integral, it diverges, so what now?

    As for the other part(s), am I right in thinking that you do something along the lines of:
    [tex] P(r>2fm) = \int^{\infty}_{2*10^{-15}} \psi(r) \psi^{*}(r) dr [/tex]
    and
    [tex] <r> = \int^{\infty}_{0} \psi(r) r \psi^{*}(r) dr [/itex]
     
  2. jcsd
  3. Jan 8, 2009 #2
    I think you shloud integrate over the whole space, so [tex]d^3x =r^2\sin\theta d\phi d\theta dr [/tex] because of the spherical geometry. However, in your case it should be [tex]r^2 dr[/tex] instead of [tex]dr[/tex] due to the radial part. I get [tex]C=\sqrt{2 \alpha}[/tex] and [tex] <r> = \frac{1}{2 \alpha}[/tex], but you should calcuate these results because I did the operations quickly.

    I hope this helps.
     
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