What Is the Normalisation Constant for a Deuteron Wavefunction?

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SUMMARY

The normalization constant for the deuteron wavefunction, given by the equation ψ(r) = C/r * e^(-αr) with α = 0.23 fm-1, is calculated using the integral C² ∫0 r-2 e-2αr dr = 1. This integral diverges, indicating a need for a different approach to determine C. The probability that the separation of the two nucleons exceeds 2 fm is calculated using P(r > 2 fm) = ∫2*10-15 ψ(r)ψ*(r) dr, and the average separation is found using = ∫0 ψ(r) r ψ*(r) dr, with the correct integration over spherical coordinates.

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Homework Statement


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.

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][/tex]
 
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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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