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Quick Normalisation Question

  • Thread starter White Ink
  • Start date
19
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1. Homework Statement
A hydrogen atom in the ground state can be described by the following wavefunction:

[tex]\Psi(r) = \frac{C}{\sqrt{4\pi}}e^{- \frac{r}{a_{0}}}[/tex]

Normalise this wavefunction.

3. The Attempt at a Solution

I did this and got:

[tex]C = \sqrt{\frac{8\pi}{a_{0}}}[/tex]

I have no way of checking this, so I was wondering if anybody could tell me whether I am right or wrong.
 
Last edited:

Answers and Replies

Dick
Science Advisor
Homework Helper
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Not quite, I don't think. Can you show us how you got that?
 
19
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[tex]\int^{\infty}_{0}\left|\Psi(r)\right|^{2}dr = 1[/tex]


[tex]\frac{C^{2}}{4\pi}\int^{\infty}_{0}e^{\frac{-2r}{a_{0}}}dr = 1[/tex]


[tex] u = \frac{2r}{a_{0}}[/tex]

[tex] du = \frac{2}{a_{0}}dr[/tex]

[tex] dr = \frac{a_{0}}{2}du[/tex]


[tex]\frac{C^{2}}{4\pi}\frac{a_{0}}{2}\int^{\infty}_{0}e^{-u}du = 1[/tex]

Using:

[tex]\int^{\infty}_{0}x^{n}e^{-x}dx = n![/tex]

[tex]\int^{\infty}_{0}e^{-u}du = 0! = 1[/tex]

Thus:

[tex]\frac{C^{2}}{4\pi}\frac{a_{0}}{2} = 1[/tex]

[tex]C^{2} = \frac{8\pi}{a_{0}}[/tex]
 
82
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Think again about the first integral. Even if the wavefunction is only dependent on r, you are still integrating over three-dimensional space. Use the appropriate volume differential for this case (spherical coordinates).

You will notice the equation will reduce to an integral in just r, but it'll have a key difference with the one you're using right now.
 
19
0
Think again about the first integral. Even if the wavefunction is only dependent on r, you are still integrating over three-dimensional space. Use the appropriate volume differential for this case (spherical coordinates).

You will notice the equation will reduce to an integral in just r, but it'll have a key difference with the one you're using right now.
I know exactly what you mean (integrating in the [tex]\theta[/tex] & [tex]\phi[/tex] directions becomes equivalent to multiplying to constant in front of the integration with respect to r by [tex]2\pi^{2}[/tex], since the wavefunction is independent of those two variables). I read over the question too quickly. Thank you for your help.
 
Dick
Science Advisor
Homework Helper
26,258
618
I know exactly what you mean (integrating in the [tex]\theta[/tex] & [tex]\phi[/tex] directions becomes equivalent to multiplying to constant in front of the integration with respect to r by [tex]2\pi^{2}[/tex], since the wavefunction is independent of those two variables). I read over the question too quickly. Thank you for your help.
In addition there will be an extra factor of r^2 in the integrand, right?
 
19
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In addition there will be an extra factor of r^2 in the integrand, right?
Now that one beats me. I can't see where an additional [tex]r^{2}[/tex] would come from.
 
Dick
Science Advisor
Homework Helper
26,258
618
dV in spherical coordinates is r^2*sin(theta)*dr*dtheta*dphi. Better look that up to make sure my use of angle names agrees with yours. And you won't get a pi^2 from the integration, do it carefully.
 
19
0
Ooops, I feel stupid now. I forgot about the metric coefficients and took my volume element to be [tex]dr[/tex][tex]d\theta[/tex] . At least I won't be making that mistake again. Thanks.
 

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