Hydrogen atom

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Homework Statement



I have a question on my quantum pset relating to calculating <p^2/2m> and <-e^2/r> for the first two spherically symmetric states of the hydrogen atom (in 3D).

The Attempt at a Solution



I started out trying to calculate the averages with [tex]\psi[/tex] ... something like, for the ground state, [tex]\psi = \frac{e^{-2r/a_0}}{\sqrt{\pi * a_0^3}}[/tex].

But then I ran into problems (when I was trying to do the <-e^2/r>) when I came up with an integral involving a term [tex]\frac{e^{-2r/a_0}}{r}[/tex]. As far as I could see, this integral sort of seems to explode at the origin / at infinity. I was wondering, should I use just the radial wavefunction part, since it has an extra r factor that would make this integral possible? I was just confused basically...
 

Answers and Replies

  • #2
nrqed
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Homework Statement



I have a question on my quantum pset relating to calculating <p^2/2m> and <-e^2/r> for the first two spherically symmetric states of the hydrogen atom (in 3D).

The Attempt at a Solution



I started out trying to calculate the averages with [tex]\psi[/tex] ... something like, for the ground state, [tex]\psi = \frac{e^{-2r/a_0}}{\sqrt{\pi * a_0^3}}[/tex].

But then I ran into problems (when I was trying to do the <-e^2/r>) when I came up with an integral involving a term [tex]\frac{e^{-2r/a_0}}{r}[/tex]. As far as I could see, this integral sort of seems to explode at the origin / at infinity. I was wondering, should I use just the radial wavefunction part, since it has an extra r factor that would make this integral possible? I was just confused basically...
You are not giving the details of what you calculated so it's hard to answer but you should not have any problem because the volume element [tex] dV = r^2 \sin \theta ~dr d\theta d\phi [/tex] provides an extra factor of r^2. Did you take this into account?
 
  • #3
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mmm, I didn't :) Thanks so much :)

I know it was smth silly I wasn't paying attention to ^_^

It's not dependent on theta or phi, so I should just multiply it by 4 pi after doing the integral in r, right?
 
  • #4
nrqed
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mmm, I didn't :) Thanks so much :)

I know it was smth silly I wasn't paying attention to ^_^

It's not dependent on theta or phi, so I should just multiply it by 4 pi after doing the integral in r, right?
That's correct. The integral over phi and theta of the angular part of the volume element gives 4 pi (you should do it once explicitly to see how it works out!)

Glad I could help.
 

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