Expectation value for Hydrogen radius

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Homework Statement


Find the expectation value for a hydrogen atom's radius if n=25 and l=0.


Homework Equations


expectation value = <f|o|f>
where f=wavefunction and o=operator

The Attempt at a Solution


So I know that to find an expectation value, you integrate over all relevant space: (f*)(o)(f)...

However I'm not sure what "f" is in this case... my book only lists hydrogen atom wavefunctions up through like n=3, but nothing past it... also what would be the operator that relates to radius??

In class I think we derived something for the H atom that shows that its wavelength has a radial and an angular part but I'm entirely confused :frown:

Also, what's the difference when the question is stated like above and when it says to find the mean radius <r> of H in a certain orbital? Isn't it pretty much the same thing?
 

Answers and Replies

  • #2
Gokul43201
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Homework Statement


Find the expectation value for a hydrogen atom's radius if n=25 and l=0.


Homework Equations


expectation value = <f|o|f>
where f=wavefunction and o=operator

The Attempt at a Solution


So I know that to find an expectation value, you integrate over all relevant space: (f*)(o)(f)...

However I'm not sure what "f" is in this case... my book only lists hydrogen atom wavefunctions up through like n=3, but nothing past it... also what would be the operator that relates to radius??
Find a table of Generalized Laguerre polynomials. Can't imagine you'd be expected to integrate 26 terms though!! :confused:

The operator you want is the position operator.

In class I think we derived something for the H atom that shows that its wavelength has a radial and an angular part but I'm entirely confused :frown:
Do you mean "wavefunction"?

Also, what's the difference when the question is stated like above and when it says to find the mean radius <r> of H in a certain orbital? Isn't it pretty much the same thing?
Pretty much.
 
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  • #3
George Jones
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Are you sure l = 0. It's much more common to take, for large n, l = n - 1, as this corresponds (for each n) to the only circular orbit in Bohr-Sommerfeld theory, and the polynomial in r then only has one term.

If it really is l = 0, then I'd use a compute package like Maple or Mathematica.
 
  • #4
Meir Achuz
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A recursion relation [Eq. (13.33) in Arfken's 2nd editon], lets you replace
r(L_n)^2 by (2n+1)(L_n)^2 in the integral for <r>.
 
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  • #5
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Oh, oops, Gokul, I did mean wavefunction, not wavelength...

And yes, my question does state l=0...

Meir Achuz- can you explain a bit more? I'm not sure what you're talking about :(

Also, what is the use of the equation:
L" + (2l+2 - rho)L' + (n - l+1)L = 0
? :-/
 
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  • #6
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The Attempt at a Solution


So I know that to find an expectation value, you integrate over all relevant space: (f*)(o)(f)...
You are on the right track

However I'm not sure what "f" is in this case... my book only lists hydrogen atom wavefunctions up through like n=3, but nothing past it...
You should probably find another textbook, which gives a general form of wavefunctions [itex] \psi_{n,l=0,m=0} (r) [/itex] for all n.

also what would be the operator that relates to radius??
In the position representation this is just multiplication by r.

In class I think we derived something for the H atom that shows that its wavelength has a radial and an angular part but I'm entirely confused :frown:
You don't need to worry about the angular part, because for l=0 there is no angular dependence.

Also, what's the difference when the question is stated like above and when it says to find the mean radius <r> of H in a certain orbital? Isn't it pretty much the same thing?
Yes, it is the same thing.

Eugene.
 
  • #7
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Eugene, I think I found the general form of wavefunctions that you're referring to, but it had a summation term which expanded from j=0 to j=n-l-1, which in my case meant j=0 to j=24, which is an awful lot of terms to deal with, so I'm just wondering if there's a different way of dealing with this... (I'm 95% sure my prof didn't want us to use computer programs)

Also, can someone explain what this does? L" + (2l+2 - rho)L' + (n - l+1)L = 0

Haha, I may be really far off but it's tempting to think that maybe that has something to do w ith the question since it has both n and l in it... or... is the generalized wavefunction (the radial part) the "solution" to this equation or something?
 
  • #8
Meir Achuz
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Meir Achuz- can you explain a bit more? I'm not sure what you're talking about :(
/
It means the answer is (2n+1)a_0
 
  • #9
George Jones
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It means the answer is (2n+1)a_0
By a hand waving argument, I expect the answer to involve n^2 (and maybe other terms) times a_0.

The classical energy goes as -1/r, and the quantum energy goes as -1/n^2.
 
  • #10
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Can anyone explain/get me started on how to get to that form of (2n+1)a_0 or another form that involves n^2?
 
  • #11
Meir Achuz
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<r>=\int r U^2.
You need integral of xL^2. Using that recursion relation in Arfken (or any other book), the integral of xL^2=(2n+1) times the integral of L^2.
Do you want a pdf of the complete solution for your teacher?
 

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