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Expectation value for Hydrogen radius

  1. Jul 26, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the expectation value for a hydrogen atom's radius if n=25 and l=0.


    2. Relevant equations
    expectation value = <f|o|f>
    where f=wavefunction and o=operator

    3. 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?
     
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  3. Jul 27, 2007 #2

    Gokul43201

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    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.

    Do you mean "wavefunction"?

    Pretty much.
     
    Last edited: Jul 27, 2007
  4. Jul 27, 2007 #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.
     
  5. Jul 27, 2007 #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>.
     
    Last edited: Jul 27, 2007
  6. Jul 28, 2007 #5
    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
    ? :-/
     
    Last edited: Jul 28, 2007
  7. Jul 29, 2007 #6
    You are on the right track

    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.

    In the position representation this is just multiplication by r.

    You don't need to worry about the angular part, because for l=0 there is no angular dependence.

    Yes, it is the same thing.

    Eugene.
     
  8. Jul 29, 2007 #7
    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?
     
  9. Jul 29, 2007 #8

    Meir Achuz

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    It means the answer is (2n+1)a_0
     
  10. Jul 29, 2007 #9

    George Jones

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    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.
     
  11. Jul 29, 2007 #10
    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?
     
  12. Jul 30, 2007 #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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