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Expectation value for electron in groundstate

  1. Oct 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the expectation value for r for an electron in the groundstate of a one-electron-atom is:
    <r>=(3/2)a[itex]_{0}[/itex]/Z



    2. Relevant equations
    Expectationvalue:
    <f(x)>=∫[itex]\psi[/itex]*f(x)[itex]\psi[/itex]dx, -∞<x>∞

    [itex]\psi[/itex][itex]_{100}[/itex]=C[itex]_{100}[/itex] exp(-Zr/a[itex]_{0}[/itex]), [itex]a_{o}\ =\ 0.5291\ \times\ 10^{-10}m , h\ =\ 6.626\ \times\ 10^{-34}\ J\ s[/itex]


    3. The attempt at a solution
    Im stuck..
     
  2. jcsd
  3. Oct 9, 2012 #2

    Ray Vickson

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    What is stopping you from doing the integral? Of course, first you must determine the correct value of C100; do you know how to do that?

    RGV
     
  4. Oct 9, 2012 #3
    Hi, yes i used that C[itex]_{100}[/itex] =(1/[itex]\sqrt{\pi}) (\frac{z}{a_{0}}[/itex])[itex]^{3/2}[/itex]

    Then i do the integral, but i must be doing it wrong cause i end up with e[itex]^{\frac{-2zr}{a_{0}}}[/itex] in the answer and also a problem of e[itex]^{∞}[/itex]
     
  5. Oct 9, 2012 #4

    vela

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    Show your work.
     
  6. Oct 10, 2012 #5
    Ok, i found a mistake, but still dont get it right.
    I write:

    <r>=[itex]\frac{1}{\pi}[/itex][itex]\frac{z_{0}^{3}}{a_{0}^{3}}[/itex]∫e[itex]^{\frac{-2Zr}{a_{0}}}[/itex]r dr

    Where the limits are from -∞ to ∞

    Integrating i let U= r, du=1, dv= e[itex]^{\frac{-2Zr}{a_{0}}}[/itex], V=[itex]\frac{a_{0}}{-2Z}[/itex]e[itex]^{\frac{-2Zr}{a_{0}}}[/itex]

    this integration leaves me with e[itex]^{∞}[/itex], or is there some trick for e[itex]^{-∞}[/itex] - e[itex]^{∞}[/itex]
     
  7. Oct 10, 2012 #6

    vela

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    The integral should be from 0 to infinity. You also have a mistake somewhere else as your expression has units of 1/length.
     
    Last edited: Oct 10, 2012
  8. Oct 10, 2012 #7
    Oh yes thats logical since its the radius, thanks. Yes and the integration by parts should be
    u=r
    du=dr
    dv=e[itex]^{\frac{-2Zr}{a_{0}}}[/itex]
    v=[itex]_{}\frac{a_{0}}{-2Z}[/itex]e[itex]^{\frac{-2Zr}{a_{0}}}[/itex] dr

    integrating [itex]\frac{z^{3}}{\pi a_{0}}[/itex]∫e[itex]^{\frac{-2Zr}{a_{0}}}[/itex] r dr from 0 to ∞ with these vaues i get
    <r>= -[itex]\frac{Za_{0}}{4\pi}[/itex]

    I wonder if my C[itex]_{100}[/itex] might be wrong..
     
  9. Oct 10, 2012 #8

    vela

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    How did you find C100?
     
  10. Oct 11, 2012 #9
    It was listed in my book, for the groundstate of an hydrogen atom so it should actually be correct.
     
  11. Oct 11, 2012 #10

    vela

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    You need to go back and review spherical coordinates so you can set up the integrals correctly. In particular, you need to fix your limits and get the volume element right.
     
  12. Oct 11, 2012 #11

    Ray Vickson

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    You should verify for yourself the value given in the book. If you cannot get the book's value, that is a signal that you may be doing something wrong---or possibly that the book made an error---but you should go with the first hypothesis unless you have overwhelming evidence to the contrary.

    Once you are able to get the correct value of C100 you will be in a better position to find <r>.

    RGV
     
  13. Oct 11, 2012 #12
    Hi, thanks alot. It really helped to use spherical coordinates :) got the correct answer now :)
     
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