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Quantum Physics: Probability density

  1. Jul 14, 2003 #1
    For the ground state of the hydrogen atom, evaluate the probabilty density psi^2(r) and the radial probability density of P(r) for the positions.

    a) r = 0
    b) r = rb




    I confused how this probability function is used. What's the technique here?

    Thanks
     
  2. jcsd
  3. Jul 14, 2003 #2

    Tom Mattson

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    In this problem, you are only being asked to evaluate functions of a single variable variable at two different points. It's no different than if you were asked to evaluate f(x) and g(x) at x1 and x2 in Algebra II, for instance.

    The wavefunction Ψ(r) should be in your book. You already stated the definition of the probability density (|Ψ|2), and the definition of the radial probability density is bound to be in your book, too. Compute those functions, and then insert r=0 and r=rb (the Bohr radius). No integration is required.
     
  4. Jul 14, 2003 #3
    Okay, what do I use for A, n, L and x?

    If I evaluate x=0 I get 0. But the answer is non-zero.
     
  5. Jul 14, 2003 #4

    Tom Mattson

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    Whoa: What are A, n, and L? Also, don't you mean r instead of x?

    If you evaluate what at r=0? The probability density is nonzero there, but the radial probability density is zero.
     
  6. Jul 14, 2003 #5
    Halliday gives:

    |Ψ|2(x) = A2Sin2(n[pi]/L * x), n = 1, 2, 3,...

    No radial coordinate.
     
  7. Jul 14, 2003 #6

    Tom Mattson

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    That's the wavefunction for a particle in a box. It isn't applicable to the hydrogen atom. You need to look up that wavefunction, which will certainly have a radial coordinate.
     
  8. Jul 14, 2003 #7
    So I want this:

    Ψ(r) = 1/(sqrt[[pi]a3/2) e-r/a

    Square that and evaulate, or can I just evaluate and squre?
     
  9. Jul 14, 2003 #8

    Tom Mattson

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    Yes, that's the one. It makes no difference what order you do the evaluating and squaring in. Don't forget that you also have to do it for the radial probability density.
     
  10. Jul 14, 2003 #9
    For some reason the answer I'm getting is no where close to the given answer: 2150(nm)-3

    I get: 6.121e^9
     
  11. Jul 14, 2003 #10

    Tom Mattson

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    And you got your answer how....?
     
  12. Jul 14, 2003 #11
    I just plugged it into my calculator.

    Since a=5.29x10^-11m the answer should be very large.

    This must not be the correct formula or something.
     
  13. Jul 14, 2003 #12

    Tom Mattson

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    You should ask yourself:
    It should be very large in what units?

    You are working in meters, and the answer was given in inverse cubic nanometers.
     
  14. Jul 14, 2003 #13
    HAHHAHA!

    That's what happens when you do physics for 10 hours straight.

    Edit: BTW, I've never seen that type of unit used before so my brain must have dismissed it.:smile:
     
    Last edited: Jul 14, 2003
  15. Jul 14, 2003 #14
    Oh BTW: Thanks for your help Tom.
     
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