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QM approximation (electron within nucleus, Griffiths 4.45b

  1. Feb 7, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the probability that an electron in the ground state of hydrogen will be found inside the nucleus?

    a) First calculate the exact answer, assuming the wave function [itex]\psi(r,\theta,\phi) = \frac{1}{\sqrt{\pi a^3}} e^{-r/a}[/itex] is correct all the way down to r=0. Let b be the radius of the nucleus

    b) Expand your result as a power series in the small number [itex]\epsilon = 2b/a[/itex] and show that the lowest-order term is the cubic: [itex]P \approx (4/3)(b/a)^3[/itex]. This should be a suitable appoximation, provided that b << a, which it is.

    3. The attempt at a solution
    Using partial integration a few times, I've gotten an answer for a) which I think is sufficient in order to solve b). [itex]P (r<b) = 1-e^{-2b/a}(2b^2/a^2 + 2b/a + 1)[/itex].

    However, I'm completely lost on b). I'm looking for pointers here on how to even attack the problem. I can rewrite P so that I get [itex]P = 1 - e^{-\epsilon}(\epsilon/2 + \epsilon + 1)[/itex] but from there I'm stumped. I think the problem is that I've mostly forgotten how to do Taylor expansion, and a quick lookup does not help me in attacking this specific problem.
     
  2. jcsd
  3. Feb 7, 2013 #2

    tiny-tim

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    Hi Saraphim! :smile:
    use the definition of ex (which is also its taylor series) …

    ex = 1 + x + x2/2 + x3/6 + …​

    then P = 1 - e-2ε(1 + 2ε + 2ε2)

    = 1 - (1 - 2ε + 4e2/2 - 8ε3/6 + …)(1 + 2ε + 2ε2)

    = … ? :smile:
     
  4. Feb 8, 2013 #3
    Thanks for your response! I have a few problems, still. First, I feel the Taylor expansion should be [itex]1 - \epsilon + \epsilon^2/2 - \epsilon^3/6 +\ ...[/itex]. Isn't that right?

    Second, I feel that I should be able to realize that the terms lower than the cubic somehow get cancelled out, but I really don't see it. :(
     
  5. Feb 8, 2013 #4

    tiny-tim

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    Good morning! :smile:
    that would be correct for e

    the question gives you e-2ε :wink:
    should cancel out now! :smile:
     
  6. Feb 8, 2013 #5
    But how is that so? In the unexpanded P without substituting in [itex]\epsilon = 2b/a[/itex], I have [itex]e^{-2b/a}[/itex].
     
  7. Feb 8, 2013 #6

    tiny-tim

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    oh, i'm sorry, i was using ε = b/a :redface:

    with ε = 2b/a,

    then P = 1 - e(1 + ε + ε2/2)

    = 1 - (1 - ε + e2/2 - ε3/6 + …)(1 + ε + ε2/2) :smile:
     
  8. Feb 8, 2013 #7
    I think I understood what I was missing. I needed to set all terms in the product with exponents higher or equal to 4 to zero. Approximation is not my string suite. Thanks for the help!
     
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