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Probability of electron location in Hydrogen atom

  1. Nov 11, 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 is within one Bohr radius of the nucleus?

    2. Relevant equations
    [tex]P_{nl}(r) = r^{2}|R_{nl}(r)|^{2}[/tex]


    3. The attempt at a solution
    Since it's an electron in the ground state of a hydrogen atom, that means n = 1, and that means it's in the s orbital, which means l = 0.

    So using the formula provided in the book for [itex]R_{10}(r)[/itex], which is [itex]\frac{2}{(a_{0})^{\frac{3}{2}}}e^{\frac{-r}{a_{0}}}[/itex]

    I just square that whole thing and get [itex]\frac{4e^{\frac{-2r}{a_{0}}}}{(a_{0})^{3}}[/itex]

    I know the value of [itex]a_{0}[/itex], but I'm not sure what r is. Is r the radius, which happens to be the same as the Bohr radius ([itex]a_{0}[/itex]) for this problem?
    I want to be able to calculate an actual number instead of having an answer with variables in it.

    Thanks.
     
  2. jcsd
  3. Nov 11, 2013 #2
    The probability is the integral of the quantity you squared. The bounds the integral should be 0 to the bohr radius... that will give you the probability of the electron being within the bohr radius. Does this make sense? Check your formula for the probability... seems as if you left off the integral entirely.
     
  4. Nov 11, 2013 #3
    Thanks. Yeah that makes sense. The book shows it with a dr on both sides without the integral sign. But yeah, the integral sign should be there.

    Now all I need to do is figure out how to properly do that integral. That's just a matter of time.

    Thanks again.
     
  5. Nov 11, 2013 #4
    Remember you can check your answer (or compute it entirely if your instructor lets you) with wolframalpha
     
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