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Hydrogen atom probability of electron inside nucleus

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data

    The ground state of the hydrogen atom has the wavefunction: [itex]\psi (r) = \frac{1}{\sqrt{\pi}} (\frac{1}{a_{0}})^{\frac{3}{2}} e^{-r/a_{0}}[/itex], where [itex]a_{0} = 0.53Å[/itex]

    a) Calculate the probability that an electron will be found inside the nucleus of the hydrogen atom, which is assumed to be [itex]10^{-15} m[/itex] in radius.

    2. Relevant equations

    [itex]P(r) = \int_{0}^{10^{-15m}}\psi^{*}(r)\psi(r) dr[/itex]

    [itex]P_{nl}(r) = \int_{r}^{r+dr}R_{nl}^{*}(r) R_{nl}(r) 4\pi r^{2} dr[/itex]

    3. The attempt at a solution

    Okay, so I tried solving this problem using the first integral above and got something ridiculous so had a bit of a read and remembered something about a volume element in the integral for the probability density. I suppose I was only considering a plane slice or something.

    Anyway, I'm having trouble understanding where the second integral comes from (I got it out of a book), in particular the factor of [itex]4\pi r^{2}[/itex] in the integral. So looking at it I'm guessing the R's refer to the radial part of the wavefunction in spherical coordinates which is the same as the first integral since it's [itex]\psi (r)[/itex]. Now the book says that the factor of [itex]4\pi r^{2}[/itex] is present because the volume enclosed between spheres of radius [itex]r[/itex] and [itex]r + dr[/itex] is given by that factor, however I can't figure it out! I've tried calculating the volume of two spheres of radius [itex]r[/itex] and [itex]r + dr[/itex] respectively and it feels like I'm sort of close but I can't get to it!

    Can anyone explain to me what it's doing there, the books explanation makes intuitive sense when you picture what you're integrating over but I don't understand how that is equivalent to this factor in the integral!

    Thank you!
     
  2. jcsd
  3. Apr 28, 2013 #2

    mfb

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    Staff: Mentor

    You don't need the second integral.
    4 pi r^2 is the surface area of a sphere, and the second integral calculates the probability to find the electron in a thin (dr) spherical shell.

    If you want to simplify the problem, you can assume that the magnitude of the wave function is constant within the nucleus.
    The result will be extremely small.
     
  4. Apr 28, 2013 #3
    I think for the problem I'm supposed to use the approximation that [itex]e^{-r/a_{0}} \approx 1[/itex] since [itex]r << a_{0}[/itex], this approach has come up many times in the past for virtually the same question in my Atomic & Nuclear Physics course and Quantum Mechanics course.

    However, my question wasn't really about solving this problem, and I understand that [itex]4 \pi r^{2}[/itex] is the surface area of a sphere and we're calculating the probability to find the electron in a thin [itex](dr)[/itex] spherical shell, however, as I said, intuitively this makes sense but I don't understand the maths of it.
     
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