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Schaum's Outline of Quantum Mechanics 8.11

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    If a hydrogen atom is in the state n = 2, l = 0, what is the probability that an electron has a value of r that is smaller than the Bohr radius.


    2. Relevant equations
    eqn (8.8)
    [tex]\psi_{200}(r, \theta, \phi) = R_{20}Y_0^0(\theta, \phi)[/tex]

    Page 305
    [tex]Y_0^0(\theta, \phi) = \frac{1}{\sqrt{4\pi}}[/tex]

    eqn (8.33)
    [tex]R_{20} = 2(2a_0)^{-3/2}(1 - \frac{r}{2a_0})e^{-r/2a_0}[/tex]


    3. The attempt at a solution
    [tex]P = \int_0^{2\pi}d\phi \int_0^{\pi}d\theta\int_0^{a_0}dr r^2|\psi(r, \theta, \phi)|^2sin(\theta)[/tex]
    [tex]= \int_0^{2\pi}d\phi \int_0^{\pi}d\theta sin(\theta) (Y_0^0(\theta, \phi))^2 \int_0^{a_0}dr r^2 4(2a_0)^{-3}(1 - \frac{r}{2a_0})^2e^{-r/a_0}[/tex]
    Now let [itex]z = r/a_0[/itex]; [itex]dr = a_0dz[/itex] then
    [tex]P = \frac{4\pi}{4\pi}\frac{4}{(2a_0)^3}\int_0^1 a_0dz a_0^2z^2(1 - \frac{z}{2})^2e^{-z}[/tex]
    [tex]= \frac{1}{2}\int_0^1 dz (z^2 - z^3 + \frac{z^4}{4})e^{-z}[/tex]
    [tex] = (-1 -z -\frac{z^2}{2} - \frac{z^4}{8})e^{-z}|_0^1[/tex]
    [tex] = 1 - \frac{21}{8e} = 0.0343[/tex]

    The book gives the answer .176

    I calculated that the probability of r being less than 4 times the Bohr radius is .176

    Placing n = 2, l = 0 in equation 8.42 gives that the mean value for r is 6 times the Bohr radius. I find it hard to believe that the electron could spend nearly 18% of the time at less than 1 Bohr radius and yet average out to 6 Bohr radii.
     
    Last edited: Oct 22, 2012
  2. jcsd
  3. Oct 22, 2012 #2

    dextercioby

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    It's + z^4/4 inside the bracket under the integral sign. And are you sure about the integrations ?
     
  4. Oct 22, 2012 #3
    Thank you. I fixed that. I am as sure about them as I can be, but if I were 100% sure, I wouldn't have posted. By the way, I also solved the integral numerically using a calculator and got the same 0.0343 result. I believe that if I am wrong, it is because I am using the wrong approach, not because of an error in calculation.
     
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