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The probability of an electron being found in a certain region - probably easy?

  1. Oct 22, 2009 #1
    the hydrogen ground state has the following wavefunction

    [tex]\psi[/tex](r,[tex]\theta[/tex],[tex]\phi[/tex]) = [tex](\frac{1}{a^3^/^2\sqrt{\pi}})e^-^r^/^a[/tex]

    where a is the bohr radius a = 5.29x10-11m
    I have to calculate the probability of finding the electron within the nucleus (radius R = 10-15m), and I am given an identity:

    [tex]\int x^2e^-^xdx = b^3/3[/tex] when this integral is evaluated between 0 and b.


    I know that the probability is given by P(r) = [tex]\int |\psi|^2 dV[/tex] = [tex]\int |\psi|^2 4\pi r^2dr[/tex] evaluating between 0 and R.

    so when I put [tex]\psi[/tex] into the probability equation, I get

    P(r) = [tex](4/a^3)\int r^2 e^-^2^r^/^a dr[/tex] and I then try to use that identity, by making the substitution that x=kr (where k = 2/a)
    so that dx = kdr and x2 = (kr)2

    which gives me [tex]k^3\int r^2 e^-^k^r dr = R^3/3[/tex]

    which yields P(r) = [tex](4/a^3)\int r^2 e^-^k^r dr = (4/a^3)R^3/3k^3 [/tex] which has units of m3.

    clearly I am going wrong because I am getting units of volume. this is frustrating me now, where am I going wrong?

    thanks.
     
  2. jcsd
  3. Oct 22, 2009 #2
    i think u used the wrong substitution....
    your integral isn't of the same form as that given in the hint.

    try [itex]y=\frac{2r}{a}[/itex]. hopefully that will help.
     
  4. Oct 22, 2009 #3
    even with your substitution this still gives me units of m3...
     
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