1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...