1. Not finding help here? Sign up for a free 30min 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!

Probability of finding an electron betweeen two spheres of radii?

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data

    For a hydrogen atom in a state with n=2 and l=0, calculate the probability of finding the electron between two spheres of radii r = 5.00a0 and r = 5.01 a0.

    2. Relevant equations

    As stated above, calculate the probability of finding the electron in the given spheres of radii.

    3. The attempt at a solution

    Maybe it's something like this?

    ∫ psi^2 (r) dr, from r = 5.00a to r = 5.01a

    = ∫ (1/32π(a^3)) * ((2 - ((r/a)^2))^2) * (e ^ (-r/a) dr, where π means pi.

    = (after complicated processes of trying to find the integral) some unsolvable identity! D=
     
  2. jcsd
  3. May 3, 2009 #2
    Short answer: yes, that's the way to do it :)

    Some hints:
    First off all, apart from an overal constant you have to find the primitive function of:
    [2-(r/a)^2]^2 exp(-r/a) = [4 - 4(r/a)^2 + (r/a)^4]exp(-r/a)

    To find this you can use the following.
    You know that:
    ∫ exp(-y b) dy = -(1/b) * exp(-y b)
    Now differentiate left and right with respect to b, twice. This gives the following:
    ∫ y^2 exp(-y b) dy = -[2(1/b)^3+y(1/b)^2] exp(-y b) + (y/b)(1/b+y)exp(-y b)

    which you can simplify a bit further, but I'll leave this up to you (also, my answer might be wrong, cause I'm being a bit sloppy with one eye on the tv ;) ). You can perfrom this trick again (differentiating twice) and that will give you an expression for the integral of y^4 exp(-yb)dy. Hopefully this puts you in the right direction.
     
  4. May 3, 2009 #3
    Thanks! But...I'm still a bit confused. So in order to find the probability of finding the electron between the two spheres of radii, I have to differentate the integral? Doesn't that cancel it out?
     
  5. May 3, 2009 #4
    No, no! That's just a trick to calculate the integral! You integrate over y, but you differentiate with respect to b. It's a neat little trick, but quite weird if you haven't seen it before. Try to work it out. You can post your results here.
     
  6. May 3, 2009 #5
    Okay, I'll try it out. But just for an extra clarification, when you put exp(-y b), you substituted (-y b) (which equals -y/b?) in for r/a, correct?

    All right, I'm going to try it now....thanks!
     
  7. May 3, 2009 #6
    All right, I tried it. =.= I came out in circles. My ending was:

    (1/(32pi(a^2))) * int((4-4u+(u^2))*(e^-u)) du, where u = r/a

    And then I got stuck. :P
     
  8. May 4, 2009 #7
    Can anyone please help by directing me in some direction? Thanks! (and I understand that xepma kindly helped me out, so thank you! But I'm still lost on where to go, because I have tried the integral, but I just can't seem to get it. =.=)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Probability of finding an electron betweeen two spheres of radii?
Loading...