• Support PF! Buy your school textbooks, materials and every day products Here!

Probability of finding an electron betweeen two spheres of radii?

  • Thread starter silversuz1
  • Start date
  • #1

Homework Statement



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.

Homework Equations



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

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=
 

Answers and Replies

  • #2
525
6
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.
 
  • #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?
 
  • #4
525
6
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.
 
  • #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!
 
  • #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
 
  • #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. =.=)
 

Related Threads for: Probability of finding an electron betweeen two spheres of radii?

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
2K
Replies
7
Views
11K
Replies
3
Views
6K
  • Last Post
Replies
3
Views
2K
Replies
2
Views
20K
Replies
6
Views
1K
  • Last Post
Replies
0
Views
2K
Top