Probability of finding an electron betweeen two spheres of radii?

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Homework Help Overview

The discussion revolves around calculating the probability of finding an electron in a hydrogen atom between two spheres of specified radii in a quantum mechanics context, specifically for the state with n=2 and l=0.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the wave function squared over a specified range to find the probability. There are attempts to derive the integral and concerns about the complexity of the calculations involved.

Discussion Status

Some participants have offered hints and guidance on how to approach the integral, while others express confusion about the differentiation technique suggested. There is an ongoing exploration of the integral's evaluation and the proper substitutions to use.

Contextual Notes

Participants note the challenge of integrating the function and the potential for misunderstanding the differentiation trick. There is acknowledgment of the complexity of the integral and the need for clarification on certain steps.

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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=
 
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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.
 
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?
 
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.
 
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!
 
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
 
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. =.=)
 

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