# Expectation value for Hydrogen radius

1. Jul 26, 2007

### physgirl

1. The problem statement, all variables and given/known data
Find the expectation value for a hydrogen atom's radius if n=25 and l=0.

2. Relevant equations
expectation value = <f|o|f>
where f=wavefunction and o=operator

3. The attempt at a solution
So I know that to find an expectation value, you integrate over all relevant space: (f*)(o)(f)...

However I'm not sure what "f" is in this case... my book only lists hydrogen atom wavefunctions up through like n=3, but nothing past it... also what would be the operator that relates to radius??

In class I think we derived something for the H atom that shows that its wavelength has a radial and an angular part but I'm entirely confused

Also, what's the difference when the question is stated like above and when it says to find the mean radius <r> of H in a certain orbital? Isn't it pretty much the same thing?

2. Jul 27, 2007

### Gokul43201

Staff Emeritus
Find a table of Generalized Laguerre polynomials. Can't imagine you'd be expected to integrate 26 terms though!!

The operator you want is the position operator.

Do you mean "wavefunction"?

Pretty much.

Last edited: Jul 27, 2007
3. Jul 27, 2007

### George Jones

Staff Emeritus
Are you sure l = 0. It's much more common to take, for large n, l = n - 1, as this corresponds (for each n) to the only circular orbit in Bohr-Sommerfeld theory, and the polynomial in r then only has one term.

If it really is l = 0, then I'd use a compute package like Maple or Mathematica.

4. Jul 27, 2007

### Meir Achuz

A recursion relation [Eq. (13.33) in Arfken's 2nd editon], lets you replace
r(L_n)^2 by (2n+1)(L_n)^2 in the integral for <r>.

Last edited: Jul 27, 2007
5. Jul 28, 2007

### physgirl

Oh, oops, Gokul, I did mean wavefunction, not wavelength...

And yes, my question does state l=0...

Meir Achuz- can you explain a bit more? I'm not sure what you're talking about :(

Also, what is the use of the equation:
L" + (2l+2 - rho)L' + (n - l+1)L = 0
? :-/

Last edited: Jul 28, 2007
6. Jul 29, 2007

### meopemuk

You are on the right track

You should probably find another textbook, which gives a general form of wavefunctions $\psi_{n,l=0,m=0} (r)$ for all n.

In the position representation this is just multiplication by r.

You don't need to worry about the angular part, because for l=0 there is no angular dependence.

Yes, it is the same thing.

Eugene.

7. Jul 29, 2007

### physgirl

Eugene, I think I found the general form of wavefunctions that you're referring to, but it had a summation term which expanded from j=0 to j=n-l-1, which in my case meant j=0 to j=24, which is an awful lot of terms to deal with, so I'm just wondering if there's a different way of dealing with this... (I'm 95% sure my prof didn't want us to use computer programs)

Also, can someone explain what this does? L" + (2l+2 - rho)L' + (n - l+1)L = 0

Haha, I may be really far off but it's tempting to think that maybe that has something to do w ith the question since it has both n and l in it... or... is the generalized wavefunction (the radial part) the "solution" to this equation or something?

8. Jul 29, 2007

### Meir Achuz

It means the answer is (2n+1)a_0

9. Jul 29, 2007

### George Jones

Staff Emeritus
By a hand waving argument, I expect the answer to involve n^2 (and maybe other terms) times a_0.

The classical energy goes as -1/r, and the quantum energy goes as -1/n^2.

10. Jul 29, 2007

### physgirl

Can anyone explain/get me started on how to get to that form of (2n+1)a_0 or another form that involves n^2?

11. Jul 30, 2007

### Meir Achuz

<r>=\int r U^2.
You need integral of xL^2. Using that recursion relation in Arfken (or any other book), the integral of xL^2=(2n+1) times the integral of L^2.
Do you want a pdf of the complete solution for your teacher?