Expectation value for Hydrogen radius

In summary: The classical energy goes as -1/r, and the quantum energy goes as... well, you get the idea.So the energy goes from -1/r to 1/r as you go from classical to quantum?Yes, that's pretty much it.
  • #1
physgirl
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Homework Statement


Find the expectation value for a hydrogen atom's radius if n=25 and l=0.


Homework Equations


expectation value = <f|o|f>
where f=wavefunction and o=operator

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 :frown:

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?
 
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  • #2
physgirl said:

Homework Statement


Find the expectation value for a hydrogen atom's radius if n=25 and l=0.

Homework Equations


expectation value = <f|o|f>
where f=wavefunction and o=operator

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

The operator you want is the position operator.

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 :frown:
Do you mean "wavefunction"?

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?
Pretty much.
 
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  • #3
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
A recursion relation [Eq. (13.33) in Arfken's 2nd editon], let's you replace
r(L_n)^2 by (2n+1)(L_n)^2 in the integral for <r>.
 
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  • #5
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
? :-/
 
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  • #6
physgirl said:

The Attempt at a Solution


So I know that to find an expectation value, you integrate over all relevant space: (f*)(o)(f)...

You are on the right track

physgirl said:
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...

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

physgirl said:
also what would be the operator that relates to radius??

In the position representation this is just multiplication by r.

physgirl said:
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 :frown:

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

physgirl said:
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?

Yes, it is the same thing.

Eugene.
 
  • #7
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
physgirl said:
Meir Achuz- can you explain a bit more? I'm not sure what you're talking about :(
/
It means the answer is (2n+1)a_0
 
  • #9
Meir Achuz said:
It means the answer is (2n+1)a_0

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
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
<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?
 

FAQ: Expectation value for Hydrogen radius

What is the expectation value for Hydrogen radius?

The expectation value for Hydrogen radius is the average distance from the nucleus to the electron in a Hydrogen atom. It is calculated using the wave function of the electron in the atom.

How is the expectation value for Hydrogen radius calculated?

The expectation value for Hydrogen radius is calculated by integrating the square of the wave function of the electron from 0 to infinity. The result is then divided by the probability of finding the electron in that region.

Why is the expectation value for Hydrogen radius important?

The expectation value for Hydrogen radius is important because it provides valuable information about the electron's position in the atom. It also helps in understanding the behavior and properties of the atom.

How does the expectation value for Hydrogen radius change with different energy levels?

The expectation value for Hydrogen radius increases as the energy level of the electron increases. This is because the electron is farther away from the nucleus and has a higher probability of being found at a larger distance.

Can the expectation value for Hydrogen radius be calculated for other atoms?

Yes, the expectation value for Hydrogen radius can be calculated for other atoms using their respective wave functions. However, it may not have the same physical significance as it does for Hydrogen, as other atoms have more complex electron configurations.

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