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<r> and <V(r)> in the Hydrogen Atom.

by dirtyaldante
Tags: <r>, <vr>, atom, hydrogen
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dirtyaldante
#1
Jan19-11, 05:22 PM
P: 10
Hi. I'm a 3rd year undergraduate studying Applied Physics and I'm having some trouble with a problem concerning the Hydrogen Atom. This is my first post so please forgive the sloppy equations. Not really used to writing this stuff out without an equation editor handy! Anyway, the problem:


1. The problem statement, all variables and given/known data

1.Determine the expectation value of the potential energy [V(r) = (e^2)/(4.pi.epsilon0.r)] in the 1s (ground state) of the Hydrogen atom.

2.What is the expectation value of r for an electron in the 1s state of the Hydrogen atom?


2. Relevant equations

<V(r)> = INT [PSI*|V(r)^|PSI]
n=1, l=0, ml=0

3. The attempt at a solution

So the method I used to solve other expectation values (<L^2>,<Lz>,<E> etc...) was to use the appropriate operator upon the square of the wavefunction in question, ie: PSI(n,l,ml).
However I can't find operators for the Potential(V(r)) or r(position I suppose) that are related to the quantum numbers n,l and ml like in the other expectation values I solved. e.g.

L^2^=l(l+1)hbar
E^=13.6/n^2
Lz^=ml.hbar

So then I would square the wavefunctions and their coefficients inside the integral and then use the operators appropriately, in this case on the PSI(1,0,0) for the 1s configuration.

Am I simply missing a fundamental operator here? Or do I have to try arrange the operators myself? Rearranging the Schrodinger equation in terms of V(r)PSI was the only thing I could think of and even then I couldn't navigate through the wavefunctions of the separate parts.

Another idea I had was to use the Laguerre polynomial to redefine the wavefunction, but I'm unsure on how to proceed after that.

Is my method just completely off for these expectation values? I have been scratching my head all day over this and would really appreciate help if anyone can offer it.
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dirtyaldante
#2
Jan19-11, 05:39 PM
P: 10
Update:
I might have the answer to the second part. If the probabilty density is P(r) and P(r)=|R(n,l)|^2 times r^2
where:
R(n,l)=Radial wavefunction defined by the laguerre polynomials
r=radius

Then the value of r for this state corresponding to the maximum chance of finding an electron is at dP(r)/dr = 0

So after some work I found that this occurs at r=4a0, where a0 is the bohr radius.

Does this mean that <r> is simply 4a0? If so how could I show this mathmatically?
dextercioby
#3
Jan19-11, 05:50 PM
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P: 11,894
I think for the 1s state the wavefunction is so simple, that the 2 integrals required for point 1 and 2 can be done directly, without resorting to complicated formulas involving associated Laguerre polynomials, or even confluent hypergeometric functions.

As for the first point, average of the potential operator in the 1s state, you can use the virial theorem in QM. Just google it, if you don't know it already or have it in your textbook.

dirtyaldante
#4
Jan19-11, 06:01 PM
P: 10
<r> and <V(r)> in the Hydrogen Atom.

OK, I'm looking at this: http://en.wikipedia.org/wiki/Virial_...ween_particles and I am trying very hard to understand what you're saying but I'm just not getting it. How do you mean "average the potential operator"? I don't have one that I can see. I thought the point was to find one? Sorry if I'm being dumb...
dextercioby
#5
Jan19-11, 06:11 PM
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The potential operator for the H-atom is simply -k/r, where by 'k' I denoted the whole bunch of constants.

So you have a 2 relations between <V> and <T>, one from the virial theorem and the other from the fact that <V>+<T>=<H> for each eigenstate of the Hamiltonian and 1s is one of them. A simple system to for <V>.
dirtyaldante
#6
Jan19-11, 06:49 PM
P: 10
Ok, I think I get what you're saying but I don't understand the reasoning behind your method or how it well help me determine the expectation value of V(r).

Can you comment on the method I proposed and how it might relate to yours?
vela
#7
Jan19-11, 11:43 PM
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PF Gold
P: 11,673
As bigubau suggested, it's easiest to just evaluate the integrals directly. The expectation value of a function f(r) is given by

[tex]\langle f(r) \rangle = \int_0^\infty f(r) p(r) \, dr[/tex]

where p(r) is the probability density. In post 2, you already said what the probability density is, namely p(r)dr = |R(r)|2r2dr, so, for example,

[tex]\langle r \rangle = \int_0^\infty r \lvert R(r) \rvert^2 r^2\,dr[/tex]

You just need to substitute for R(r) now and evaluate the resulting integral.
dirtyaldante
#8
Jan20-11, 07:42 AM
P: 10
That's exactly what I need! Matches one of the useful integrals supplied too. Awesomeness. Thanks!
dirtyaldante
#9
Jan20-11, 09:13 AM
P: 10
Solutions for anybody who cares:

<V(r)> = (1/8.pi.epsilon0) sqrt(a0)
<r> = (3/2) a0

where a0 is the bohr radius = 5.2917710^−11

If somebody wants to double check my solutions they are more than welcome to but I think we can agree they look pretty good.

Thanks for the help guys!
vela
#10
Jan20-11, 07:18 PM
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PF Gold
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Your expectation value for V(r) isn't correct. For one thing, the units don't work out. Also, some constant factors are missing as well.
dirtyaldante
#11
Jan21-11, 08:47 AM
P: 10
Aha, you're right of course. I dropped the e^2 by mistake. That fixs the units.
vela
#12
Jan21-11, 03:14 PM
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PF Gold
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Not quite. The units of <1/r> should be 1/length, so it'll be proportional 1/a0, not sqrt(a0).
dirtyaldante
#13
Jan22-11, 08:02 AM
P: 10
Yep, I'm an idiot. Messed up the integration. Pretty sure i got it right THIS time though...

<V(r)>=e^2/(pi.epsilon0.a0)
vela
#14
Jan22-11, 10:14 PM
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PF Gold
P: 11,673
I got

[tex]\langle V(r) \rangle = -\frac{1}{4\pi\epsilon_0} \frac{e^2}{a_0}[/tex]
dirtyaldante
#15
Jan23-11, 12:11 PM
P: 10
Forgot to type the 4 and neglected the minus like I always do. Old habits die hard.


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