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Expected value for potential energy (quantum)

by Sapper6
Tags: energy, expected, potential, quantum
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Sapper6
#1
Nov4-10, 12:13 PM
P: 11
1. The problem statement, all variables and given/known data
The radial distribution factor for a 1s orbital given: R10
Calculate the expected value for potential energy of a He atom in the ground state.


2. Relevant equations

i understand the integral math where I solve down to <1/r> = z/a

but now, how do i use the V(r) = Z/(4(pi)E) = k (1/r) equation


3. The attempt at a solution

i know Z is 2 for He and E is in coulombs and I need to end with joules, but I am just stumped on this part.
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vela
#2
Nov4-10, 12:49 PM
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What do you mean by
Quote Quote by Sapper6 View Post
but now, how do i use the V(r) = Z/(4(pi)E) = k (1/r) equation.
Sapper6
#3
Nov4-10, 12:54 PM
P: 11
I am trying to find the expected potential energy. The equation I have is:

Z
--------- x (1/r) = V(r)
4*(pi)*E

Z is charge which equals 2 for Helium (number of protons) and I would sub in (2/a) for 1/r

i am assuming i would use Bohr's radius here where ao= 5.291x10^10m

but I don't know how to find expected potential energy after I solved the radial integral for <1/r>.. especially what is E here

vela
#4
Nov4-10, 01:14 PM
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Expected value for potential energy (quantum)

OK, you're missing some pretty basic stuff, which is why your confusion seemed so strange to me. First, the potential is

[tex]V(r) = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r}[/tex]

You should recognize that expression from basic from basic electromagnetics. It's not E in the expression, but ϵ0, the permittivity of free space. Second, the expectation value of the potential energy is

[tex]\langle V(r) \rangle = \langle \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r}\rangle[/itex]

If you already have <1/r>, you pretty much have the answer.
Sapper6
#5
Nov4-10, 01:24 PM
P: 11
i am having trouble following the units. aren't you missing an e^2 term?

the units should go:

V(r) = Ze^2/(4(pi)e0) * <1/r>

Ze^2 = C^2 e0 = C/(V*m) 1/r = 1/m

So, V(r) = C*V = J

if my <1/r> is actually (2/a), i think i substitute in the following:

Z= 2 (no units for Helium)
a= 5.29 x 10^10 m
what is eo?

i have a constant that is written eo= 8.854 x 10^12 F/m and i know F is 9.648C/mol

can you please walk me through the substitution with eo?
vela
#6
Nov4-10, 01:38 PM
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Quote Quote by Sapper6 View Post
i am having trouble following the units. aren't you missing an e^2 term?
Yup, you're right. I fixed the previous post.
the units should go:

V(r) = Ze^2/(4(pi)e0) * <1/r>

Ze^2 = C^2 e0 = C/(V*m) 1/r = 1/m

So, V(r) = C*V = J

if my <1/r> is actually (2/a), i think i substitute in the following:

Z= 2 (no units for Helium)
a= 5.29 x 10^10 m
what is eo?

i have a constant that is written eo= 8.854 x 10^12 F/m and i know F is 9.648C/mol

can you please walk me through the substitution with eo?
A farad is a coulomb per volt, so ϵ0=8.854x1012 C/(V m), which are the units you had above. A farad is definitely not a coulomb per mole.

You're thinking of the Faraday constant F, the amount of charge in one mole of electrons (ignoring the sign), which is not the same thing as a farad, also denoted by F, the unit of capacitance.
Sapper6
#7
Nov4-10, 01:49 PM
P: 11
thank you, i understand now, i appreciate the help
sandbeda
#8
Jan18-11, 05:25 PM
P: 5
Can somebody set this straight for me (I have the same problem)? All you need to do is substitute in values? If so, where does the value of r come from, the expectation value for r, <r>?
vela
#9
Jan19-11, 12:17 AM
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To find <1/r>, you have to do the integral

[tex]\int \psi^*(\vec{r})\left(\frac{1}{z}\right)\psi(\vec{r})\,d^3\vec{r}[/tex]

where ψ is the wavefunction for the state.
Mancho
#10
Mar30-11, 05:31 AM
P: 7
I have a question about the last expression, I mean the integral for the expectation value. If we are given potential V(x) and ground state energy E0, and corresponding eigenvector U0 , is it possible to calculate <V(x)> without knowing wavefunction?

Another question is: because V(x) is a linear operator can we assume that
[tex]
\int\psi^*V(x)\psi dx=\left|\psi\right|^2\int{V(x)dx} ?
[/tex]
vela
#11
Mar30-11, 06:34 AM
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Quote Quote by Mancho View Post
I have a question about the last expression, I mean the integral for the expectation value. If we are given potential V(x) and ground state energy E0, and corresponding eigenvector U0 , is it possible to calculate <V(x)> without knowing wavefunction?
Perhaps. In the case of the simple harmonic oscillator, you certainly can. Did you have a specific problem in mind?
Another question is: because V(x) is a linear operator can we assume that
[tex]
\int\psi^*V(x)\psi dx=\left|\psi\right|^2\int{V(x)dx} ?
[/tex]
Nope. I don't see how you got that from the linearity of V(x).
Mancho
#12
Mar30-11, 07:02 AM
P: 7
Quote Quote by vela View Post
Perhaps. In the case of the simple harmonic oscillator, you certainly can. Did you have a specific problem in mind?
Yes, I was given V(x)=1/(cosh(x-pi/2)) where 0<x<pi; I had to calculate ground state energy numerically and I got it, as well as eigenvector. then I was simply asked to calculate <V(x)>, but I couldn't get an idea how I could get it without knowing wavefunction. I tried to analytically solve Schrodinger equation with this potential but it seems too complicated, that's why I think maybe there is some other way.

Quote Quote by vela View Post
Nope. I don't see how you got that from the linearity of V(x).
OK, seems I misunderstood.
vela
#13
Mar30-11, 02:04 PM
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You should start a new thread with your problem and showing your work so far.


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