Expected value for potential energy (quantum)

  1. 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.
     
  2. jcsd
  3. vela

    vela 12,733
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    What do you mean by
     
  4. 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
     
  5. vela

    vela 12,733
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    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.
     
    Last edited: Nov 4, 2010
  6. 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?
     
  7. vela

    vela 12,733
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    Yup, you're right. I fixed the previous post.
    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.
     
  8. thank you, i understand now, i appreciate the help
     
  9. 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>?
     
  10. vela

    vela 12,733
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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.
     
  11. 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]
     
  12. vela

    vela 12,733
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    Perhaps. In the case of the simple harmonic oscillator, you certainly can. Did you have a specific problem in mind?
    Nope. I don't see how you got that from the linearity of V(x).
     
  13. Yes, I was given V(x)=1/(cosh(x-pi/2))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.

    OK, seems I misunderstood.
     
  14. vela

    vela 12,733
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    You should start a new thread with your problem and showing your work so far.
     
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