Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Perturbation of hydrogen energy due to nucleus

  1. Feb 28, 2005 #1
    Consider the ground state of the hydrogen atom. Estimate the correction [tex]\frac{\Delta E}{E_1s} [/tex] caused by the finite size of the nucleus. Assume that it is a unifromly charged shell with radius b and the potential inside is given by [tex]\frac{-e^2}{4\pi \epsilon b}[/tex]

    Calculate the first order energy eorrection to the ground state and expand in [tex]\frac{b}{a_0}[/tex]. Keep the leading term and observe [tex]\frac{\Delta E}{E_1s}[/tex] for b = 10^-15m.

    Ok, I need help in constructing the interaction W (or H'). Once I get that, I would then calculate the expectation value of it by sandwiching it between [tex]\psi_1s[/tex]. Is this correct and how would I construct the interaction?

    Here is what I have so far

    H0 = (p^2)/2m - e^2/r and H = H0 for r > r0

    H = (p^2)/2m -e^2/(4pi epsilon b) = H0 + H' for r < r0

    Then I would solve for H' and use the perturbation equation. Is this correct ?

    Last edited: Feb 28, 2005
  2. jcsd
  3. Feb 28, 2005 #2
    Ok, i tried it and it is not making any sense. H' somehow does not depend on r. What am I doing wrong?

  4. Feb 28, 2005 #3
    Anything guys...whatever you can suggest would be great.

  5. Mar 1, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    The perturbation is a constant,indeed...The radius of the nucleus is a constant.And because the [tex] \psi_{1,0,0} (r,\theta,\phi) [/tex] is normalized,the integration will be trivial.

  6. Mar 1, 2005 #5
    Are the steps I used correct?
  7. Mar 1, 2005 #6
    Ok, I think I made a mistake. H' does in fact depend on r.

    H' = H0 + e^2/r - e^2/(4pi epsilon b)

    So I am going to get a constant term plus a term that depends on r so there will be some dependence. Where am I slipping up?

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook