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

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