1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: Griffiths QM 6.22

  1. Feb 20, 2015 #1
    1. The problem statement, all variables and given/known data

    Starting from

    [tex]E^1_{fs} = \left<n l m_l m_s| (H_r + H_{so})| n l m_l m_s \right>[/tex]

    and using

    [tex]E_r^1 = -\frac{(E_n)^2}{2mc^2}\left[\frac{4n}{l+1/2} - 3\right][/tex]


    [tex]H_{so} = \frac{e^2}{8\pi\epsilon_0}\frac{S\cdot L}{m^2c^2r^3}[/tex]


    [tex]\left<\frac{1}{r^3}\right> = \frac{1}{l(l+1/2)(l+1)n^3a^3}[/tex]

    where 'a' is the Bohr radius,


    [tex]\left<S \cdot L\right> = \left<S_x\right>\left<L_x\right> + \left<S_y\right>\left<L_y\right> + \left<S_z\right>\left<L_z\right>[/tex]


    [tex]E_{fs}^1 = \frac{13.6}{n^3}\alpha^2\left[\frac{3}{4n} - \left(\frac{l(l+1) - m_lm_s}{l(l+1/2)(l+1}\right)\right] [/tex]

    2. Relevant equations

    As above.

    3. The attempt at a solution

    The first bit of this is extremely straight-forward. I substitute the appropriate values and use the S.L stuff above to get

    [tex]E^1_{fs} = \left<n l m_l m_s| (H_r + H_{so})| n l m_l m_s \right> = \frac{-(E_n)^2}{2mc^2}\left[\frac{4n}{l+l/2} - 3\right] + \frac{e^2}{8\pi\epsilon_0}\frac{\hbar^2m_lm_s}{m^2c^2}\frac{1}{l(l+1/2)(l+1)n^3a^3}[/tex]

    but from here on I seem to run into a stumbling block. I just don't how to get out any value of 13.6 [eV]. I know I can reexpress

    [tex]E_n = \frac{-\alpha^2mc^2}{2n^2} = \frac{-m}{2n^2}\left(\frac{e^2}{4\pi\epsilon_0\hbar}\right)^2[/tex]

    but I am not seeing how to move forward. Trying to use this on the the E_n term in the RH of the above I get


    which seems to get me nowhere.

    Could someone give me a kick in the right direction? I'm not seeing where to go with all this mathturbation.
  2. jcsd
  3. Feb 20, 2015 #2
    13.6 eV is just the result of all the constants in the [itex]E_n[/itex] term. Off the top of my head, I think you can write :
    [tex] E_n = \frac{-13.6}{n^2} [/tex]
    But do check this
  4. Feb 23, 2015 #3
    You are right. Looking that up helped a lot. The formula was on the front jacket of the book and I saw a few other definitions (equations) that helped me solve the problem. Thanks a lot.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted