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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]

    and

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

    and

    [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,

    and

    [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]


    derive

    [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

    [tex]\frac{\alpha^4mc^2}{8n^4}[/tex]

    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.
     
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