# Homework Help: Griffiths QM 6.22

1. Feb 20, 2015

### snatchingthepi

1. The problem statement, all variables and given/known data

Starting from

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

and using

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

and

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

and

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

where 'a' is the Bohr radius,

and

$$\left<S \cdot L\right> = \left<S_x\right>\left<L_x\right> + \left<S_y\right>\left<L_y\right> + \left<S_z\right>\left<L_z\right>$$

derive

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

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

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

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

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

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

$$\frac{\alpha^4mc^2}{8n^4}$$

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. Feb 20, 2015

### barefeet

13.6 eV is just the result of all the constants in the $E_n$ term. Off the top of my head, I think you can write :
$$E_n = \frac{-13.6}{n^2}$$
But do check this

3. Feb 23, 2015

### snatchingthepi

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.