Fine structure, hydrogen atom, principal quantum number 3

[fluidistic]

Homework Statement

The level n=3 for atoms with 1 electron have the states [itex]3s_{1/2}[/itex], [itex]3p_{1/2}[/itex], [itex]3p_{3/2}[/itex], [itex]3d_{3/2}[/itex], [itex]3d_{5/2}[/itex]. If we ignore the spin-orbit coupling these states are degenerated. Calculate the degeneration due to the the spin-orbit coupling for the levels 3p and 3d for the hydrogen atom.

Homework Equations

[itex]E_{n,j}=E_n \left [1+\frac{Z^2 \alpha ^2}{n} \left ( \frac{1}{j+1/2}-\frac{3}{4n} \right ) \right ][/itex]. I've found this formula in books and in wikipedia. My professor gave us a slightly different formula so I'm confused which one to use.
His formula: [itex]E_{n,j}=E_n - \frac{Z^4 \alpha ^2}{n^3}E_0 \left ( \frac{2}{2j+1} -\frac{3}{4n} \right )[/itex].

The Attempt at a Solution

I reach, using the book/wikipedia formula that for the states with j=5/2, [itex]\Delta E \approx -2.24 \times 10 ^{-6}eV[/itex].
For the states with j=3/2, [itex]\Delta E \approx -6.70 \times 10 ^{-6}eV[/itex].
For the states with j=1/2, [itex]\Delta E \approx -2.46 \times 10 ^{-5}eV[/itex].
I have the feeling that my numbers are somehow wrong but I am not sure.
Could someone confirm my results? I tried to check up these values in some book and on the Internet but I didn't find them.
Thanks for any input!
Edit: Hmm I think both formula are equivalent.

 

[jbsteele]

Right? Because the second one is just the first one with the factor E_0=m_e c^2 \alpha ^2/2. Therefore I think my results are correct. Right? A:Your calculation is correct, although if you are going to use $\alpha$ you should use the fine structure constant, which is $\alpha \approx 1/137$.The difference in energy between the 3 levels is significant enough that it has been measured and is due to the spin-orbit coupling, which is a relativistic effect.The fact that the 3p level is higher in energy than the 3d level is known as the Lamb shift.