# Spin-orbit coupling constant for rubidium

I have come across the following question in my course notes:

The $5s\to 5p$ transition in rubidium is split into two components with wavelengths of 780nm and 795nm respectively. For the $5p$ state, calculate the spin-orbit coupling constant $a$.

I have calculated the spin orbit splitting energy to be $E=\frac{3}{2} a$, calculated the energy of the 795nm component to be $E=\dfrac{hc}{\lambda}=2.5\times 10^{-19} J$, suggesting that $a=1.67\times 10^{-19}J$. I'm not convinced by this answer as it gives a very large value for the magnetic field of the electron due to its orbital motion.

Is the method I'm using correct, or does something seem to be out of place?

• Show us the calculation and then we can check for errors. – Chris Gerig Dec 8 '12 at 7:35

The spin-orbit interaction is responsible for splitting the $5p$ level in two, hence the two identified $5s \rightarrow 5p$ wavelengths. To calculate the splitting magnitude, you want to be looking at the difference $\Delta E$ between the two transition energies.
By the way, I'm seeing the coupling constant typically characterized in inverse centimeters: $$a = \frac{2}{3} \frac{\Delta E}{hc}$$