Spin-orbit coupling constant for rubidium

Question

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?

Answer 1

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: $$ a = \frac{2}{3} \frac{\Delta E}{hc} $$