- #1

kelly0303

- 579

- 33

$$\delta\nu = K\frac{m_1m_2}{m_1-m_2}+F\delta<r^2>$$

where ##m_{1,2}## are the masses of the 2 isotopes, ##\delta<r^2>## is the change in the mean square charge radius between the 2 isotopes and K and F are some parameters having to do with the electronic transition that is considered. I am a bit confused about how the spin-orbit coupling comes into play. For example, assume that we ignore the spin orbit coupling for now, and we have a transition from an S to a P state. The parameters of this transition are ##\delta\nu_{S-P}## (which we measure) and ##F_{S-P}## and ##K_{S-P}## (which are usually calculated numerically). If we account for the spin orbit coupling (assume it is of the form ##A S\cdot L##), the P state will get split, say, into ##P_{1/2}## and ##P_{3/2}##. Now we have 2 isotope shifts: ##\delta\nu_{S-P_{1/2}}## and ##\delta\nu_{S-P_{3/2}}## and a value of K and F for each of the 2. Is there any relationship between the ##\delta\nu_{S-P}## and ##\delta\nu_{S-P_{1/2}}## and ##\delta\nu_{S-P_{3/2}}##? Or between ##K_{S-P}## and ##K_{S-P_{1/2}}## and ##K_{S-P_{3/2}}## and same for F? Thank you!