Answer to Question #118516 in Atomic and Nuclear Physics for Moel Tariburu

Question

Explain why the spectroscopic term symbol for Boron in the ground state is 2P3/2.

Answer 1

Electron configuration of Boron is $Z=1s^22s^22p^1$

For ground state$2p^1, s=\frac{1}{2} and$ $l=1$

The possible value of J is $J=l+s=1+\frac{1}{2}=\frac{3}{2}$

And $L=l=1 ,L=1$ this means $L=P$

Spectroscopic term symbol formula is

$^{2s+1}L_J=^{(2(\frac{1}{2})+1)}L_\frac{3}{2}=^2P_\frac{3}{2}$