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

Question #118516
Explain why the spectroscopic term symbol for Boron in the ground state is 2P3/2.
1
Expert's answer
2020-05-27T10:32:32-0400

Electron configuration of Boron is Z=1s22s22p1Z=1s^22s^22p^1

For ground state2p1,s=12and2p^1, s=\frac{1}{2} and l=1l=1

The possible value of J is J=l+s=1+12=32J=l+s=1+\frac{1}{2}=\frac{3}{2}

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

Spectroscopic term symbol formula is

2s+1LJ=(2(12)+1)L32=2P32^{2s+1}L_J=^{(2(\frac{1}{2})+1)}L_\frac{3}{2}=^2P_\frac{3}{2}


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