Boron Ground State

Boron, with $\bgroup\color{black}$Z=5$\egroup$ has the 1S and 2S levels filled. They add up to $\bgroup\color{black}$j=0$\egroup$ as do all closed shells. The valence electron is in the 2P state and hence has $\bgroup\color{black}$\ell=1$\egroup$ and $\bgroup\color{black}$s={1\over 2}$\egroup$ . Since the shell is not half full we couple to the the lowest $\bgroup\color{black}$j=\vert\ell-s\vert={1\over 2}$\egroup$ . So the ground state is $\bgroup\color{black}$^2P_{1\over 2}$\egroup$ .

$\bgroup\color{black}$m_\ell$\egroup$	e
1	$\bgroup\color{black}$\uparrow$\egroup$
0
-1
$\bgroup\color{black}$s=\sum m_s={1\over 2}$\egroup$
$\bgroup\color{black}$\ell=\sum m_\ell=1$\egroup$

Jim Branson 2013-04-22