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
\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