 Solution

Neglect hyperfine structure and use the electronic weak-field Zeeman splitting regime, where ${\mu }_{B}B$ is small compared with the spin-orbit interaction splitting. The sodium levels are $3s{}^2{S}_{1/2}$, $3p{}^2{P}_{1/2}$ and $3p{}^2{P}_{3/2}$. The Landé g-factor formula givesFor example, inserting $L=1,S=1/2,J=1/2$ gives $3/2+(3/4-2)/(2\cdot 3/4)=2/3$. Each sublevel with magnetic quantum number $m_J$ shifts by $g_J{m}_J{\mu }_{B}B$. Thus the lower level has shifts $\pm {\mu }_{B}B$, the upper $J=1/2$ level has shifts $\pm {\mu }_{B}B/3$, and the upper $J=3/2$ level has shifts $\pm 2{\mu }_{B}B$ and $\pm 2{\mu }_{B}B/3$.

The electric-dipole selection rule allows $\Delta J=0,\pm 1$ except $0\leftrightarrow 0$, and $m_u-m_l=0,\pm 1$. Both fine structure branches are allowed. The emitted photon energy isAll the allowed components of the sodium D-line Zeeman pattern are listed below; the last column is the photon shift in units of ${\mu }_{B}B$:These are four components of D1 and six of D2. Their strengths need not be equal; the table and arrows specify allowed transitions, not intensities.

The $3p{}^2{P}_{3/2}$ level lies above $3p{}^2{P}_{1/2}$, so D2 has the larger zero-field photon energy. ^[ref] In the stated weak-field regime its most positive shift is $5{\mu }_{B}B/3$. Since $\lambda =hc/E_{\gamma }$, the shortest wavelength is therefore produced byThe tempting $m_u=3/2\to m_l=-1/2$ pair has a larger difference of level shifts but is forbidden because $|m_u-m_l|=2$.