= Solution
Neglect <hyperfine structure> and use the electronic weak-field <Zeeman splitting> regime, where $\mu_BB$ is small compared with the <spin-orbit interaction> splitting. The <sodium> levels are $3s\,{}^2S_{1/2}$, $3p\,{}^2P_{1/2}$ and $3p\,{}^2P_{3/2}$. The <Landé g-factor> formula gives
$$
\boxed{g(3s\,{}^2S_{1/2})=2,\qquad g(3p\,{}^2P_{1/2})=\frac23,\qquad g(3p\,{}^2P_{3/2})=\frac43.}
$$
For example, inserting $L=1,S=1/2,J=1/2$ gives $3/2+(3/4-2)/(2\cdot3/4)=2/3$. Each sublevel with <magnetic quantum number> $m_J$ shifts by $g_Jm_J\mu_BB$. Thus the lower level has shifts $\pm\mu_BB$, the upper $J=1/2$ level has shifts $\pm\mu_BB/3$, and the upper $J=3/2$ level has shifts $\pm2\mu_BB$ and $\pm2\mu_BB/3$.
The <electric-dipole selection rule> allows $\Delta J=0,\pm1$ except $0\leftrightarrow0$, and $m_u-m_l=0,\pm1$. Both <fine structure> branches are allowed. The emitted <photon> <energy> is
$$
E_\gamma=E_u^{(0)}-E_l^{(0)}+\mu_BB(g_um_u-2m_l).
$$
All the allowed components of the <sodium D-line Zeeman pattern> are listed below; the last column is the <photon> shift in units of $\mu_BB$:
$$
\begin{array}{c|r|r|r}
\text{upper term}&m_u&m_l&\delta E_\gamma/(\mu_BB)\\ \hline
{}^2P_{1/2}&-1/2&-1/2&+2/3\\
&-1/2&+1/2&-4/3\\
&+1/2&-1/2&+4/3\\
&+1/2&+1/2&-2/3\\ \hline
{}^2P_{3/2}&-3/2&-1/2&-1\\
&-1/2&-1/2&+1/3\\
&-1/2&+1/2&-5/3\\
&+1/2&-1/2&+5/3\\
&+1/2&+1/2&-1/3\\
&+3/2&+1/2&+1
\end{array}
$$
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.
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2018-sodium-zeeman.png]
{title=All ten electronic dipole transitions of the weak-field sodium D doublet}
The $3p\,{}^2P_{3/2}$ level lies above $3p\,{}^2P_{1/2}$, so D2 has the larger zero-field <photon> <energy>. https://physics.nist.gov/PhysRefData/Handbook/Tables/sodiumtable5.htm[NIST, <Energy levels> of Neutral <Sodium>, 3p fine-structure ordering]{ref} In the stated weak-field regime its most positive shift is $5\mu_BB/3$. Since $\lambda=hc/E_\gamma$, the shortest <wavelength> is therefore produced by
$$
\boxed{3p\,{}^2P_{3/2},\ m_J=+\frac12\ \longrightarrow\ 3s\,{}^2S_{1/2},\ m_J=-\frac12.}
$$
The tempting $m_u=3/2\to m_l=-1/2$ pair has a larger difference of level shifts but is forbidden because $|m_u-m_l|=2$.
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