The sodium D lines are due to ${}^{2}P_{1/2} \rightarrow {}^{2}S_{1/2}(\Delta E_1)$ and ${}^{2}P_{1/2} \rightarrow {}^{2}S_{1/2}(\Delta E_2)$ transitions. The splitting due to spin-orbit coupling in ${}^{2}p$ state of the sodium atom is
(a)$\Delta E_2+\Delta E_1$
(b)$\Delta E_2-\Delta E_1$
(c)$\dfrac{\Delta E_2+\Delta E_1}{2}$
(d)$\dfrac{\Delta E_2-\Delta E_1}{2}$
Answer
Answer: B ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
The D lines are ²P₃/₂ → ²S₁/₂ (D₂) and ²P₁/₂ → ²S₁/₂ (D₁); both end on the same level, so the spin–orbit splitting of the 3p state is their difference, ΔE₂ − ΔE₁ (17 cm⁻¹).