 Solution

The limit near 121.8 nm lies about $hc/(121.8\mathrm{nm})\simeq 10.18$ eV above its lower term. Subtracting from the approximately $11.88$ eV ground-state ionization energy puts that lower term near $1.70$ eV: the $3d{}^2{D}_{3/2,5/2}$ pair. The first upper levels are the already identified $4p{}^2{P}_{1/2,3/2}$ pair. Their allowed decays areThe fourth possible pairing, $P_{1/2}\to D_{5/2}$, has $|\Delta J|=2$ and is forbidden as an electric dipole transition. The three lines therefore arise from doublet terms, not a spin-triplet term: “triplet series” counts three spectral lines.

For example, the first two wavelengths together with the $4p_{3/2}$ energy give $E(3d_{3/2})\simeq 1.693$ eV and $E(3d_{5/2})\simeq 1.700$ eV. The longer 854.2 nm wavelength from the same upper level places its lower D level higher. The 866.2 nm transition independently returns the lower D energy within the rounding precision. Increasing the upper principal quantum number gives $np{}^{2}P\to 3d{}^{2}D$ triplets with two nearby lower-term ionization limits near 121.8 nm. The arrows are included in b; their emission direction is P to D, not the reverse. The independent tabulation confirms the D-term position and ordering.^[ref]