Calculate the photon energy (in kJ/mol) for the single Li emission and the two Na emission wavelengths. To accomplish this, first calculate the energy in units of /photon from Equation (4) and then multiply the result by Avogadro's number to express the energy in /mol of photons. Lastly, divide by 1000 to convert this result to units of kl/mol. Next, use the photon energies to determine the valence orbital energies for both Li and Na. For lithium, the transition is from the 2p- to the 2s-orbital, and the 2s-orbital energy is -520.3 kJ/mol. Use this to find the energy of the 2p-orbitals in Li. For sodium, the higher-energy photon is emitted when the electron drops from one of the 3p-orbitals to the 3s-orbital, while the lower energy photon is emitted when the electron drops from one of the 3d-orbitals to one of the 3p-orbitals. Use these facts, along with the known energy of the 3s-orbital (-495.8 kJ/mol), to find the energies of the 3p- and 3d-orbitals.

Here's what I get  

Explanation:

(a) Lithium

i) Energy of transition line

The strong line of Li appears at 670.8 nm.

The formula for the energy of a photon is

E = hc/λ

If λ = 670.8 nm, then  

ii) Energy of Li(2p) orbital

(b) Sodium

i) Energies of transition lines

The two emission lines of Na are at 589.0 nm and 589.6 nm.

As before, we can calculate the energy of the first line as

ΔE₁ = 3.373 × 10⁻¹⁹ J/photon = 203.1 kJ/mol.

The energy of the second line is 3.369 × 10⁻¹⁹ J/photon = 202.9 kJ/mol.

ii) Energies of the sodium orbitals

The premise in your question is wrong. The two lines do not arise from 3d ⟶ 3p and 3p ⟶ 3s transitions.

Instead, the 3p orbitals are split into two energy levels by spin-orbit coupling. Both lines come from 3p ⟶ 3s transitions. These levels are 203.1 kJ/mol and 202.9 kJ/mol above the 3s level. I have diagrammed the situation in the figure below.

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