Analysis of the helium spectrum: para- and ortho-helium

Theory

Helium has two electrons. Their total spin is either \(S = 0\) (singlet, para-helium) or \(S = 1\) (triplet, ortho-helium). Electric-dipole transitions obey \(\Delta S = 0\), so the two systems behave almost like two different elements. The ground state 1¹S has no triplet partner, because the Pauli principle forbids 1s² with parallel spins.

When one electron is excited to \(n\ell\), the inner electron screens the nucleus to an effective charge close to 1. The levels are then hydrogen-like, corrected by a quantum defect \(\delta_\ell\):

$$ E_{n\ell} = E_I - \frac{hcR_{He}}{(n-\delta_\ell)^2},\qquad \sigma = \frac1\lambda = T_{2P} - \frac{R_{He}}{(n-\delta_D)^2} .$$

For D states the outer electron hardly penetrates the core, so \(\delta_D \approx 0.002\). A plot of σ against \(1/n^2\) for the nD → 2P series is then a straight line with slope \(-R_{He}\) and intercept equal to the term value \(T_{2P}\), the binding energy of the 2P level.

Singlet–triplet splitting

The spatial wavefunction is symmetric for singlets and antisymmetric for triplets. Triplet electrons avoid each other, feel less Coulomb repulsion, and lie lower by twice the exchange integral, \(\Delta E = 2K\). From the two series limits, \(\Delta E(2P) = T_{2^3P} - T_{2^1P}\).

Procedure

1. Calibrate the grating with Hg 546.07 nm (left and right readings).
2. Switch to the He lamp and measure every visible line on both sides: yellow 587.6, red 667.8 and 706.5, green 501.6 and 492.2, blue 471.3 and 447.1, violet 438.8 and 402.6, and so on.
3. Assign each line with the Grotrian diagram (click a transition).
4. Plot σ_vac against 1/n² for both D series and fit straight lines.

What the analysis gives you

• \(R_{He}\) for each system. Expected: \(R_\infty/(1+m_e/M_\alpha)\) = 109 722.3 cm⁻¹.
• Term values \(T(2^1P)\) ≈ 27 176 cm⁻¹ and \(T(2^3P)\) ≈ 29 224 cm⁻¹, so the exchange splitting is about 2048 cm⁻¹ (0.254 eV).
• The ionisation energy \(E_I = E(2^1P) + T(2^1P)\), using the 2¹P excitation energy 21.218 eV known from the 58.4 nm resonance line (a UV line, not visible here). Expected: 24.587 eV.
• The effective quantum number \(n^* = \sqrt{R/T_n}\) for each line, and hence the quantum defect.

Precautions

• The helium discharge has several bright red and yellow lines close together. Make sure you are on the right one by checking the wavelength predicted from your calibration.
• The violet lines (402.6, 396.5, 388.9 nm) are faint. Darken the room (here: they are rendered dimmer) and open the slit slightly.

Viva questions

Why are there no transitions between singlets and triplets?

The electric-dipole operator does not act on spin, so ΔS = 0. Spin–orbit coupling mixes the states only weakly in a light atom like He. The intercombination line 2³P→1¹S (59.1 nm) exists, but it is about 10⁷ times weaker.

Why is the triplet lower in energy?

The antisymmetric spatial wavefunction vanishes when r₁ = r₂, which reduces the electron–electron repulsion (the exchange integral K > 0). This is Hund's first rule.

Why is δ largest for S states?

Low-ℓ orbitals penetrate the 1s core, where the outer electron sees more nuclear charge, so they are bound more strongly. That extra binding is the quantum defect.

Why is 2³S metastable?

It cannot decay to 1¹S by an allowed transition: both ΔS = 0 and Δℓ = ±1 are violated. Its lifetime is about 7800 s.