A Zeeman triplet that cannot exist at 0.05 T
A student reports resolving the normal Zeeman pattern of an Fe I line at λ = 500.0 nm (atomic mass 56 u) in a 0.05 T electromagnet. The record shows three components equally spaced by 0.00010 nm, with the central component unshifted. The source is a hollow-cathode lamp at 400 K, Doppler broadening is the dominant linewidth, and the monochromator plus slits are quoted at an instrumental FWHM of 0.0004 nm. The student argues that the three-line pattern proves the transition is a singlet and that the g-factor is 1. No check of the expected magnitude of the splitting was made before the run.
1. Calculate the normal Zeeman splitting expected at 0.05 T in both frequency and wavelength, and compare with the reported 0.00010 nm.
Read the worked answer
Normal Zeeman splitting (g = 1, Δm = ±1): Δν = μB B / h = eB/(4πm) = (13.996 GHz T^−1) × 0.05 T = 0.700 GHz.
Convert to wavelength at 500.0 nm: Δλ = λ²Δν/c = (5.000e−7)²(6.998e8)/(2.998e8) = (2.500e−13)(2.334) = 5.83e−13 m ≈ 0.00058 nm.
Comparison: the reported spacing is 0.00010 nm, a factor of ≈ 6 smaller than the prediction. Equivalently, the reported spacing corresponds to Δν = 0.12 GHz, which would require B ≈ 0.12/13.996 ≈ 8.6 mT, not 50 mT.
So either the field was ~6× lower than displayed, or the observed triplet is not the normal Zeeman pattern at all. The claim "three lines ⇒ singlet ⇒ g = 1" is circular: without a magnitude check, three equally spaced bumps are equally consistent with an etalon fringe pattern, a mis-set field, or a contamination line.
2. Is the triplet resolvable at all under these conditions? Quantify.
Read the worked answer
Doppler width (1/e half-width in frequency): Δν_D = (1/λ)·√(2kT/M) √(2kT/M) = √(2 × 1.381e−23 × 400 / (56 × 1.6605e−27)) = √(1.105e−20 / 9.30e−26) = √(1.188e5) = 345 m s^−1 Δν_D = 345 / 5.000e−7 = 6.9e8 Hz ≈ 0.69 GHz
In wavelength: Δλ_D = λ(v/c) = 500.0 nm × (345/2.998e8) ≈ 0.00058 nm.
Now compare the three numbers at 0.05 T:
- Zeeman splitting: 0.00058 nm
- Doppler width: 0.00058 nm - identical
- Instrumental FWHM: 0.0004 nm
With splitting comparable to (not ≫) the Doppler width, the σ and π components merge into a single unresolved profile; there is no way to see three resolved peaks. Independently, the spectrometer cannot resolve 0.0001 nm structure at a 0.0004 nm slit width. So the observation is impossible on two independent grounds, before any question of the transition's multiplet structure arises.
Rule of thumb: to resolve components, the splitting must exceed ~3× the Doppler width and ~2× the instrumental width.
3. What field would be needed to resolve the normal Zeeman triplet, and name two experimental changes that would help just as much as a bigger magnet.
Read the worked answer
Require the splitting to be ~5× the Doppler width (5 × 0.69 = 3.45 GHz) for clean separation: B = 3.45 GHz / 13.996 GHz T^−1 ≈ 0.25 T. If instead one demands the splitting exceed the instrumental width 0.0004 nm, the required splitting in wavelength is 0.0004 nm → Δν = 0.48 GHz → B ≈ 0.035 T; but the Doppler width dominates, so this weaker criterion is misleading - the Doppler limit sets the real floor.
Experimental alternatives, each attacking a different broadening term:
- Cool the source (liquid-N2 hollow cathode, or a discharge at ~77 K). Since Δν_D ∝ √T, dropping from 400 K to 80 K reduces the Doppler width by √5 ≈ 2.24, gaining the equivalent of a >5× field increase.
- Use a heavier species - the same ∝ 1/√M scaling; a mass-200 atom halves the Doppler width relative to Fe.
- Replace the grating monochromator with a Fabry–Pérot étalon or a scanning interferometer to move the instrumental limit far below 0.0004 nm.
The deeper point: the student's "three equally spaced lines" should first have been compared with λ²μB B/(hc), the one number that fixes what is physically possible at the stated field.