Verification of Moseley's law with X-ray fluorescence
Aim: to record the characteristic K- and L-series X-ray spectra of a set of elements with an energy-dispersive detector, verify Moseley's law \(\sqrt{E}\propto (Z-\sigma)\), and determine the screening constants \(\sigma_K\) and \(\sigma_L\).
XRF spectrometer
Multichannel analyser: energy spectrum
Click a peak to fit a Gaussian on a linear background. Drag across the plot to zoom; double-click to reset the zoom.
Observation table
Moseley plot: √(E/Ry) against Z
Results
Theory
When a K-shell vacancy is created (by the primary X-rays), an L electron fills it and emits a Kα photon. Moseley (1913) found that the square root of the frequency is linear in the atomic number. In Bohr's model, with the nuclear charge screened by σ,
$$ E_{K\alpha} = Ry\,(Z-\sigma_K)^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right),\qquad \sqrt{\frac{E}{Ry}} = \sqrt{\tfrac34}\,(Z-\sigma_K),\qquad \sigma_K\approx 1 ;$$ $$ E_{L\alpha} = Ry\,(Z-\sigma_L)^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right),\qquad \sqrt{\frac{E}{Ry}} = \sqrt{\tfrac{5}{36}}\,(Z-\sigma_L),\qquad \sigma_L\approx 7.4 .$$A straight-line fit of \(\sqrt{E/Ry}\) against \(Z\) gives a slope, to compare with \(\sqrt{3/4} = 0.866\) or \(\sqrt{5/36} = 0.373\), and an intercept giving σ = −intercept/slope.
This was the first proof that the atomic number, not the atomic weight, orders the periodic table. It predicted the missing elements 43, 61, 72 and 75.
Physics built into the simulated spectrum
- Characteristic lines appear only if the tube voltage exceeds the absorption edge (e.g. Sn K needs more than 29.2 kV). Try lowering the kV.
- Kα₁/Kα₂ (2:1) are unresolved, so the fitted centroid lies about 7–35 eV below Kα₁.
- Rayleigh and Compton scattered Mo Kα/Kβ tube lines, a bremsstrahlung background, and Si escape peaks at E − 1.74 keV.
- Detector resolution FWHM = √(N² + 2.355²·F·ε·E), with Fano factor F = 0.115 and ε = 3.65 eV. Statistics are Poisson.
Procedure
- Mount an element foil, set 35 kV / 0.5 mA, and acquire for 60 s of live time.
- Identify the Kα and Kβ peaks (and the Lα/Lβ peaks for heavy elements). Click a peak to fit it, then add it to the table with the correct line label.
- Repeat for at least 8 elements spanning Z = 20–50 for K lines, and the heavy elements for L lines.
- Plot √(E/Ry) against Z for each series and fit straight lines.
Error analysis
The centroid uncertainty from the Gaussian fit is \(\approx \sigma_{peak}/\sqrt{N_{net}}\), typically below 1 eV for strong peaks. The scatter about Moseley's line is much larger. It is physical: relativistic corrections grow as \((Z\alpha)^2\), so the data curve slightly upward and the fitted slope exceeds the Bohr value.
Viva questions
Why σ ≈ 1 for Kα?
When the K-shell vacancy is filled, the L electron sees the nucleus screened by the one remaining 1s electron, so Z_eff ≈ Z − 1.
Why is σ_L much larger?
An M electron filling an L vacancy is screened by the two remaining K electrons and the seven remaining L electrons.
Why no K lines from Sn at 25 kV?
The primary photons must have more energy than the K-shell binding energy (29.2 keV) to create a vacancy. The maximum photon energy equals the tube voltage (Duane–Hunt limit).
Why does Kβ have higher energy than Kα?
Kβ comes from the M shell (n = 3 → 1), which is further from the K shell in energy than L (n = 2 → 1).