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Physics IV course sequence · Interactive lesson

Inner-shell X-rays and Moseley's law

Knock an electron out of the innermost shell and the hole closes again in about a femtosecond, sometimes by emitting a photon. Those photons carry the sharpest fingerprint an element has, and reading them is how the periodic table got its ordering and how an electron microscope names what it is looking at.

UPIV-11.09 · Lesson 9 of 9 in this unit. Position is not a completion score.
Starting model: Inner-shell X-rays and Moseley's lawX-ray tube output against photon energy. The sloping curve is the Kramers bremsstrahlung continuum, falling to zero at the Duane-Hunt cut-off eV₀; riding on it are Kα at 17.2 keV and the weaker Kβ, drawn only once the tube voltage clears the K edge at 20.7 keV. All three energies are screened hydrogenic estimates, with σ = 1, 2 and 3 for Kα, Kβ and the edge, so they sit a few percent off measured values at middle Z and up to 10% off at the heavy end. Line heights use a compressed scale; the continuum is linear.anode Z = 42 tube 70 kVcut-off 70 keV, λₘᵢₙ = 0.0177 nmovervoltage V/EK = 3.38K edge 20.7 keVKα 17.2 keVKβ 19.4 keVintensity / arb.0photon energy / keV100
Start with a picture. Change the model in the lab below.
1

Build the model

Knock out a 1s electron and the atom announces its atomic number.

The key idea

Moseley's law states that the square root of a characteristic X-ray frequency rises linearly with atomic number, √ν = A(Z − σ), where σ is one screening constant standing in for every other electron in the atom.

Example. For the Kα line σ = 1, so E = 10.204 eV × (Z − 1)². Copper, Z = 29, gives 10.204 × 784 = 8.00 keV against a measured 8.05 keV — a 0.6% error from a one-electron formula applied to a 29-electron atom.

The complete physical model and assumptions

Strip a 1s electron out of a heavy atom and the thirty or forty electrons outside barely matter. The hole sits inside every other shell, so an electron falling into it from n = 2 travels in an almost pure Coulomb field of charge Z − 1: the nucleus, minus the one 1s partner still sitting there. That is the entire content of Moseley's law. A hydrogenic formula written for one electron, with Z replaced by Z − σ and a single number σ standing in for the whole rest of the atom, reproduces the Kα frequencies Moseley measured, aluminium to silver, to within about three percent. The payoff is that the square root of frequency plotted against atomic number is a straight line, so one line in a spectrum reads off Z with no chemistry at all, and Z rather than atomic weight is what orders the periodic table. The cost is that σ is a fitted stand-in. It carries no information about configuration, it conceals the spin-orbit doublet that splits every K line in two, and it ignores relativity, so the straight line bends upward past Z of about 50. And the photon is never guaranteed: the same vacancy can be filled by ejecting an outer electron instead, and in light atoms that Auger channel wins more than nine times out of ten.

1Two spectra come off one anode

Point 70 keV electrons at a metal anode and two different things come back. Most of the beam energy simply heats the anode, lost in collisions with atomic electrons, but electrons braking in the nuclear fields radiate a broad continuum — bremsstrahlung — whose intensity per unit photon energy Kramers wrote as I(E) ∝ Z(Eₘₐₓ − E). It stops dead at Eₘₐₓ = eV₀, the Duane-Hunt limit, because no photon can carry more than the whole kinetic energy of one electron; at 60 kV that is 60 keV, or λₘᵢₙ = 20.7 pm. Standing on that continuum are a few narrow lines whose energies do not move when you change the voltage. Raise V₀ and the cut-off slides to higher energy while the whole continuum brightens as V₀²; the lines only grow taller. Change the anode metal and the lines jump somewhere else entirely. That single observation — position set by the target, not by the beam — is what tells you the lines come from the atom's own level structure and not from the collision.

2Why a one-electron formula works on a fifty-electron atom

A K vacancy is a hole in the innermost shell of an atom that may have fifty electrons outside it. An electron dropping from n = 2 into that hole spends its whole journey inside every other shell, where the enclosed charge is the nucleus plus the one 1s electron still present: Z − 1. So the transition is hydrogenic, and E(Kα) = (3/4)(13.606 eV)(Z − 1)². Copper: 10.204 × 28² = 8.00 keV, measured 8.05. Molybdenum: 10.204 × 41² = 17.15 keV, measured 17.48. Tungsten: 10.204 × 73² = 54.4 keV, measured 59.3. The error runs from 0.6% at Z = 29 to 8.3% at Z = 74, and at the heavy end it is not a screening problem — it is relativity. The 1s electron of tungsten moves at roughly Zα = 74/137 = 0.54 c, so it is bound more deeply than a non-relativistic formula allows, and Moseley's straight line curves upward.

3Moseley's straight line, and what it settled

Moseley photographed the K and L lines of about forty elements, aluminium to gold, in 1913-14 and plotted √ν against atomic number. The Kα points fell on a straight line crossing the axis at Z = 1, with slope √((3/4)cR∞) = 4.97 × 10⁷ Hz^½. Two consequences followed at once. First, the abscissa is an integer you can count, so the ordering principle of the periodic table is nuclear charge and not atomic weight — which puts cobalt before nickel even though cobalt is the heavier of the two, 58.93 against 58.69. Second, a gap in the line is a missing element whose line position you can predict in advance; the holes at Z = 43, 61, 72 and 75 were later filled by technetium, promethium, hafnium and rhenium. Chemistry had ordered the elements by similarity and mass. One photographic plate replaced that with counting.

4You must make the vacancy first: the K edge is the threshold

No line appears until something creates the hole. The tube must supply at least the K-shell binding energy EK, which is the same energy as the K absorption edge — the step in the attenuation curve where photons suddenly become able to eject a 1s electron. For molybdenum EK = 20.0 keV, so a 15 kV tube produces bremsstrahlung and no K lines at all, however long you run it. Above threshold the characteristic intensity grows roughly as (V₀ − VK)1.6, which is why analytical work uses an overvoltage of two to three rather than sitting just above the edge. The edge always lies above every line in its own series: Kβ is the 3p → 1s jump, and the series limit, ∞ → 1s, is the edge itself. Molybdenum: Kα 17.5 keV, Kβ 19.6 keV, edge 20.0 keV.

5Selection rules decide which lines exist, and the doublet

Inner-shell transitions obey the same electric-dipole rules as optical ones: Δl = ±1 and Δj = 0, ±1. That is why the strong K lines come from 2p and 3p, and why there is no 2s → 1s line at all, even though the 2s electron is right there and the energy would be within a few percent of Kα. The same rules predict the doublet. Spin-orbit coupling splits the 2p level into 2p₃/₂ and 2p₁/₂, giving Kα₁ and Kα₂ with intensities in the ratio of their degeneracies 2j + 1, that is 4:2, or 2:1. The splitting climbs steeply with Z — 20 eV in copper, 105 eV in molybdenum, 1.34 keV in tungsten — so a spectrometer that shows one copper Kα line will cleanly resolve two in tungsten.

6In light atoms the photon is the minority channel

Filling the hole does not have to make a photon. The atom can hand the released energy to another electron and eject it instead: an Auger electron, leaving a doubly ionised atom. The two channels compete, and the fluorescence yield ωK is the photon's share. Auger rates are almost independent of Z while the radiative rate climbs roughly as Z⁴, so ωK ≈ Z⁴/(Z⁴ + 9.0 × 10⁵) to within a few percent above Z ≈ 20. Aluminium manages 0.036, thirty-six photons per thousand vacancies; copper 0.44; tungsten 0.96. This is the practical ceiling on X-ray fluorescence analysis of light elements, and the reason Auger spectroscopy is the technique of choice for them: what does not leave as a photon leaves as an electron, and an electron cannot escape from more than about a nanometre of solid, so it reports on the surface.

2

Key equations

Read the units, assumptions and limits before substituting.

Moseley's law for the Kα line

√ν = A(Z − σ), A = √(3cR∞/4) = 4.967 × 10⁷ Hz^½

One measured wavelength returns an integer: the line names the element with no chemistry involved.

ν in Hz, Z the atomic number; this A with σ = 1 is the Kα line. Each line has its own A and σ: Lα takes A = √(5cR∞/36) with σ ≈ 7.4

Kα photon energy

E(Kα) = (3/4) R∞hc (Z − 1)² = 10.204 eV × (Z − 1)²

Molybdenum, Z = 42: 10.204 × 41² = 17.15 keV, against 17.48 keV measured.

R∞hc = 13.606 eV; the 3/4 is 1/1² − 1/2², the n = 2 → 1 jump

Lα photon energy

E(Lα) = (5/36) R∞hc (Z − 7.4)² = 1.890 eV × (Z − 7.4)²

Tungsten, Z = 74: 1.890 × 66.6² = 8.38 keV, within 0.2% of the measured 8.40 keV.

5/36 = 1/2² − 1/3²; σ = 7.4 stands for the K and L electrons already in place

Duane–Hunt short-wavelength limit

λₘᵢₙ = hc/(eV₀) = 1.2398 nm⋅kV / V₀

The continuum stops where one electron gives a single photon everything it has: 60 kV → 20.7 pm.

V₀ the tube voltage in kV, λₘᵢₙ in nm; hc = 1239.84 eV⋅nm. No Z appears.

K fluorescence yield

ωK = Γrad/(Γrad + ΓAuger) ≈ Z⁴/(Z⁴ + 9.0 × 10⁵)

Zinc, Z = 30: 0.474 predicted, 0.474 measured. Aluminium radiates only 3.6% of its K vacancies.

Dimensionless, 0 to 1; Γ are level widths in eV. The fit holds above Z ≈ 20.

Kramers bremsstrahlung continuum

I(E) dE ∝ Z(Eₘₐₓ − E) dE, Eₘₐₓ = eV₀

The background a characteristic line has to stand above, and why heavy anodes at high kV are bright.

Intensity per unit photon energy from the anode; integrating gives total ∝ Z V₀²
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

42
20 to 80; steps of 1.
70 kV
10 to 100 kV; steps of 5.

Try this. Set Z to 74 and step the voltage up from 65 kV: only continuum until the tube passes the 69 keV K edge, then a Kα spike that climbs with the overvoltage. Sweep Z instead and both lines slide right as (Z − 1)² while the cut-off holds still and the continuum brightens.

Live results

Kα ENERGY
17.15 keV
K-EDGE THRESHOLD
20.69 keV
CUT-OFF WAVELENGTH
0.0177 nm
OVERVOLTAGE V/EK
3.38

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics IV modelX-ray tube output against photon energy. The sloping curve is the Kramers bremsstrahlung continuum, falling to zero at the Duane-Hunt cut-off eV₀; riding on it are Kα at 17.2 keV and the weaker Kβ, drawn only once the tube voltage clears the K edge at 20.7 keV. All three energies are screened hydrogenic estimates, with σ = 1, 2 and 3 for Kα, Kβ and the edge, so they sit a few percent off measured values at middle Z and up to 10% off at the heavy end. Line heights use a compressed scale; the continuum is linear.anode Z = 42 tube 70 kVcut-off 70 keV, λₘᵢₙ = 0.0177 nmovervoltage V/EK = 3.38K edge 20.7 keVKα 17.2 keVKβ 19.4 keVintensity / arb.0photon energy / keV100
X-ray tube output against photon energy. The sloping curve is the Kramers bremsstrahlung continuum, falling to zero at the Duane-Hunt cut-off eV₀; riding on it are Kα at 17.2 keV and the weaker Kβ, drawn only once the tube voltage clears the K edge at 20.7 keV. All three energies are screened hydrogenic estimates, with σ = 1, 2 and 3 for Kα, Kβ and the edge, so they sit a few percent off measured values at middle Z and up to 10% off at the heavy end. Line heights use a compressed scale; the continuum is linear.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Anode atomic number Z: 42; Tube accelerating voltage: 70 kV

Baseline 42; choose 20 to 80, in steps of 1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The commonest error is to treat the Kα energy and the K absorption edge as one number, then set the tube voltage to the Kα energy and wonder why no line appears. Kα is what is released when an electron falls from n = 2 into a 1s hole; EK is what it costs to lift a 1s electron out of the atom altogether, and it is larger by the binding energy of the L shell. EK is the limit that Kα, Kβ, Kγ converge on from below, so it lies above every line in the series. For molybdenum: Kα 17.5 keV, Kβ 19.6 keV, edge 20.0 keV. Run that anode at 18 kV and you get the continuum and the soft L lines near 2.3 keV, but not one K photon. The gap is exploited in reverse too: zirconium's edge at 18.0 keV falls between molybdenum's two lines, so a zirconium foil absorbs the Kβ and passes the Kα.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

An unknown metal foil bombarded by electrons emits a strong characteristic line at wavelength 154.2 pm. Using E(Kα) = 10.204 eV × (Z − 1)², identify the element.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

A tungsten-anode tube (Z = 74, K edge 69.5 keV) runs at 100 kV. Find the shortest wavelength it emits, estimate the Kα energy from Moseley's law and compare it with the measured 59.3 keV, and estimate what fraction of the K vacancies produce a photon rather than an Auger electron.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

A spectrometer records Kα centroids of 6.400 keV from iron (Z = 26) and 17.44 keV from molybdenum (Z = 42). A third sample gives 22.10 keV. Fit √E = a(Z − σ) to the two standards, identify the unknown, and say how far the fit can be trusted.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

A molybdenum-anode tube (Z = 42, Kα 17.5 keV, Kβ 19.6 keV, K edge 20.0 keV) is run at 18 kV. What comes out of it?
7

Review & continue

Moseley's law states that the square root of a characteristic X-ray frequency rises linearly with atomic number, √ν = A(Z − σ), where σ is one screening constant standing in for every other electron in the atom.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.