X-ray Tube Spectrum

What comes out of an X-ray tube

An X-ray tube accelerates electrons through a few tens of kilovolts and stops them in a metal anode. What comes off is two things at once: a continuous spectrum from electrons braking in the field of the nuclei — Bremsstrahlung — and the characteristic lines of whatever the anode is made of.

The continuum has a hard edge on the short-wavelength side, and it is the one number here you can write down exactly:

$${\lambda }_{\text{min}}=\frac{hc}{eU}$$

• h — the Planck constant, which turns a frequency into an energy
• c — the speed of light; hc/λ is therefore the energy of one photon of wavelength λ
• e — the elementary charge, the charge carried by one electron
• U — the accelerating voltage across the tube, the number you set on the generator; eU is the kinetic energy one electron picks up crossing it

An electron cannot emit a photon carrying more energy than it has. So the shortest wavelength in the beam is set by the accelerating voltage and by nothing else — not by the anode, not by the current. Everything a diffraction experiment uses sits somewhere to the right of that edge.

Everything below is drawn for the tube set on the left. Change the anode and the lines move while the continuum stays put; change the voltage and the edge moves; change the current and the whole spectrum scales, which is the one effect that is a pure multiplier.

The spectrum

Mo anode (Z = 42) at 50 kV. The continuum starts at 0.2480 Å and peaks at 0.3720 Å.

The curve is Kramers’ law (1923): the energy an idealised thick target radiates per unit frequency falls linearly to zero at the limit, I(ν) ∝ Z(ν₀ − ν), and drawn per unit wavelength that is I(λ) ∝ (λ/λ_min − 1)/λ³. Its maximum falls at exactly 1.5λ_min — substitute u = λ/λ_min and the derivative of (u − 1)/u³ vanishes at u = 3/2, whatever the voltage. Counted in photons rather than energy, each carrying hc/λ, the same law is (u − 1)/u² and peaks at 2λ_min instead, so where the continuum “peaks” depends on what is counted; this curve and the I ∝ ZU²i row are both energy. A real tube’s continuum falls away faster than this on the long side, because the anode reabsorbs its own soft radiation and the beryllium window absorbs more of it, and neither is in the formula. The short side is not affected: λ_min itself is exact.

The picture draws one Kα line where the table lists two. Kα₁ and Kα₂ are 0.00429 Å apart, against a frame about 1.0 Å wide — about one part in 225, so at the scale of a whole tube spectrum they are one line and no amount of redrawing separates them. The position marked is the weighted mean, which is what a detector that cannot resolve them measures. You see the doublet in a diffraction pattern instead, at high angle, where Bragg’s law spreads the two apart in 2θ.

The characteristic lines are drawn where they fall, not how tall they are. Their heights relative to each other are meaningful — Kα₁ is twice Kα₂, which this site recovers from its own wavelength constants rather than being told — but their height relative to the continuum is not computed here: it depends on how far the voltage overshoots the edge, on the fluorescence yield and on the target’s self-absorption, and this site holds none of those. In a real diffraction experiment Kα towers over the background by a factor of some hundreds, which is the only reason the method works at all.

The characteristic lines, and whether this voltage reaches them

A K line appears only once the beam can knock a 1s electron out of the anode altogether, which takes the K absorption edge energy: 20.000 kV for Mo. Below it the continuum is all there is, however long the exposure. Until August 2026, this site shipped no edge energies and gave a bound instead — the Kβ photon energy, 19.609 keV, which the edge must exceed because Kβ only has to reach the M shell while the edge has to reach the continuum. That bound was low by 1.95% here, and the shortfall grows with atomic number as the M shell pulls away from the continuum.

All three lines switch on together, at the same voltage: they all begin with a hole in the K shell, so what has to be paid for is the ionisation and not the transition. The Kα₁ : Kα₂ height of 2.00 : 1 is not typed in anywhere — it is solved back out of this site’s own three wavelength constants for Mo, whose Kα value is the weighted mean of the other two. That is a consistency check on the three constants rather than a measurement of the ratio: the mean was formed with this weighting, so recovering it proves the three numbers agree with each other. What makes the ratio 2 : 1 in the first place is the row above — L₃ holds four electrons and L₂ two.

Where the lines come from: one atom of Mo

A characteristic line is two events, and only the first one costs the voltage. An electron from the beam knocks a 1s electron clean out of the anode atom; the atom is then a K-shell hole with an outer electron directly above it, and one falls in. The photon carries away exactly the difference between the two levels, so the energy of each line is the height of the level it fell from — which is what lets the picture be drawn to scale from nothing but this site’s own wavelengths. Kα is 2p → 1s and Kβ is 3p → 1s. 2s never appears, though it lies between them: a transition has to change the orbital angular momentum by one, and 2s → 1s does not.

The 2p level is not one level. Spin-orbit coupling splits it into 2p_3/2 and 2p_1/2, 105.1 eV apart for Mo — and that separation is measured here, as the difference between the Kα₁ and Kα₂ photon energies, not looked up. Two starting levels, one destination, two lines. Their 2 : 1 intensity is the ratio of how many electrons are available to fall: 2p_3/2 holds 4 and 2p_1/2 holds 2. That is the same 2 : 1 this page recovers further down from the three shipped wavelengths alone — a count of electrons and a piece of arithmetic on a weighted mean, arriving at one number by roads that share nothing.

The energy column is not a second calculation: with the K shell taken as the zero of the scale, each level’s height is its line’s photon energy, which is how the picture above is placed. The 3p row has no electron count because Kβ’s height is not predicted from occupancy alone — the line drawn here is Kβ_1,3, a 3p → 1s transition from a core level, and it competes for that vacancy with the satellites beside it: Kβ₅ from 3d and Kβ₂ from the valence band, whose intensity depends on the chemical bonding rather than on a shell occupancy. This site does not claim a figure for any of them.

Power, and where nearly all of it goes

Kramers also estimates what fraction of the beam power leaves as X-rays: η ≈ 1.1 × 10⁻⁹ Z U, with U in volts. For this tube that is 0.231%, so of 1,500 W going in, about 3.5 W comes out as X-rays and the rest — 1,497 W — is heat the anode has to lose. The textbook figure of “1% X-rays, 99% heat” is a tungsten tube at 100 kV; a crystallography tube runs a light anode at a third of the voltage, so it is nearer a tenth of a per cent. That is why the anode is water-cooled, why a rotating anode exists at all, and why a microfocus source can manage on 30 W.

That last row is a ratio, not a measurement: the law is written with a proportionality and carries no constant, so it compares two settings of the same tube and says nothing about photons per second. Doubling the current doubles it; doubling the voltage quadruples it, and shifts the whole continuum to shorter wavelengths at the same time.

What was used