X-ray Tube Spectrum

The spectrum

Mo anode (Z = 42) at 50 kV. The continuum starts at 0.2480 Å and peaks at 0.4959 Å.

The curve is Kramers’ law (1923), I(λ) ∝ (λ/λ_min − 1)/λ², which is the spectrum of an idealised target: every electron stops in one step, and nothing absorbs the photons on the way out. Its maximum falls at exactly 2λ_min — substitute u = λ/λ_min and the derivative of (u − 1)/u² vanishes at u = 2, whatever the voltage. A real tube peaks nearer 1.5λ_min, because the anode reabsorbs its own soft radiation and the beryllium window absorbs more of it — both of which eat the long side of the curve and neither of which 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 — and this site ships no edge energies and predicts none. What it can give is a bound that needs no extra data: the edge lies above Kβ, because Kβ only has to reach the M shell while the edge has to reach the continuum. So 19.609 kV is a voltage the tube must exceed — a necessary condition, not a sufficient one. The true figure for Mo is a little higher; for the lighter anodes the gap is under a per cent, and it widens 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.

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