In short

An X-ray is a photon with wavelength roughly 0.01 to 10 nanometres — between ultraviolet and gamma rays. You make one by accelerating an electron through a voltage V, smashing it into a heavy metal target (tungsten, molybdenum, copper), and catching whatever photon emerges.

The emitted spectrum has two parts superposed:

  1. A continuous spectrum (bremsstrahlung) — electrons decelerate in the target and radiate, across all wavelengths above a sharp lower cutoff set by energy conservation:
\lambda_\text{min} = \frac{hc}{eV}.

At 50 kV, \lambda_\text{min} = 0.0248 nm. This is the Duane–Hunt limit.

  1. A line spectrum (characteristic X-rays) — sharp peaks sitting on top of the continuum, at wavelengths that depend only on the target element. The strongest are the line (L-shell electron fills a K-shell hole) and the line (M-shell fills a K-shell hole). They obey Moseley's law:
\sqrt{f} = a(Z - b),

where f is the line frequency, Z is the atomic number of the target, and a, b are constants (for Kα, a \approx 4.96\times10^7 Hz^{1/2}, b \approx 1).

The rest of X-ray physics — medical imaging, airport security, computed tomography (CT), X-ray diffraction of crystals, stellar X-ray astronomy — is applied Duane–Hunt and applied Moseley.

Walk into the radiology department at AIIMS Delhi on any weekday. The technologist slides a film cassette behind a patient's hand, presses a button for a fraction of a second, and out comes an image that a physician can read — black where soft tissue let the beam through, white where bone stopped it. What lit that film is a beam of photons whose wavelength is about ten thousand times shorter than visible light. Each photon carries roughly ten thousand times the energy of a photon of yellow light. That is why it can pass through three centimetres of flesh without being absorbed, and why it gets stopped by a millimetre of calcified bone: energy enough to interact with inner electrons, not low enough to be scattered by outer ones.

These photons did not exist five minutes before the exposure. They were made, on demand, inside a sealed glass tube the size of a mango. A hot filament in one end of the tube boils off electrons. A potential difference of 50,000 volts (or more) between the filament and a chunk of tungsten at the other end accelerates those electrons up to about 40 per cent of the speed of light. When an electron slams into the tungsten, most of its kinetic energy becomes heat in the metal. But a small fraction — a percent or two — gets radiated away as a photon. That photon is an X-ray.

The spectrum of X-rays coming out of such a tube has two components layered on top of each other, and they have completely different origins. One is a smooth, broad hump that depends only on the tube voltage — push more energy in and the hump extends to shorter wavelengths. The other is a set of sharp, narrow spikes whose wavelengths depend only on which element the target is made of — switch from tungsten to molybdenum and the spikes jump to new wavelengths. In this article you will derive where each component comes from, from scratch.

The X-ray tube — the apparatus that makes the photons

Before the physics, the hardware. An X-ray tube is a two-terminal vacuum tube:

Schematic of an X-ray tubeA horizontally oriented evacuated glass envelope. On the left, a hot tungsten filament labelled cathode glows. On the right, an angled tungsten anode labelled target. A high voltage V is applied between them, minus on the cathode, plus on the anode. An accelerated electron travels from cathode to anode and on impact produces a burst of X-ray photons leaving the tube through a side window. Cooling water channels behind the anode carry away waste heat.evacuated glass envelopecathode (hot filament)tungsten anode (target)+high voltage supply: V (~ 30–150 kV)e⁻ accelerated through VX-ray photons
A Coolidge-style X-ray tube. A hot tungsten filament (the cathode) releases electrons by thermionic emission. A high voltage $V$ accelerates each electron across the evacuated gap. The electron crashes into a tungsten anode, tilted at roughly 20° so the emerging X-ray beam exits sideways through a thin beryllium window. Only about 1% of the deposited energy becomes X-rays; the remaining 99% becomes heat, which is why the anode is water-cooled and often rotates to distribute the thermal load.

The one number that matters for the photon energies is the accelerating voltage V. An electron crossing a potential difference V gains kinetic energy

K = eV,

Why: work done by the electric field on charge e moving through potential difference V is eV. All of that work becomes kinetic energy of the electron (the tube is evacuated, so nothing slows it down en route).

and slams into the target with that energy. At V = 50 kV, K = 50 keV = 8.01\times10^{-15} J, about 6\times10^4 times a visible-light photon's energy. The photon that comes out can carry at most this much energy — the electron has nothing more to give — and that is the entire content of the Duane–Hunt cutoff that you will derive next.

The continuous spectrum — bremsstrahlung and the Duane–Hunt cutoff

Most of the X-ray spectrum is a smooth, broad continuum. The physics behind it has a German name — bremsstrahlung — that literally means "braking radiation." Here is what happens:

An electron moving at high speed enters the tungsten. It does not hit the nucleus directly — the nucleus is 10^5 times smaller than the atom. Instead, it passes close to a tungsten nucleus and gets deflected by the Coulomb force. During the deflection the electron decelerates (and accelerates transversely), and any accelerated charge radiates electromagnetic waves (a fact you saw in the photoelectric effect article — it is the same radiation-by-acceleration story as in a radio antenna, just at a million times higher frequency).

The electron can lose anywhere from zero to all of its kinetic energy in a single encounter. If it barely grazes the nucleus, it radiates a low-energy photon. If it brakes hard, it radiates a high-energy photon. A typical electron makes many such encounters before coming to rest, radiating a photon of some energy at each one. The cumulative result is photons of every energy up to — but not beyond — the electron's initial eV.

The upper bound on photon energy is absolute. An electron whose total kinetic energy is eV cannot radiate a photon carrying more than eV — that would violate conservation of energy. Nothing else is conserved strictly in a single encounter (the struck atom can recoil, electrons can lose any fraction of their energy, etc.), so photons at all lower energies are possible. The maximum photon energy is therefore

E_\text{max} = eV.

Step 1. Relate photon energy to wavelength using E = hf and f = c/\lambda.

E_\text{max} = hf_\text{max} = \frac{hc}{\lambda_\text{min}}.

Why: shorter wavelength means higher frequency and higher energy. The maximum-energy photon is the minimum-wavelength photon.

Step 2. Set E_\text{max} = eV and solve for \lambda_\text{min}.

\frac{hc}{\lambda_\text{min}} = eV \;\Longrightarrow\; \boxed{\;\lambda_\text{min} = \frac{hc}{eV}\;}

Why: the electron's entire kinetic energy becomes one photon in the extreme case. This happens with vanishingly small probability, but the bremsstrahlung continuum runs right up to (and stops at) this wavelength. Beyond it, the spectrum is cleanly zero — you cannot extract a photon carrying more energy than you put into an electron.

Step 3. Plug in numbers for V = 50 kV. Use the handy combination hc = 1240 eV·nm.

\lambda_\text{min} = \frac{hc}{eV} = \frac{1240\ \text{eV·nm}}{50\,000\ \text{eV}} = 0.0248\ \text{nm} = 24.8\ \text{pm}.

Why: the hc = 1240 eV·nm shortcut (seen already in photons) collapses the calculation to one division. At 50 kV the shortest wavelength is 24.8 picometres — about 200 times shorter than the smallest visible-light wavelength. That is why X-rays can resolve features at the scale of the distances between atoms in a crystal.

The relation \lambda_\text{min} = hc/(eV) is called the Duane–Hunt limit or the Duane–Hunt law, after the 1915 measurement that confirmed it. It is the cleanest quantitative test of the photon picture of light outside the photoelectric effect. At a given tube voltage V, no photon emerges at a wavelength shorter than hc/(eV). Measure \lambda_\text{min} by dispersing the X-ray beam with a crystal, and you get a direct determination of Planck's constant — the ratio eV/\lambda_\text{min} = hc is a quantity made out of measurable tube parameters and the resulting cutoff wavelength, with h as the only unknown.

The characteristic spectrum — why there are sharp lines on top

Measure the actual X-ray spectrum carefully and you find the smooth bremsstrahlung hump is decorated with a few sharp, tall spikes. A typical tungsten spectrum has a spike at 20.9 pm (the Kα line) and another at 18.4 pm (Kβ). Switch the target to copper and the spikes jump to 154 pm and 139 pm. Molybdenum: 71 pm and 63 pm. Same shape each time, wavelengths different.

What is producing these lines? The answer is the same Bohr-style energy-level physics that produced the hydrogen spectrum, except now the atom is a heavy element with 29 or 42 or 74 electrons, and the transitions happen deep in the inner shells rather than in the outer ones.

Here is the two-step process:

  1. Knocking out an inner electron. An incoming 50 keV electron has enormous energy compared to the outer-shell electrons of a tungsten atom (whose binding energies are a few eV). It can reach right down and eject a K-shell electron (binding energy about 70 keV for tungsten — but this is slightly less than the 50 keV... see below). For a lighter element like copper (K binding 8.98 keV), a 50 keV electron easily knocks out a K electron.

  2. Filling the vacancy. The hole in the K-shell is filled by an electron dropping from a higher shell (L, M, N, ...). When it drops, the atom emits a single photon whose energy equals the difference in binding energies:

h f = E_\text{higher} - E_\text{lower}.

Why: inner electrons feel almost the full nuclear charge because they are closer to the nucleus than outer electrons. When an L-shell electron drops into a K-shell vacancy, the energy of the photon released is tens of keV — deep X-ray territory. The characteristic lines are exactly the hydrogen-like Bohr transitions, scaled up by the near-full nuclear charge an inner electron sees.

This gives you a naming convention straight out of spectroscopy notation:

The Greek letter indicates how far the electron fell (α for one level, β for two, γ for three). The letter in front (K, L, M) indicates which shell the final state is in.

Origin of characteristic X-raysA schematic atom with a dense nucleus at the centre and three circular electron shells labelled K (closest), L (middle), M (farthest). An incoming high-energy electron from the left collides with and ejects a K-shell electron, leaving a vacancy. A curved arrow shows an L-shell electron dropping into the K vacancy, accompanied by a photon labelled K-alpha leaving the atom. A second arrow shows an M-shell electron dropping into the K vacancy, labelled K-beta.nucleus (+Ze)KLMincoming 50 keV e⁻ejects K electronL→K: Kα photon emittedM→K: Kβ photon
The two-step origin of characteristic X-rays. An incoming high-energy electron knocks out a K-shell (inner) electron. The vacancy is filled by an L-shell electron dropping in, releasing a **Kα** photon. If the vacancy is filled by an M-shell electron instead, a **Kβ** photon is released. The photon energies are entirely set by the binding-energy differences of the target element — not by the tube voltage.

Moseley's law — why \sqrt{f} is linear in Z

In 1913, the same year Bohr published his hydrogen atom, the Manchester experimentalist Henry Moseley measured characteristic X-ray wavelengths for every element he could get his hands on — from aluminium to gold, crossing most of the periodic table. When he plotted \sqrt{f_{K\alpha}} versus Z (atomic number), he got a straight line:

\sqrt{f_{K\alpha}} = a(Z - b),

with b \approx 1 (so Z - b is almost, but not quite, the atomic number itself). This is Moseley's law, and it is derivable from Bohr's model with one small modification.

Step 1. For a hydrogen-like ion of nuclear charge Ze, the Bohr energy levels are

E_n = -\frac{13.6\ \text{eV}\cdot Z^2}{n^2}.

Why: you derived this in the Bohr's model article. Replacing e^2 with Ze^2 in the Coulomb attraction scales the energy by Z^2.

Step 2. A tungsten K-shell electron does not quite see the full nuclear charge Z e. There is one other K-shell electron sitting at roughly the same distance from the nucleus, and on average it screens one unit of the nuclear charge. The electron of interest effectively experiences a reduced charge (Z-1)e. This inner screening is the small correction Moseley's b \approx 1 captures.

E_n \approx -\frac{13.6\ \text{eV}\cdot (Z - 1)^2}{n^2}.

Why: Pauli exclusion puts two electrons (opposite spins) into the K shell. When one is present as a spectator, it reduces the effective nuclear charge seen by its partner by about one unit. This is a first-order approximation; the exact screening is slightly less than one because the spectator electron is not a rigid shell.

Step 3. For the Kα transition (n_i = 2 \to n_f = 1) the photon energy is the difference:

h f_{K\alpha} = E_{n=2} - E_{n=1} = 13.6\ \text{eV}\cdot (Z-1)^2 \cdot \left(\frac{1}{1} - \frac{1}{4}\right) = 13.6\ \text{eV}\cdot (Z-1)^2 \cdot \frac{3}{4}.
h f_{K\alpha} = 10.2\ \text{eV}\cdot (Z - 1)^2.

Why: this is the Bohr transition formula applied to inner electrons, with the screening correction. Because outer electrons are far away and are shielded by inner electrons, they make almost no difference to the K-shell energetics — which is why the characteristic X-ray wavelength depends only on Z and is almost completely insensitive to the chemical state of the atom.

Step 4. Solve for \sqrt{f_{K\alpha}}.

f_{K\alpha} = \frac{10.2\ \text{eV}}{h}\cdot (Z - 1)^2,
\sqrt{f_{K\alpha}} = \sqrt{\frac{10.2\ \text{eV}}{h}}\cdot (Z - 1) = a(Z - 1).

Why: the square root turns the (Z-1)^2 dependence into a linear one. The constant a = \sqrt{10.2\ \text{eV}/h} evaluates to 4.96\times 10^7 Hz^{1/2}.

Compare to Moseley's experimental form \sqrt{f} = a(Z - b): you get b = 1 exactly, from the one-electron screening assumption, and a = 4.96\times 10^7 Hz^{1/2} from the Rydberg energy. Both match Moseley's measurements almost perfectly. (The real b for Kα lines is closer to 1.1 across the periodic table — a 10% correction that comes from finer details of the screening.)

Moseley's result was more than a pretty formula. It gave atomic number a physical meaning — before 1913, elements were ordered in the periodic table by atomic weight, which occasionally put elements out of chemical order (cobalt is slightly heavier than nickel but chemically comes before it). Moseley's Z is the charge on the nucleus, full stop. It is an integer, and every integer from 1 upward corresponds to exactly one element. Gaps in his plot identified elements that had not yet been discovered (promethium, rhenium). Characteristic X-rays are not just a physics result; they remade the periodic table.

Explore the spectrum — the tube voltage in your hands

Drag the slider below from 20 kV to 120 kV and watch the bremsstrahlung cutoff wavelength \lambda_\text{min} = hc/(eV) shrink as the tube voltage rises. The red curve plots \lambda_\text{min} against tube voltage; the two horizontal dashed lines mark the tungsten Kα (20.9 pm) and Kβ (18.4 pm) characteristic wavelengths — the cutoff only crosses these once V is high enough to eject a K-shell electron in the first place.

Interactive: tube voltage and the Duane-Hunt cutoffA plot with tube voltage from 0 to 130 kilovolts on the horizontal axis and minimum wavelength from 0 to 70 picometres on the vertical axis. A red curve shows lambda-min equals 1240 divided by V, dropping steeply from about 60 picometres at 20 kilovolts to about 12 picometres at 100 kilovolts. A draggable red point slides along the voltage axis. Two horizontal dashed lines mark the tungsten K-alpha line at 20.9 picometres and K-beta line at 18.4 picometres.tube voltage V (kV)minimum wavelength λ_min (pm)204060801001200204060Kα (20.9 pm)Kβ (18.4 pm)drag the red point on the V axis
Drag the red point along the horizontal axis to change the tube voltage $V$. The red curve is the Duane-Hunt relation $\lambda_\text{min} = 1240/V$ (with $V$ in kV giving $\lambda$ in pm). The cutoff wavelength falls off as $1/V$. The two horizontal dashed lines mark tungsten's Kα (20.9 pm) and Kβ (18.4 pm) characteristic wavelengths. Medical diagnostic X-rays typically run at 50–120 kV; dental X-rays at 60–70 kV. Below $V \approx 70$ kV a tungsten tube cannot even ionise the K shell, so the characteristic lines do not appear — only the continuum.

Worked examples

Example 1: Duane-Hunt cutoff for a 40 kV dental X-ray

A dental X-ray machine runs at V = 40 kV. Compute the shortest wavelength emitted and the maximum photon energy.

Duane-Hunt cutoff at 40 kVA sketch of the bremsstrahlung continuum rising from zero at lambda equals 31 picometres, the Duane-Hunt cutoff for 40 kilovolts, peaking around 45 picometres, and falling off at longer wavelengths. The sharp cutoff on the short-wavelength side is highlighted in accent colour.λ (pm)intensity316090120λ_min = 31 pm
Bremsstrahlung from a 40 kV tube. The smooth red curve rises sharply at $\lambda_\text{min} = 31$ pm — no photons appear at shorter wavelengths.

Step 1. Use the Duane-Hunt formula \lambda_\text{min} = hc/(eV).

\lambda_\text{min} = \frac{hc}{eV}.

Why: at the cutoff, the incoming electron converts all its kinetic energy eV into a single photon of energy hc/\lambda. No photon can carry more energy than the electron brought in.

Step 2. Plug in numbers using the hc = 1240 eV·nm shortcut.

\lambda_\text{min} = \frac{1240\ \text{eV·nm}}{40\,000\ \text{eV}} = 0.0310\ \text{nm} = 31.0\ \text{pm}.

Why: the voltage V = 40 kV in the denominator gives electron energy eV = 40\,000 eV. The hc shortcut — 1240 eV·nm — is the product of Planck's constant and the speed of light when you want the answer in nanometres.

Step 3. Compute E_\text{max}.

E_\text{max} = eV = e \cdot 40\,000\ \text{V} = 40\ \text{keV} = 6.4 \times 10^{-15}\ \text{J}.

Why: 1 eV = 1.602\times 10^{-19} J, so 40 keV = 40\,000\cdot 1.602\times 10^{-19} = 6.41 \times 10^{-15} J. Or just keep it in electron-volts; X-ray physicists almost always do.

Result. \lambda_\text{min} = 31.0 pm, E_\text{max} = 40 keV.

What this shows. A dental X-ray photon can carry up to 40 keV of energy — about the ionisation energy of the iodine K-shell electron, and well above the 1–2 eV binding energies of typical biological molecules. That is why X-rays damage DNA (enough energy per photon to break any chemical bond) and why the radiographer stands behind a lead screen during exposure. The wavelength 31 pm is about a third the spacing of atoms in a typical salt crystal — perfect for diffracting off crystalline solids but still roughly fifty thousand times shorter than visible light.

Example 2: Identify an unknown target from its Kα line

An X-ray tube of unknown target material produces a characteristic line at wavelength \lambda_{K\alpha} = 154 pm. Use Moseley's law to identify the element.

Moseley plotA plot of the square root of K-alpha frequency versus atomic number Z for several elements. Data points at Z equals 13 (aluminium), 26 (iron), 29 (copper), 42 (molybdenum), and 74 (tungsten) fall on a straight line with slope a and x-intercept at Z equals 1. A dotted horizontal line marks the square root frequency corresponding to 154 picometres, and a vertical dotted line drops to the x-axis at Z equals 29.atomic number Z√f (Hz^½)Al (13)Fe (26)Cu (29)Mo (42)W (74)√f = 1.39×10⁹Z = 29 ⇒ Cu
Moseley's plot of $\sqrt{f_{K\alpha}}$ against atomic number $Z$. The measured line at $\lambda = 154$ pm corresponds to $\sqrt{f} = 1.39\times10^9$ Hz$^{1/2}$, which falls at $Z = 29$ — copper.

Step 1. Frequency from wavelength.

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\ \text{m/s}}{154\times10^{-12}\ \text{m}} = 1.95\times 10^{18}\ \text{Hz}.

Why: f = c/\lambda for any electromagnetic wave. Keep wavelength in metres to get frequency in hertz.

Step 2. Moseley's formula for Kα with a = 4.96\times 10^7 Hz^{1/2} and b = 1:

\sqrt{f} = a(Z - 1).

Why: the screening correction is that one K-shell partner electron reduces the effective nuclear charge by 1, giving b = 1 in the Moseley form.

Step 3. Compute \sqrt{f} and solve for Z.

\sqrt{f} = \sqrt{1.95\times 10^{18}} = 1.396\times10^9\ \text{Hz}^{1/2}.
Z - 1 = \frac{\sqrt{f}}{a} = \frac{1.396\times10^9}{4.96\times10^7} = 28.15.
Z = 29.15.

Why: Z is an integer (every element has an integer atomic number), so Z \approx 29. The small excess over 29 is the b \neq 1 higher-order correction — in practice you round to the nearest integer.

Step 4. Look up Z = 29: it is copper. (The periodic-table row: 1 H, 2 He, 3 Li, ... 28 Ni, 29 Cu, 30 Zn.)

Result. The target is copper. Measured \lambda_{K\alpha} = 154 pm agrees with the handbook value for Cu Kα (154.06 pm) to three significant figures.

What this shows. Two measurements — a single wavelength and a wall chart of the periodic table — are enough to identify an unknown element. This technique is called X-ray fluorescence (XRF) and is the standard method for chemical analysis in industry: point an X-ray beam at a sample, read off which characteristic lines come back, and the list of elements present in the sample pops out. A portable XRF gun is how the BARC-developed field kit used for ore sampling at Jaduguda mines distinguishes uranium-bearing ore from waste rock in seconds — one photon's wavelength, one integer Z, one element identified.

Example 3: Photon counts per electron at 100 kV

A Coolidge X-ray tube runs at V = 100 kV with a tube current I = 20 mA. Assume (generously) that 1% of the electron beam energy converts to X-ray photons, with an average photon energy of 40 keV. Estimate the number of X-ray photons emitted per second.

Step 1. Electron beam power.

P_\text{beam} = V \cdot I = 10^5\ \text{V}\cdot 2\times 10^{-2}\ \text{A} = 2000\ \text{W}.

Why: electrical power is voltage times current. The 20 mA tube current is the rate of electron charge flowing onto the anode, and the 100 kV accelerates each electron through that potential difference.

Step 2. X-ray emission power (1% of beam power).

P_X = 0.01 \cdot P_\text{beam} = 20\ \text{W}.

Why: most of the electron beam energy (99%) goes into heating the anode. Only a small fraction becomes radiation. This is why X-ray tubes need heavy water cooling and why hospital tubes often have rotating anodes to spread the thermal load.

Step 3. Energy per photon in joules.

E_\gamma = 40\ \text{keV}\cdot 1.602\times10^{-19}\ \text{J/eV}\cdot 10^3 = 6.41\times 10^{-15}\ \text{J}.

Step 4. Photon rate.

\frac{\text{photons}}{\text{second}} = \frac{P_X}{E_\gamma} = \frac{20}{6.41\times 10^{-15}} = 3.1\times 10^{15}\ \text{photons/s}.

Why: total radiated power divided by energy per photon is the number of photons emitted per second. Three million billion photons per second — about 6 nanomoles of photons — is a typical X-ray tube output.

Result. About 3\times 10^{15} photons per second.

What this shows. Even at millihole-per-second photon fluxes, a single radiograph at AIIMS exposes the patient to tens of millions of photons passing through the body in a fraction of a second. Of those, most go through; a few percent are absorbed — enough to form an image, far fewer than would harm tissue at the low exposure times used in diagnostic radiography (around 100 ms).

Applications — why this physics matters beyond the lab

X-rays get used everywhere for two reasons. First, the wavelength is comparable to the spacing between atoms in solids (0.1–0.3 nm), so X-rays diffract off crystals — this is the foundation of X-ray crystallography, which determined the structure of table salt in 1913, DNA in 1953, and most of the proteins in the protein databank since 1960. Second, the photon energy is high enough to pass through materials that stop visible light (flesh, luggage walls, bearing housings) but not high enough to penetrate thick lead or dense bone, so X-rays give you shadow images of interiors.

Indian science and medicine use X-rays at every scale:

Common confusions

If you are here for JEE physics — Duane-Hunt, characteristic lines, Moseley's law, basic applications — you have all you need. What follows is the layer for readers who want the quantum-electrodynamics edges, the Compton-scattering connection, the full derivation of bremsstrahlung intensity, and a peek at synchrotron sources.

The spectral shape of the bremsstrahlung continuum — Kramers' formula

The Duane-Hunt argument tells you where the continuum stops, but not the shape of it. The full shape, derived by H. A. Kramers in 1923 with classical electrodynamics plus a quantum cutoff, is

I(\lambda)\,d\lambda \propto \frac{Z}{\lambda^2}\left(\frac{\lambda}{\lambda_\text{min}} - 1\right)d\lambda \qquad (\lambda > \lambda_\text{min}),

and I = 0 for \lambda < \lambda_\text{min}. The intensity rises from zero at the cutoff, peaks near \lambda \approx 1.5\,\lambda_\text{min}, and falls off slowly at long wavelengths. The total radiated power is proportional to Z V^2: higher Z anode or higher tube voltage both boost output, with voltage having a quadratic effect.

Modern QED (quantum electrodynamics) calculations refine Kramers' formula with relativistic and screening corrections, but Kramers' 1923 shape is accurate to about 10% across the X-ray band. You can see the 1/\lambda^2 decay in the interactive spectrum above.

Compton scattering — where the X-ray photon meets the electron again

Fire an X-ray photon at a loose electron. You expect the photon to scatter elastically, the way light from a slide projector bounces off a whiteboard. In fact the scattered photon has a longer wavelength than the incident one, by an amount

\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta),

where \theta is the scattering angle and h/(m_e c) = 2.43 pm is the Compton wavelength of the electron. This is the Compton effect (1923), and it was the direct experimental confirmation that X-ray photons carry momentum p = h/\lambda — that they are quantum particles, not classical waves. You will see the derivation in the Compton scattering article; the relevant fact for X-ray imaging is that some fraction of diagnostic X-rays passing through a patient lose energy and change direction by Compton scattering inside soft tissue. That scattered radiation degrades image contrast (it is what anti-scatter grids remove) and delivers some of the dose to the radiologist's hands if they are not shielded.

Synchrotron X-rays — an alternative source

A relativistic electron moving in a circular storage ring radiates strongly in the forward direction — this is synchrotron radiation. It is bremsstrahlung's cousin (both are acceleration radiation), but the continuous bending in a storage ring is a much cleaner, brighter, more tunable, and more coherent source than a hot filament. Modern third-generation synchrotrons (ESRF in France, APS in Illinois, SPring-8 in Japan, and Indus-2 at RRCAT Indore) produce X-ray beams up to 10^9 times brighter than a rotating-anode laboratory tube.

The full synchrotron spectrum of an electron of energy E curving in a magnetic field B peaks at a critical wavelength

\lambda_c = \frac{4\pi}{3}\frac{R}{\gamma^3}, \qquad \gamma = E/(m_e c^2),

where R is the storage-ring radius. For Indus-2, E = 2.5 GeV, giving \gamma = 4900 and a critical wavelength of about 0.16 nm — in the hard X-ray band. The radiation comes out in a narrow forward cone of half-angle 1/\gamma \approx 0.2 mrad — essentially a laser beam of X-rays — which is why it can be focused onto submicron spots and used to image single protein crystals the size of a speck of dust.

X-ray absorption — Beer's law with the K-edge

When an X-ray beam passes through a material of thickness x, its intensity falls exponentially:

I(x) = I_0 e^{-\mu x},

where \mu is the linear attenuation coefficient of the material at that wavelength. The interesting feature is that \mu jumps sharply upward whenever the photon energy crosses a shell binding energy — at iodine, for instance, the K-edge sits at 33.2 keV, where the photon becomes just energetic enough to ionise a K-shell electron. Contrast agents in medical imaging exploit this: iodine contrast injected into blood vessels shows up dark on X-ray film because the photon energy of diagnostic beams straddles the iodine K-edge, making iodine a ferocious absorber in that band while leaving surrounding tissue almost unchanged.

You can derive \mu from first principles (it involves photoelectric cross-section, Compton cross-section, and pair production for very hard X-rays), but the short version is: \mu \sim Z^4 / E^3 in the photoelectric regime below 100 keV. Denser, heavier-Z materials absorb more strongly, which is why bones (calcium, Z = 20) show up bright on a chest film while fat and air are transparent.

Why characteristic X-rays are not exactly monochromatic

A real Kα line is not a single wavelength but a narrow doublet, Kα_1 and Kα_2, separated by a few parts per thousand. The splitting arises from spin-orbit coupling in the 2p subshell: the 2p_{3/2} and 2p_{1/2} states have slightly different energies, and both can drop to the 1s vacancy, producing two distinct Kα lines. High-resolution X-ray spectroscopy resolves these fine-structure splittings and uses them to identify chemical bonding environments — the Kα line of an atom in a molecule is shifted slightly from its position in the free element, by an amount called the chemical shift, which XPS (X-ray photoelectron spectroscopy) at every materials-science department in India exploits routinely.

Where this leads next