--- title: X-Rays and the Compton Effect module: Origins of the Quantum moduleNumber: 1 lessonNumber: 3 order: 103 summary: > X-rays are short-wavelength electromagnetic waves produced when fast electrons are braked in a target, and their diffraction by crystals lets Bragg's law measure atomic spacings. Compton then scattered X-rays off electrons and found the wavelength shifted by an amount that only a photon carrying momentum hf/c could explain, closing the case for the particle nature of light. topics: [Origins of the Quantum] sources: - book: Tipler & Llewellyn ref: "Ch. 3 — Quantization of Charge, Light, and Energy; §3-4 X Rays and the Compton Effect, Derivation of Compton's Equation" - book: Tipler & Mosca ref: "Ch. 34 §34-2 The Compton Effect" draft: false --- The [photoelectric effect](/quantum-mechanics/old-quantum-theory/the-photoelectric-effect-and-the-photon) showed that light carries energy in quanta $hf$. It did not show that a quantum also carries momentum, or that it can bounce off a particle and conserve both energy and momentum like an ordinary collision. Compton's 1923 experiment did. The stage was set by X-rays — radiation discovered almost by accident, shown to be electromagnetic waves of very short wavelength, and by then a routine tool for probing crystals. ## The nature of X-rays Roentgen found in 1895 that a cathode-ray tube emits penetrating "rays" from the point where the electron beam strikes the glass or a metal target. The rays pass through materials opaque to light, darken photographic film, and are not deflected by magnetic fields. Their origin is classical: an electron slammed to a halt in the target is a decelerating charge, and an accelerated charge radiates. > **Definition (Bremsstrahlung).** The continuous X-ray spectrum emitted when > fast electrons are decelerated ("braked") in the electric fields of target > atoms. The name is German for "braking radiation." An electron that loses all > its kinetic energy in a single stop emits one photon of the maximum possible > energy. The wavelengths, inferred from slight diffraction broadening, lie near $0.01$–$0.10\ \text{nm}$ — a thousand times shorter than visible light and comparable to the spacing of atoms in a crystal. A modern X-ray tube accelerates electrons through tens of kilovolts onto a tungsten anode. $$ % caption: An X-ray tube. Electrons boiled off a hot filament are accelerated % across a high voltage onto a metal target, where they brake and emit X-rays. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % glass envelope \draw[black, thick] (-3.4,-1.4) rectangle (3.4,1.6); % filament (cathode) \draw[acc, very thick] (-3.0,0.4) .. controls (-2.6,0.7) and (-2.6,-0.1) .. (-3.0,0.2); \node[anchor=east, font=\scriptsize, text=acc] at (-3.1,0.1) {hot wire}; \node[anchor=north, font=\scriptsize] at (-3.0,-0.15) {cathode}; % electron beam \draw[->, black, thick] (-2.5,0.3) -- (1.7,0.3); \node[anchor=south, font=\scriptsize] at (-0.4,0.35) {electron beam}; % anode / target \draw[acc, very thick] (2.0,-0.9) -- (2.0,1.1); \draw[acc, very thick] (2.0,0.6) -- (2.6,1.0); \node[anchor=west, font=\scriptsize, text=acc] at (2.05,-0.9) {tungsten target}; \node[anchor=west, font=\scriptsize] at (2.4,-0.4) {anode}; % emitted x rays \draw[->, black] (2.1,0.6) -- (3.0,1.4); \draw[->, black] (2.15,0.45) -- (3.2,0.9); \draw[->, black] (2.15,0.7) -- (2.9,1.9); \node[anchor=south, font=\scriptsize] at (3.2,1.5) {X-rays}; % high voltage source \draw[black] (-3.0,0.0) -- (-3.0,-2.1) -- (2.0,-2.1) -- (2.0,-0.9); \draw[black] (-0.75,-2.1) -- (-0.75,-1.85); \draw[black, very thick] (-0.5,-2.1) -- (-0.5,-1.7); \node[anchor=north, font=\scriptsize] at (-0.6,-2.15) {high voltage $V$}; \end{tikzpicture} $$ The spectrum has two parts. A **continuous** background is the bremsstrahlung. On top of it sit sharp **characteristic lines** whose wavelengths are fixed by the target element (labeled $K_\alpha$, $K_\beta$ historically). The continuous part has a sharp short-wavelength cutoff $\lambda_m$ that depends only on the tube voltage, not on the target. $$ % caption: X-ray tube spectrum. A continuous bremsstrahlung hump carries sharp % characteristic lines set by the target; the short-wavelength cutoff depends only % on the accelerating voltage. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (7.4,0) node[anchor=north east] {wavelength}; \draw[->, black] (0,0) -- (0,4.2) node[anchor=south east, align=left] {intensity}; % continuous bremsstrahlung starting sharply at cutoff \draw[acc, thick, smooth] plot coordinates {(1.3,0)(1.6,1.4)(2.1,2.1)(2.9,2.2)(3.9,1.8)(5.1,1.15)(6.4,0.6)(7.1,0.42)}; % cutoff marker \draw[black, dashed] (1.3,0) -- (1.3,1.6); \node[anchor=south, font=\scriptsize] at (1.3,1.6) {cutoff}; \node[anchor=north, font=\scriptsize] at (1.3,0) {shortest}; % characteristic lines (two spikes) \draw[black, very thick] (3.3,2.05) -- (3.3,3.9); \draw[black, very thick] (4.0,1.75) -- (4.0,3.1); \node[anchor=south, font=\scriptsize, text=black] at (3.3,3.9) {characteristic}; \node[anchor=south, font=\scriptsize, text=black] at (4.15,3.1) {lines}; \end{tikzpicture} $$ Einstein read the cutoff as an inverse photoelectric effect. The most energetic photon appears when an electron surrenders its entire kinetic energy $eV$ to a single photon. With the few-eV work function negligible against a $20{,}000\ \text{eV}$ electron, $$ eV \approx h f_m = \frac{hc}{\lambda_m} \;\Rightarrow\; \lambda_m = \frac{hc}{eV} = \frac{1240}{V}\ \text{nm} \ (V\ \text{in volts}). $$ This is the **Duane-Hunt rule**. It gives another independent route to $h/e$, and it is target-independent because it depends only on the electron's energy, not on which atom stops it. The continuous spectrum was thus explained by the quantum hypothesis; the sharp lines are a fingerprint of the atom's inner shells and wait for the [nuclear atom](/atomic-physics/early-models-and-old-quantum-theory/x-ray-spectra-franck-hertz). ## Bragg diffraction Because their wavelength matches atomic spacing, X-rays diffract off the regular planes of a crystal. Waves reflected from successive parallel planes travel path lengths differing by $2d\sin\theta$, where $d$ is the plane spacing and $\theta$ the glancing angle. Constructive interference — a bright diffracted beam — requires that difference to be a whole number of wavelengths. $$ % caption: Bragg reflection from two crystal planes at glancing angle 30 degrees. % The ray reflecting off the lower plane travels an extra path equal to twice the % spacing times the sine of the angle; a bright beam forms when that is a whole % number of wavelengths. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.15] \definecolor{acc}{HTML}{4A6FA5} % two atomic planes \draw[black] (-3.3,1.4) -- (3.3,1.4); \draw[black] (-3.3,0.0) -- (3.3,0.0); \foreach \x in {-2.8,-2.1,-1.4,-0.7,0.0,0.7,1.4,2.1,2.8}{ \fill[black] (\x,1.4) circle (2.0pt); \fill[black] (\x,0.0) circle (2.0pt); } \node[anchor=west, font=\scriptsize] at (3.35,1.4) {upper plane}; \node[anchor=west, font=\scriptsize] at (3.35,0.0) {lower plane}; % plane spacing d \draw[<->, acc] (-3.1,1.4) -- (-3.1,0.0); \node[anchor=east, font=\scriptsize, text=acc] at (-3.15,0.7) {spacing $d$}; % ray A: incident to P1=(0,1.4) and reflected \draw[->, black, thick] (-2.165,2.65) -- (0,1.4); \draw[->, black, thick] (0,1.4) -- (2.165,2.65); % ray B: parallel, down to P2=(0,0) and reflected \draw[->, black, thick] (-2.6,1.5) -- (0,0); \draw[->, black, thick] (0,0) -- (2.6,1.5); % wavefront perpendiculars from P1 to ray B (dashed) \draw[acc, dashed] (0,1.4) -- (-0.606,0.35); \draw[acc, dashed] (0,1.4) -- (0.606,0.35); % extra-path segments highlighted \draw[acc, very thick] (-0.606,0.35) -- (0,0); \draw[acc, very thick] (0,0) -- (0.606,0.35); \node[anchor=west, font=\scriptsize, text=acc] at (0.75,0.25) {extra path}; % glancing-angle arc at P2, labelled in the open space below \draw[acc] (-0.7,0) arc (180:150:0.7); \node[anchor=north, font=\scriptsize, text=acc] at (-1.5,-0.2) {glancing angle}; \draw[acc] (-1.5,-0.12) -- (-0.55,0.16); \end{tikzpicture} $$ > **Theorem (Bragg condition).** X-rays of wavelength $\lambda$ reflected from > crystal planes of spacing $d$ interfere constructively when > $$ > 2 d \sin\theta = m\lambda, \qquad m = 1, 2, 3, \dots > $$ > A Bragg spectrometer rotates crystal and detector together to hold the incident > and scattered angles equal; measuring the angles that brighten the beam gives > $\lambda$ if $d$ is known, or $d$ if $\lambda$ is known. Bragg's law is the workaday tool of X-ray spectroscopy. Compton used exactly such a spectrometer to measure the wavelength shift that carries his name. ## The Compton effect Scattered X-rays had been noticed to come out "softer" — longer in wavelength, more easily absorbed — than the incident beam. A wave cannot do this: an oscillating electron driven by an incident wave re-radiates at the same frequency. Compton treated the scattering instead as a **collision** between a photon and a single free electron, with the photon carrying both energy and momentum: $$ E = hf = \frac{hc}{\lambda}, \qquad p = \frac{hf}{c} = \frac{h}{\lambda}. $$ The momentum relation $p = E/c$ is the massless limit of the relativistic [energy-momentum relation](/relativity/foundations/relativistic-momentum-energy) $E^2 = (pc)^2 + (mc^2)^2$. The photon strikes a stationary electron, transfers part of its energy and momentum to it, and recoils at reduced frequency. $$ % caption: Compton scattering as a collision. The incident photon transfers % energy and momentum to a stationary electron; the photon scatters through an % angle with longer wavelength, and the electron recoils. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % incident photon \draw[->, acc, very thick] (-3.4,0) -- (-0.2,0); \node[anchor=south, font=\scriptsize, text=acc] at (-2.0,0.05) {incident photon}; % target electron at origin \fill[black] (0,0) circle (2.4pt); \node[anchor=north east, font=\scriptsize] at (0.0,-0.1) {electron at rest}; % scattered photon at angle \draw[->, acc, very thick] (0,0) -- (2.7,1.7); \node[anchor=south west, font=\scriptsize, text=acc] at (1.6,1.05) {scattered photon}; % recoil electron \draw[->, black!70, very thick] (0,0) -- (2.5,-1.35); \node[anchor=north west, font=\scriptsize, text=black!70] at (1.4,-0.8) {recoil electron}; % scattering-angle arc \draw[black] (0.9,0) arc (0:32:0.9); \node[font=\scriptsize] at (1.15,0.28) {scattering angle}; % baseline dashed \draw[black, dashed] (0,0) -- (3.0,0); \end{tikzpicture} $$ ### Deriving the wavelength shift Let the incident photon have momentum $p_1 = h/\lambda_1$ along $x$, the scattered photon $p_2 = h/\lambda_2$ at angle $\theta$, and the recoil electron momentum $p_e$ at angle $\varphi$. Conservation of momentum in the two directions: $$ p_1 = p_2\cos\theta + p_e\cos\varphi, \qquad 0 = p_2\sin\theta - p_e\sin\varphi. $$ Isolating the electron terms and squaring to eliminate $\varphi$: $$ p_e^2 = p_1^2 + p_2^2 - 2 p_1 p_2 \cos\theta. $$ Conservation of energy uses the relativistic electron energy $E_e = \sqrt{(p_e c)^2 + (mc^2)^2}$: $$ h f_1 + mc^2 = h f_2 + E_e \;\Rightarrow\; E_e = (hf_1 - hf_2) + mc^2. $$ Squaring $E_e^2 = (p_e c)^2 + (mc^2)^2$ and substituting both results, the $(mc^2)^2$ and quadratic photon terms cancel, leaving $$ (hf_1 - hf_2)\, mc^2 = h f_1\, h f_2\,(1 - \cos\theta). $$ Dividing by $h f_1 f_2 mc^2$ and using $\lambda = c/f$ turns frequencies into wavelengths: > **Theorem (Compton's equation).** A photon scattered through angle $\theta$ by > a free electron shifts wavelength by > $$ > \Delta\lambda = \lambda_2 - \lambda_1 = \frac{h}{mc}\,(1 - \cos\theta). > $$ > The shift is independent of the incident wavelength. The constant > $$ > \lambda_C = \frac{h}{mc} = \frac{hc}{mc^2} > = \frac{1240\ \text{eV}\cdot\text{nm}}{5.11 \times 10^5\ \text{eV}} > = 0.00243\ \text{nm} > $$ > is the **Compton wavelength** of the electron. The shift ranges from zero at forward scattering ($\theta = 0$) to a maximum of $2\lambda_C = 0.00486\ \text{nm}$ at backscattering ($\theta = 180^\circ$). $$ % caption: Compton wavelength shift (in nm) versus scattering angle in degrees. % It vanishes in the forward direction and grows to twice the Compton wavelength, % 0.00486 nm, straight back, tracing the factor one minus cosine. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (8.0,0); \draw[->, black] (0,0) -- (0,3.9) node[anchor=south east, align=left] {wavelength\\shift (nm)}; % 1 - cos theta curve, theta 0..180 mapped to x 0..7 \draw[acc, thick, smooth] plot coordinates {(0,0)(0.7,0.1)(1.4,0.38)(2.1,0.82)(2.8,1.35)(3.5,1.9)(4.2,2.42)(4.9,2.82)(5.6,3.05)(6.3,3.15)(7.0,3.18)}; % axis ticks (degrees, no degree glyph) \node[anchor=north, font=\scriptsize] at (0,0) {0}; \node[anchor=north, font=\scriptsize] at (3.5,0) {90}; \node[anchor=north, font=\scriptsize] at (7.0,0) {180}; \node[anchor=north, font=\scriptsize] at (4.0,-0.55) {scattering angle (degrees)}; \draw[black, dashed] (3.5,0) -- (3.5,1.9); \draw[black, dashed] (0,3.18) -- (7.0,3.18); \node[anchor=east, font=\scriptsize] at (-0.05,3.18) {0.00486}; \node[anchor=east, font=\scriptsize] at (-0.05,1.9) {0.00243}; \end{tikzpicture} $$ ### Two peaks in the data Compton scattered the $0.0711\ \text{nm}$ $K_\alpha$ line of molybdenum off graphite and measured the scattered wavelength with a Bragg spectrometer. At each angle he found not one peak but two: a shifted peak at $\lambda_2$, and an unshifted peak at the original $\lambda_1$. $$ % caption: Scattered-intensity spectra at increasing angle. Each shows an % unshifted peak from photons that rebound off whole atoms and a shifted peak from % photons scattered by nearly free electrons; the gap widens with angle. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % three stacked mini-spectra \foreach \y/\shift/\lab in {2.6/0.2/{small angle}, 1.3/0.7/{medium angle}, 0.0/1.2/{large angle}}{ \draw[black] (0,\y) -- (5.6,\y); % unshifted peak at x=1.4 \draw[black, thick] (1.4,\y) -- (1.4,\y+0.75); % shifted peak \draw[acc, thick] ({1.4+\shift},\y) -- ({1.4+\shift},\y+0.6); \node[anchor=west, font=\scriptsize] at (5.7,\y+0.3) {\lab}; } \node[black, anchor=south, font=\scriptsize] at (1.4,3.4) {unshifted}; \node[acc, anchor=south west, font=\scriptsize] at (2.6,3.2) {shifted}; \node[anchor=north, font=\scriptsize] at (2.8,-0.1) {wavelength}; \end{tikzpicture} $$ The unshifted peak comes from photons that scatter off tightly bound inner electrons. Bound so firmly that the whole atom recoils, the effective mass in Compton's formula is the atomic mass — roughly $10^4$ electron masses — so $\Delta\lambda \approx h/(Mc)$ is negligible. The shifted peak comes from the outer, nearly free electrons, where the electron mass applies and the full shift appears. The measured variation of $\Delta\lambda$ with angle matched $(h/mc)(1-\cos\theta)$ exactly. ## Worked examples **Shortest X-ray from a picture tube.** Electrons accelerated through $25\ \text{kV}$ striking the tube face produce a Duane-Hunt cutoff $$ \lambda_m = \frac{1240}{V} = \frac{1240}{25{,}000} = 0.050\ \text{nm}. $$ These penetrate matter effectively, which is why such tubes carry shielding. **A Compton measurement.** Suppose a $\theta = 120^\circ$ scattering shifts the wavelength by $1.5\%$. From Compton's equation, $$ \Delta\lambda = \lambda_C\,(1 - \cos 120^\circ) = 0.00243\,(1.5) = 0.003645\ \text{nm}. $$ Since $\Delta\lambda = 0.015\,\lambda_1$, $$ \lambda_1 = \frac{0.003645}{0.015} = 0.243\ \text{nm}. $$ At $\theta = 75^\circ$ the same incident line scatters to $$ \lambda_2 = \lambda_1 + \lambda_C\,(1 - \cos 75^\circ) = 0.243 + 0.00243(0.741) = 0.245\ \text{nm}. $$ The fractional shift is tiny, which is why Compton scattering is seen with X-rays and gamma rays — where $\lambda_1$ is small enough that $\Delta\lambda/\lambda_1$ is measurable — and not with visible light. ## Evidence for the photon By 1923 light had three particle-like behaviors on record. The photoelectric effect showed quantized energy exchange; the Compton effect showed quantized momentum exchange in a clean two-body collision obeying relativistic conservation laws; and pair production, the conversion of a photon into an electron-positron pair, shows a photon converting entirely to matter. | Interaction | Photon behaves as | Conserved quantities | Signature | | --- | --- | --- | --- | | Photoelectric effect | quantum of energy $hf$ | energy | threshold frequency, no time lag | | Compton effect | particle with momentum $hf/c$ | energy and momentum | wavelength shift $\propto (1-\cos\theta)$ | | Pair production | quantum converting to mass | energy, momentum, charge | electron-positron pair above $2mc^2$ | None of these erases the wave behavior — diffraction and the Bragg condition are irreducibly wave phenomena, and the same X-rays that scatter as photons also diffract as waves in the very spectrometer that measures the scattering. Light is both, and which face it shows depends on the experiment. Matter turns out to be the same way: the next module gives particles a [wavelength](/quantum-mechanics/matter-waves/de-broglie-waves-and-electron-diffraction) and builds the [wave-particle duality](/quantum-mechanics/matter-waves/the-uncertainty-principle) into a single framework.[^tl-xray][^tl-compton] [^tl-xray]: Tipler & Llewellyn, §3-4 — Roentgen's discovery, bremsstrahlung, the Bragg condition $2d\sin\theta = m\lambda$, and the Duane-Hunt cutoff $\lambda_m = hc/eV$ as an inverse photoelectric effect. [^tl-compton]: Tipler & Llewellyn, §3-4 and "Derivation of Compton's Equation" — relativistic collision of a photon and free electron giving $\Delta\lambda = (h/mc)(1-\cos\theta)$, Compton wavelength $0.00243\ \text{nm}$, and the two-peak molybdenum-on-graphite data.