Compton Scattering

Q: What is the Compton wavelength?

It is $h / (m_e c)$, about $2.426 \times 10^{-12}$ m, or 2.426 picometers. It is set by the electron's mass and by the constants $h$ and $c$. The largest shift happens at 180 degrees, where it equals twice the Compton wavelength.

In one sentence: When an X-ray bounces off an electron, it gives away some of its energy, so its wavelength gets longer, and the size of that change depends only on the angle it bounces through.

$\lambda' - \lambda = \frac{h}{m_e c}\left(1 - \cos\theta\right)$

Read it out loud: “Lambda prime minus lambda equals h over m sub e times c, times the quantity one minus cosine theta. Lambda is the wavelength before the collision, and lambda prime is the wavelength after it. The small dash after lambda is called prime. H is Planck's constant, m sub e is the mass of the electron, and c is the speed of light. Theta, said THAY-tuh, is the scattering angle, the angle the photon turns through. Cosine is a function of an angle that gives a number from minus 1 to 1.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Powers of ten, like $10^{-12}$ and $10^{-31}$: Very large and very small numbers are written as a short number times a power of ten. The small raised number, the exponent, counts how many times to multiply by 10, and a minus sign means divide instead. So $10^{-12}$ is a decimal point followed by 11 zeros and then a 1, which is a trillionth. One picometer is $10^{-12}$ m, and one nanometer is $10^{-9}$ m. When you divide powers of ten, subtract their exponents. This is called scientific notation.
X-rays and gamma rays: X-rays and gamma rays are forms of light whose wavelengths are far shorter than visible light. X-ray wavelengths are roughly 0.01 to 10 nm. Their photons each carry much more energy than a visible photon. Medical X-ray images are made with them, and gamma rays come from radioactive decay and from some of the most energetic events in space.
Photon energy and the relation $E = hc/\lambda$: A photon’s energy is set by its wavelength through $E = hc / \lambda$ , where $h$ is Planck’s constant and $c$ is the speed of light. A shorter wavelength means a higher energy. So if a photon gives energy away, its wavelength must get longer. The X-ray in this page’s example has a wavelength of 0.0700 nm, and each of its photons carries about 17,700 eV.
Electron volt (eV) and the kilo prefix (keV): An electron volt, written eV, is the energy one electron gains when it is pushed through a voltage of 1 volt. It equals $1.602176634 \times 10^{-19}$ J, a convenient size for photon energies. The prefix kilo, written k, means one thousand. So 17.7 keV is 17,700 eV, and a kilo-electron volt is 1,000 times the energy of one electron volt.
Angles, degrees and cosine: An angle is measured in degrees, with a full turn being 360 degrees and a right angle being 90 degrees. The cosine of an angle, written $\cos\theta$ , is a number that depends on the angle. $\cos 0^\circ = 1$ , $\cos 90^\circ = 0$ and $\cos 180^\circ = -1$ . The values move smoothly between these. A calculator must be set to degree mode, or it will treat 90 as 90 radians and give a wrong answer.
Free electron and scattering: Scattering means that a particle or a wave is knocked off its original path by a collision with another particle. A free electron is an electron that is not held by an atom, so it can recoil when hit. The electron’s mass, $m_e = 9.1093837 \times 10^{-31}$ kg, is written with a power of ten because the electron is extremely light.

What each symbol means

Symbol	Meaning
$\lambda$	The wavelength of the X-ray or gamma-ray photon before it hits the electron, in meters (m). In the example it is 0.0700 nm, which is $7.00 \times 10^{-11}$ m, far shorter than any visible wavelength.
$\lambda'$	The wavelength after scattering, in meters. The small dash, called prime, means after. It is always equal to or longer than $\lambda$ , because the photon gives some of its energy to the electron.
$\theta$	The scattering angle, in degrees: the angle between the photon’s original direction and its direction after the collision. An angle of 0 degrees means the photon goes straight on, and 180 degrees means it bounces straight back.
$h$	Planck’s constant, $6.62607015 \times 10^{-34}$ J s, with the unit joule seconds, which is energy multiplied by time. It is a fixed number of nature, and in this formula it is combined with the electron’s mass and the speed of light.
$m_e$	The mass of the electron, $9.1093837 \times 10^{-31}$ kg, measured in kilograms. The electron is extremely light, so its mass sets the size of the Compton shift. A heavier particle would shift the wavelength by a much smaller amount.
$c$	The speed of light in empty space, $2.99792458 \times 10^{8}$ m/s, exact by definition. In this formula it combines with the electron’s mass and Planck’s constant into one fixed length, the Compton wavelength of the electron.

The formula piece by piece

$\theta$: Start with the angle, which sets how far the photon turns. An angle of 0 degrees means the photon goes straight on and nothing changes. Larger angles mean a bigger turn and a bigger wavelength change.
$\cos\theta$: The cosine of that angle. It is 1 at 0 degrees, 0 at 90 degrees, and $-1$ at 180 degrees, so it runs smoothly from one extreme to the other as the photon turns from straight on to straight back.
$1 - \cos\theta$: Subtract the cosine from 1. This shift factor is 0 for a photon that goes straight on and 2 for a photon that bounces straight back, so it measures how much of a turn happened.
$\frac{h}{m_e c}$: The Compton wavelength of the electron: Planck’s constant divided by the electron’s mass times the speed of light. It is a fixed length of about $2.426 \times 10^{-12}$ m, or 2.426 picometers, set by the electron alone.
$\lambda' - \lambda$: The change in wavelength: the new wavelength minus the old one. It is never negative for this process. It depends on the angle and the electron, but not on how long the original wavelength was.
$\lambda' - \lambda = \frac{h}{m_e c}\left(1 - \cos\theta\right)$: The whole formula in one sentence: the increase in an X-ray’s wavelength equals the Compton wavelength of the electron times one minus the cosine of the scattering angle.

What the formula is doing, and why

What goes in. An X-ray or gamma-ray photon with wavelength $\lambda$ , a free electron at rest, and the angle $\theta$ through which the photon turns. What happens. The photon hits the electron and gives it some energy and momentum. The photon leaves with less energy, so its wavelength grows to $\lambda'$ . What comes out. The change $\lambda' - \lambda$ , which is called the Compton shift.

Where it came from. In 1923 Arthur Compton measured X-rays that had bounced off electrons and found that some came out with longer wavelengths. Classical wave theory could not explain that change. Treating the X-ray as a photon that collides with an electron explains it, and this formula gives the size of the change. The result is one of the standard pieces of evidence that light carries momentum like a particle.

Why does the photon lose energy? Think of a billiard collision. The photon gives some energy and momentum to the electron, so less is left for the photon. Because $E = hc / \lambda$ , less energy means a longer wavelength. The electron recoils, moving away from the impact with the energy the photon gave up.

Why this shape? The factor $\frac{h}{m_e c}$ is the Compton wavelength, which is fixed by the electron alone. The factor $1 - \cos\theta$ is zero when the photon goes straight on, and it reaches its largest value, 2, when the photon bounces straight back at 180 degrees. So the biggest possible shift is twice the Compton wavelength.

What the formula does not say. The shift does not depend on the starting wavelength. An X-ray of 0.07 nm and one of 0.5 nm get the same absolute shift at the same angle. For X-rays the shift is a noticeable fraction of the wavelength, but for visible light it is far too small to see. The formula assumes a free electron at rest, so it is an approximation for electrons held in atoms.

An everyday picture

Think of a cue ball hitting a resting ball on a pool table. The cue ball slows down and leaves at an angle, and some of its energy moves into the ball it hit. The photon plays the cue ball, and the electron plays the ball it hit. In the same way that a cue ball which gives away energy moves more slowly, a photon that gives away energy leaves with a longer wavelength.

Worked example: An X-ray bouncing off an electron at 90 degrees

An X-ray with wavelength $\lambda = 0.0700$ nm, which is $7.00 \times 10^{-11}$ m, scatters off a free electron at rest and turns through $\theta = 90^\circ$ . A right angle is 90 degrees. We use the Compton wavelength of the electron, $\frac{h}{m_e c} = 2.426 \times 10^{-12}$ m, which equals 0.002426 nm.

Find the shift factor. The angle is 90 degrees, and $\cos 90^\circ = 0$ , because the cosine of a right angle is zero. The shift factor is therefore $1 - \cos 90^\circ = 1 - 0 = 1$ . A right angle means the photon turns sideways, which is a medium-sized turn.
Multiply the Compton wavelength by the shift factor. The shift is $\lambda' - \lambda = (2.426 \times 10^{-12}\ \text{m}) \times 1 = 2.426 \times 10^{-12}$ m. Multiplying by 1 leaves the number unchanged, so the shift equals the full Compton wavelength.
Convert the shift to nanometers. One nanometer is $10^{-9}$ m. Dividing $2.426 \times 10^{-12}$ m by $10^{-9}$ m subtracts the exponents, $-12 - (-9) = -3$ , so the shift is $2.426 \times 10^{-3}$ nm. That is 0.002426 nm, or 2.426 picometers, where a picometer is a trillionth of a meter.
Add the shift to the original wavelength. The new wavelength is $\lambda' = 0.0700$ nm $+ 0.002426$ nm $= 0.072426$ nm, which rounds to 0.07243 nm. The photon came out longer, as the formula predicts for a collision in which the photon gives away energy.
Find the percent change. The shift divided by the original wavelength is $\frac{0.002426}{0.0700} \approx 0.0347$ . Multiplying by 100 gives about 3.5 percent. For an X-ray that is a noticeable change, which is why it can be measured, even though the shift is only a few picometers.
Find the photon energies before and after. Using $E = hc / \lambda$ with $hc = 1239.84$ eV nm, the photon energy before the collision is $\frac{1239.84\ \text{eV nm}}{0.0700\ \text{nm}}$ , which is about 17,712 eV, or 17.7 keV. After the collision it is $\frac{1239.84\ \text{eV nm}}{0.072426\ \text{nm}}$ , which is about 17,119 eV, or 17.1 keV. The photon lost about 593 eV to the electron.

Answer: The scattered X-ray has a wavelength of about 0.07243 nm, about 3.5 percent longer than before. The electron took away about 593 eV, which is about 0.59 keV. Classical wave theory predicts no such change at this angle, so the change is evidence for photons.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Using degrees where the calculator is in radians mode. Check the mode, because $\cos 90^\circ = 0$ , but $\cos 90$ read as radians is about $-0.448$ .
Thinking the scattered photon can have a shorter wavelength. The photon gives energy to the electron, so the new wavelength is equal to or longer than the old one.
Assuming the shift depends on the starting wavelength. For a given angle, the change $\lambda' - \lambda$ is the same for any X-ray, because it depends only on the electron and the angle.
Mixing meters and nanometers. The Compton wavelength is $2.426 \times 10^{-12}$ m, and the X-ray wavelength is 0.0700 nm. Convert both to the same unit before subtracting, because mixed units give an answer that is wrong by a large factor.

Where you will meet it

Gamma-ray and X-ray detectors, where the Compton effect sets the energy that the detector records for a scattered photon.
Radiation shielding calculations, since Compton scattering is one of the main ways gamma rays lose energy in matter.
Astrophysics, where high-energy photons can gain energy by scattering off fast electrons, the reverse of the example on this page.
Physics teaching, where the Compton shift is standard evidence that light behaves as a particle with momentum.

Questions people ask

What is the Compton wavelength?

It is $h / (m_e c)$ , about $2.426 \times 10^{-12}$ m, or 2.426 picometers. It is set by the electron’s mass and by the constants $h$ and $c$ . The largest shift happens at 180 degrees, where it equals twice the Compton wavelength.

Why does the photon lose energy?

Energy and momentum must both be conserved in the collision. The electron gains some momentum and kinetic energy, so the photon has less energy left, and therefore a longer wavelength.

Does visible light show this effect?

The absolute shift is the same size, about 2.4 picometers, at any wavelength. But a visible wavelength is about 500 nm, which is more than 200,000 times longer than the shift. The change is far too small to see, so the effect matters mainly for X-rays and gamma rays.

Photon Momentum: A photon has no mass, yet it still carries a small push, so light can push on objects, and light with a shorter wavelength pushes harder for each photon.
Planck-Einstein Relation (E = hf): E = hf says each packet of light carries energy in proportion to its frequency: higher frequency means more energy per packet, and brightness does not change it.
Photoelectric Effect: Light can knock electrons out of a metal, but only if each tiny packet of light, called a photon, carries enough energy to pay the cost of leaving, and any extra energy becomes the electron’s speed.