F. Revolutions in Modern Physics: Quantum Mechanics and Special Relativity · Quantum MechanicsPhoton Momentum and the Compton Effect
F2.2 F3.1 F2.1 F1.2About 70 minutes
Learning goals- I can calculate the momentum of a photon using p = h/λ.
- I can describe Compton's X-ray scattering experiment and explain why it is evidence for the particle model of light.
- I can use the Compton equation to find the change in wavelength of a scattered photon.
- I can apply conservation of energy and momentum to photon-electron collisions and explain radiation pressure and solar sails.
Words to know
photon momentum
The momentum carried by a photon: p = h/λ = E/c, even though a photon has no mass.
Compton effect
The increase in wavelength of X-rays or gamma rays scattered by electrons, explained as a collision between a photon and an electron.
Compton wavelength
h/(mₑc) = 2.43 × 10⁻¹² m, the scale of the wavelength shift in Compton scattering.
scattering angle (θ)
The angle between the photon's original direction and its direction after scattering.
radiation pressure
The pressure exerted on a surface by light because photons carry momentum.
solar sail
A large, thin, reflective sheet that propels a spacecraft using the momentum of sunlight.
Can light push?
A particle that carries energy usually carries momentum too. If light is made of photons, do photons have momentum, even though they have no mass? Einstein's special relativity (see the unit Special Relativity) says yes: for anything moving at the speed of light, E = pc. Combining with E = hc/λ gives
p = E/c = h/λ
Shorter wavelength means more momentum. A visible photon carries an incredibly small momentum, but there are a lot of photons. Comets' dust tails point away from the Sun partly because sunlight pushes on the dust. In 2010 the Japanese spacecraft IKAROS unfurled a thin sail about 14 m across and showed that sunlight alone could steer and accelerate it in interplanetary space.
Remember from the unit Momentum and Collisions
Momentum p = mv for ordinary objects. In a collision, the total momentum of the system is conserved, and in an elastic collision kinetic energy is conserved too. In two dimensions, conserve the x- and y-components separately. Compton's experiment applies exactly these ideas to a photon hitting an electron.
Worked exampleExample 1: Momentum of a visible photon
Find the momentum of a 500 nm photon. How fast would an electron have to move to have the same momentum?
- p = h/λ = 6.63 × 10⁻³⁴ J·s ÷ 5.00 × 10⁻⁷ m = 1.33 × 10⁻²⁷ kg·m/s.
- For the electron: v = p/m = 1.33 × 10⁻²⁷ ÷ 9.11 × 10⁻³¹ = 1.46 × 10³ m/s.
p = 1.33 × 10⁻²⁷ kg·m/s, the same as an electron moving at about 1.5 km/s. Tiny for a baseball, but not negligible for an electron.
Compton's experiment (1923)
Arthur Compton aimed a beam of X-rays of a single wavelength at a block of graphite (carbon) and measured the wavelength of X-rays scattered at different angles. A wave model predicts that the electrons would simply vibrate at the incoming frequency and re-radiate X-rays of the same wavelength.
Instead, at each angle Compton found a scattered beam with a longer wavelength, and the shift grew with the scattering angle. He explained it by treating the X-ray as a particle (a photon) that collides with a nearly free electron, like a cue ball hitting another ball:
- The electron recoils and takes some kinetic energy, so the photon loses energy. Less energy means lower frequency and longer wavelength.
- Energy and momentum are both conserved in the collision.
Working through the algebra (with relativistic energy for the electron) gives the Compton equation:
Δλ = λ′ − λ = (h / mₑc)(1 − cos θ)
The constant h/(mₑc) = 2.43 × 10⁻¹² m (2.43 pm) is called the Compton wavelength of the electron. Compton's measurements matched this prediction, and he shared the 1927 Nobel Prize in Physics for the discovery.
Worked exampleExample 2: Scattering at 90°
X-rays of wavelength 0.0710 nm are scattered by graphite. Find the wavelength of the X-rays scattered at 90°.
- Δλ = (h/mₑc)(1 − cos 90°) = (2.43 × 10⁻¹² m)(1 − 0) = 2.43 × 10⁻¹² m = 0.00243 nm.
- λ′ = λ + Δλ = 0.0710 nm + 0.00243 nm = 0.0734 nm.
0.0734 nm, about 3% longer, which Compton could measure with an X-ray spectrometer.
Worked exampleExample 3: Energy given to the electron
A 0.0500 nm X-ray photon bounces straight back (θ = 180°) from an electron. Find the photon's energy before and after, and the kinetic energy of the recoiling electron.
- Δλ = (2.43 pm)(1 − cos 180°) = (2.43 pm)(2) = 4.86 pm = 0.00486 nm. So λ′ = 0.0549 nm.
- Before: E = 1240 eV·nm ÷ 0.0500 nm = 24.8 keV. After: E′ = 1240 ÷ 0.0549 = 22.6 keV.
- By conservation of energy, the electron gains E − E′ = 24.8 − 22.6 = 2.2 keV.
The photon drops from 24.8 keV to 22.6 keV; the electron recoils with about 2.2 keV of kinetic energy.
Why not with visible light?
The largest possible shift is only about 4.9 pm. For a 500 nm visible photon that is a change of about 0.001%, far too small to notice; also, electrons in matter are bound to atoms with a few eV of energy, comparable to the photon's own energy. For X-rays (wavelengths near 0.01 to 0.1 nm and energies of thousands of eV) the shift is a few percent and the electrons behave as if they were free. That is why the Compton effect needed X-rays.
The Compton effect matters in medicine. In X-ray imaging and radiation therapy, many photons are Compton-scattered in the body; this scattered radiation is one reason technologists shield patients and stand behind barriers.
Radiation pressure
When light is absorbed, each photon gives its momentum p to the surface. When light is reflected straight back, the momentum reverses, so the surface receives 2p. For a beam of power P:
- force when fully absorbed: F = P/c
- force when fully reflected: F = 2P/c
Near Earth, sunlight delivers about 1.36 × 10³ W/m² above the atmosphere. On a perfectly reflecting 200 m² sail facing the Sun, F = 2(1.36 × 10³)(200) ÷ (3.00 × 10⁸) ≈ 1.8 × 10⁻³ N. That is roughly the weight of several grains of rice, but it acts continuously with no fuel, and over months it can change a spacecraft's speed considerably.
Just for funRiddle: what has no mass, never stops moving, and can still push a spacecraft?
A photon. It can push you around without weighing anything.
Hands-on activityFeel conservation of momentum in a collision
You need: two coins of the same size (for example two loonies) and a smooth table.
- Put one coin (the "electron") at rest. Flick the other (the "photon") so it hits the first slightly off-centre.
- Watch the directions: the flicked coin goes off at an angle θ and the other recoils at a different angle on the other side of the original line.
- Repeat with a more direct hit. The flicked coin is deflected through a larger angle, and the target coin moves off faster.
- Sketch one collision: the original path, both final paths and the angle θ.
What to look for: the larger the deflection angle of the incoming coin, the more energy it gives to the target and the slower it moves afterwards. In Compton scattering, a photon cannot slow down (it always moves at c), so losing energy shows up as a longer wavelength instead. The coins are only an analogy: the photon has no mass, and the full Compton formula needs relativity.
Your turn
1Find the momentum of an X-ray photon with λ = 0.10 nm.
p = h/λ = 6.63 × 10⁻³⁴ ÷ 1.0 × 10⁻¹⁰ = 6.6 × 10⁻²⁴ kg·m/s.
2Find the momentum of a 1.0 MeV gamma-ray photon. (Use p = E/c.)
E = 1.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 1.6 × 10⁻¹³ J. p = 1.6 × 10⁻¹³ ÷ 3.00 × 10⁸ = 5.3 × 10⁻²² kg·m/s.
3Find the Compton shift for θ = 60°.
Δλ = (2.43 pm)(1 − cos 60°) = (2.43)(0.5) = 1.21 pm (1.21 × 10⁻¹² m).
4X-rays of 0.0200 nm scatter at 60°. Find the scattered wavelength.
λ′ = 0.0200 + 0.00121 = 0.0212 nm.
5At what scattering angle is there no Compton shift? At what angle is it greatest?
No shift at θ = 0° (photon not deflected). Greatest at θ = 180° (photon bounces straight back), Δλ = 4.86 pm.
6Why does the scattered photon have a longer wavelength, not a slower speed?
Photons always travel at c. When the photon loses energy to the electron, its energy E = hf drops, so its frequency drops and its wavelength increases.
7A 2.0 W laser beam is completely absorbed by a target. What force does it exert?
F = P/c = 2.0 ÷ 3.00 × 10⁸ = 6.7 × 10⁻⁹ N.
More practice: warm-up
1Write the formula for the momentum of a photon.
2Who discovered the increase in X-ray wavelength on scattering?
3What is the value of h/(mₑc)?
4Which photon has more momentum, red or blue?
Blue (shorter wavelength)
5Does a reflecting surface or an absorbing surface feel a bigger push from the same light?
A reflecting surface (twice as much: 2P/c vs P/c)
More practice: on your own
1X-rays of wavelength 0.0500 nm scatter at 120°. Find the scattered wavelength and the energy lost by the photon.
Δλ = 2.43(1 − cos 120°) = 2.43(1.5) = 3.64 pm. λ′ = 0.0536 nm. E = 1240 ÷ 0.0500 = 24.8 keV; E′ = 1240 ÷ 0.0536 = 23.1 keV. Lost ≈ 1.7 keV (to the electron).
2At what angle does the Compton shift equal 1.00 pm?
1 − cos θ = 1.00 ÷ 2.43 = 0.412, so cos θ = 0.588 and θ ≈ 54°.
3Find the force on a perfectly reflecting 25 m² sail facing the Sun near Earth (1.36 × 10³ W/m²).
F = 2IA/c = 2(1.36 × 10³)(25) ÷ 3.00 × 10⁸ = 2.3 × 10⁻⁴ N.
4Explain why Compton's results are evidence for the particle model of light.
A wave would scatter at the same wavelength. The measured shift, and how it depends on angle, exactly match a collision in which a photon with energy hf and momentum h/λ hits an electron and both energy and momentum are conserved. That is particle behaviour.
5Why is the Compton effect hard to see with visible light?
The maximum shift (4.9 pm) is about 0.001% of a visible wavelength, too small to detect; also visible photons' energies (a few eV) are similar to the binding energies of electrons, so the electrons do not act as free targets.
Challenge
1A 0.100 nm photon collides head-on with a free electron at rest and bounces straight back. Use conservation of momentum to find the momentum of the recoiling electron.
λ′ = 0.100 + 0.00486 = 0.10486 nm. Take the original direction as positive. Before: p = h/λ = 6.63 × 10⁻²⁴ kg·m/s. After, the photon moves backward: p′ = −h/λ′ = −6.323 × 10⁻²⁴ kg·m/s. Electron: pₑ = p − p′ = 6.63 × 10⁻²⁴ + 6.323 × 10⁻²⁴ = 1.30 × 10⁻²³ kg·m/s (forward).
2A spacecraft of mass 100 kg has a 200 m² perfectly reflecting sail facing the Sun near Earth (force about 1.8 × 10⁻³ N). Ignoring gravity, how much would its speed increase in 30 days?
a = F/m = 1.8 × 10⁻³ ÷ 100 = 1.8 × 10⁻⁵ m/s². t = 30 × 86 400 = 2.59 × 10⁶ s. Δv = at = 47 m/s (about 170 km/h), with no fuel at all.
3Show that the units of h/(mₑc) are metres.
h is in J·s = kg·m²/s. mₑc is in kg·m/s. Divide: (kg·m²/s) ÷ (kg·m/s) = m. ✓
Quiz
1. What is the momentum of a photon with wavelength λ?
2. In the Compton effect, the scattered X-ray photon has
3. For which scattering angle is the Compton shift largest?
4. Why was the Compton effect considered strong evidence for photons?
5. A 0.0300 nm X-ray is scattered at 90°. The scattered wavelength is about
6. A light beam of power P is completely reflected straight back from a mirror. The force on the mirror is
Big idea (F2.2, F3.1): photons carry momentum p = h/λ. In the Compton effect an X-ray photon collides with an electron like a particle, losing energy and gaining wavelength. Light can even push a spacecraft.
How to help: the coin activity is a good way to discuss collisions; ask your teen why the photon gets a longer wavelength rather than slowing down. If they have taken the Momentum and Collisions unit, have them explain which quantities are conserved.
Common difficulties:
- Thinking massless means no momentum. For light, p = h/λ = E/c.
- Using degrees vs radians incorrectly in cos θ (set the calculator to degrees).
- Adding Δλ in picometres to λ in nanometres without converting (1 pm = 0.001 nm).
- Forgetting that a reflected beam gives twice the push of an absorbed one.