Compton Scattering: Study Guide (Physics 2 15.6)

A photon that bounces

In Topic 15.5 a photon gave all its energy to an electron and disappeared. In Compton scattering a photon gives only part of its energy to an electron and carries on, in a new direction.

Fire a beam of X-rays at a target whose outer electrons are only loosely held, so you can model them as free electrons at rest. Place a detector at an angle θ to the original beam direction. The detector finds X-rays with a longer wavelength than the incoming beam. The bigger the angle θ, the bigger the increase in wavelength. Arthur Compton reported this in 1923, and the change is called the Compton effect or Compton shift.

What a wave model predicts

Treat the X-rays as a classical wave. The wave’s oscillating electric field shakes the electron at the wave’s own frequency. A shaking charge radiates at the frequency it is shaken at. So the scattered wave should have the same frequency and wavelength as the incoming wave, at every angle. The measured shift does not fit this model.

What a particle model predicts

Treat the X-ray as a stream of photons. Each photon hits one electron, like one ball hitting another. The electron recoils and takes some kinetic energy. The photon must leave with less energy:

E′ = hf′ < E = hf, so f′ < f and λ′ > λ.

The more the photon is deflected, the more energy it hands over, just as a ball that bounces straight back pushes a target harder than a glancing blow. This matches the measurements, so Compton scattering is strong evidence that light comes in discrete energy packets, photons, that behave like particles in collisions.

The momentum of a photon

A photon has no mass, but it does carry momentum:

p = h/λ = E/c

A shorter wavelength means a larger momentum. For a 50.0 pm X-ray photon, p = (6.63 × 10⁻³⁴ J·s) ÷ (50.0 × 10⁻¹² m) = 1.33 × 10⁻²³ kg·m/s. Do not use p = mv for a photon; it has no mass, and it always moves at c in a vacuum.

When the photon’s energy goes down, its frequency goes down, its wavelength goes up and its momentum goes down. Its speed stays at c.

Conservation laws in the collision

The photon and the electron form an isolated system during the collision, so two conservation laws apply.

Energy. The electron starts at rest. The energy the photon loses becomes the electron’s kinetic energy:

hc/λ = hc/λ′ + K_e

Momentum. Momentum is a vector, so conserve each component. Take the incoming photon along +x. The scattered photon leaves at angle θ above the x axis, and the electron recoils at angle α below it (Figure 1).

• x: h/λ = (h/λ′) cos θ + p_e cos α
• y: 0 = (h/λ′) sin θ − p_e sin α

The y equation tells you something useful without any numbers: before the collision there is no y-momentum, so if the photon goes up, the electron must go down. The electron never moves off along the photon’s new direction.

The Compton shift equation

Solving the energy and momentum equations together (with the relativistic form of the electron’s energy, which is beyond this course) gives the shift in wavelength:

Δλ = λ′ − λ = (h / m_e c)(1 − cos θ)

where m_e = 9.11 × 10⁻³¹ kg is the electron mass and θ is the angle between the photon’s old and new directions. You do not need to derive it; you need to be able to use it and explain its features.

The constant h/(m_e c) has units of length. With the course values:

h/(m_e c) = (6.63 × 10⁻³⁴) ÷ [(9.11 × 10⁻³¹)(3.00 × 10⁸)] = 2.43 × 10⁻¹² m = 2.43 pm

Scattering angle θ	1 − cos θ	Δλ
0° (photon carries straight on)	0	0
60°	0.5	1.21 pm
90°	1	2.43 pm
180° (photon bounces straight back)	2	4.85 pm

Features to notice:

• Δλ depends only on θ. It does not depend on the starting wavelength or on the brightness of the beam.
• Δλ rises with θ from 0 at 0° to a maximum of 2h/(m_e c) at 180°.
• The shift is tiny. For a 50 pm X-ray, a 4.85 pm shift is about 10% of the wavelength. For 500 nm visible light it is about 0.001%, far too small to notice. That is why the effect is studied with X-rays.
• Shorter starting wavelengths lose a bigger fraction of their energy, because the same Δλ is a bigger fraction of a small λ.

Worked example 1: energy shared in a collision

Question. An X-ray photon of wavelength 50.0 pm scatters off a free electron at rest. The photon is detected at θ = 60°. Find (a) the scattered wavelength, (b) the energies of the photon before and after, and (c) the kinetic energy of the electron.

1. (a) Δλ = (2.43 pm)(1 − cos 60°) = (2.43 pm)(0.5) = 1.21 pm. So λ′ = 50.0 pm + 1.21 pm = 51.2 pm.
2. (b) Before: E = hc/λ = (6.63 × 10⁻³⁴)(3.00 × 10⁸) ÷ (50.0 × 10⁻¹²) = 3.98 × 10⁻¹⁵ J. In electron volts: 3.978 × 10⁻¹⁵ ÷ 1.60 × 10⁻¹⁹ = 2.49 × 10⁴ eV = 24.9 keV.
3. After: E′ = hc/λ′ = (1.989 × 10⁻²⁵ J·m) ÷ (51.21 × 10⁻¹² m) = 3.88 × 10⁻¹⁵ J = 24.3 keV.
4. (c) Energy conservation: K_e = E − E′ = 3.978 × 10⁻¹⁵ − 3.884 × 10⁻¹⁵ = 9.42 × 10⁻¹⁷ J = 0.589 keV (589 eV).

Interpretation. The photon loses only 2.4% of its energy. Its frequency drops from 6.00 × 10¹⁸ Hz to 5.86 × 10¹⁸ Hz, and its speed stays at c. Keep unrounded values for E and E′: K_e is a small difference between two large numbers, so rounding them early ruins the answer.

Worked example 2: momentum in two dimensions

Question. The same 50.0 pm photon instead scatters at θ = 90°, so it leaves along +y. Find the magnitude and direction of the electron’s momentum, and check that energy is conserved.

1. Scattered wavelength: Δλ = 2.43 pm × (1 − cos 90°) = 2.43 pm, so λ′ = 52.43 pm.
2. Photon momenta: p = h/λ = 6.63 × 10⁻³⁴ ÷ 50.0 × 10⁻¹² = 1.326 × 10⁻²³ kg·m/s (along +x). p′ = h/λ′ = 6.63 × 10⁻³⁴ ÷ 52.43 × 10⁻¹² = 1.265 × 10⁻²³ kg·m/s (along +y).
3. Conserve x-momentum: the scattered photon has no x-component, so the electron’s x-momentum is p_ex = 1.326 × 10⁻²³ kg·m/s.
4. Conserve y-momentum: total y-momentum is zero, so p_ey = −1.265 × 10⁻²³ kg·m/s (downward).
5. Magnitude: p_e = √(1.326² + 1.265²) × 10⁻²³ = 1.83 × 10⁻²³ kg·m/s.
6. Direction: tan α = 1.265 ÷ 1.326, so α = 43.6° below the +x axis.
7. Energy check: K_e = hc/λ − hc/λ′ = 1.84 × 10⁻¹⁶ J (1.15 keV). From the momentum, p_e²/(2m_e) = (1.832 × 10⁻²³)² ÷ (2 × 9.11 × 10⁻³¹) = 1.84 × 10⁻¹⁶ J.

Interpretation. The two values of K_e agree to about 0.1%. The tiny difference comes from relativity: the electron moves at about 7% of the speed of light, so ½mv² is very slightly off. The electron’s momentum (1.83 × 10⁻²³) is larger than the incoming photon’s (1.33 × 10⁻²³). That is not a contradiction: momentum is a vector, and the electron’s downward component cancels the photon’s upward component.

Comparing scattering situations

Questions often ask you to compare without full calculation.

• Two angles, same beam. Larger θ gives larger Δλ, lower E′ and larger K_e.
• Two beams, same angle. The shift Δλ is the same. The beam with the shorter wavelength has more energy per photon and loses a larger fraction of it. For θ = 90°, a 20 pm photon loses about 11% of its energy, while an 80 pm photon loses about 3%.
• Factor of change. Going from θ = 90° to θ = 180° doubles Δλ, because 1 − cos θ goes from 1 to 2.

It also helps to compare this topic with the last one, because exam questions often mix them.

	Photoelectric effect (15.5)	Compton scattering (15.6)
What happens to the photon	Absorbed completely	Survives with less energy, in a new direction
Electron	Bound in a material; needs φ to escape	Treated as free and at rest
Main equation	K_max = hf − φ	Δλ = (h/m_e c)(1 − cos θ)
Conservation laws used	Energy	Energy and momentum
Typical radiation	Visible or ultraviolet	X-rays

Common misconceptions

• “The photon slows down.” It always travels at c in a vacuum. Lower energy means lower frequency and longer wavelength.
• “A photon has no mass, so it has no momentum.” A photon has momentum p = h/λ. Using p = mv for a photon is wrong.
• “The scattered photon has a shorter wavelength.” It has lost energy, so its wavelength is longer.
• “Δλ is bigger for longer wavelengths.” Δλ depends only on θ. The fractional change is bigger for shorter wavelengths.
• “The electron moves off in the same direction as the scattered photon.” The y-momentum must cancel, so the electron moves to the opposite side of the original line.
• Confusing Compton scattering with the photoelectric effect. In the photoelectric effect the photon is absorbed completely. In Compton scattering it survives with less energy.
• Rounding E and E′ before subtracting. The electron’s energy is a small difference, so keep all figures until the end.

Where this leads

Compton scattering completes the evidence that light behaves as particles with energy and momentum. Topic 15.7 turns to the nucleus: in fission, fusion and nuclear decay, conservation laws again decide what can happen. Test yourself with the practice questions, then use the revision notes and the topic checklist to consolidate. For the photon model of absorption, look back at the photoelectric effect.