IB Physics HL Topic 5 — Quantum Physics Paper 1 & 2 Δλ = (h/m_ec)(1 − cosθ) ~17 min read

Compton Scattering

The photoelectric effect proved light can act like a particle. Compton scattering nails it down completely: it shows a photon behaving exactly like a billiard ball. Fire a high-energy X-ray photon at a stationary electron and the two collide, obeying conservation of energy and momentum just like two snooker balls. The photon bounces off with less energy — and therefore a longer wavelength — and the electron recoils away. Only a particle can transfer energy and momentum in a collision like this, so Compton scattering is the ultimate evidence for the particle nature of light.

📘 What you need to know

Compton scattering: a high-energy photon (X-ray or gamma) collides with an orbital electron, causing an increase in the photon’s wavelength and the ejection of the electron
The photon transfers some energy & momentum to the electron, so the photon loses energy
Less photon energy means a longer wavelength (since E = hc/λ)
The photon is deflected; the electron recoils in a different direction, conserving momentum
The Compton formula: Δλ = (h/m_ec)(1 − cosθ)
Δλ = λ_f − λ_i is the increase in wavelength; θ is the photon’s scattering angle
h/m_ec is a constant called the Compton wavelength (≈ 2.43 × 10⁻¹² m)
The bigger the scattering angle, the bigger the wavelength shift; maximum shift is at θ = 180°
At θ = 0° there is no change in wavelength; the formula assumes the electron is initially at rest

A photon-electron collision

Picture a moving photon striking a stationary electron. Because a photon carries both energy (E = hf) and momentum (p = h/λ), the collision behaves like any particle collision — energy and momentum are both conserved, shared between the two after impact.

The incident photon (blue, short wavelength) strikes the electron, scatters off at angle θ with a longer wavelength (red, less energy), and the electron recoils away — energy and momentum both conserved.

Photon in
energy E_i

collision
(energy + momentum
conserved)

Photon out
lower E_f

Longer λ
+ recoil electron

Here’s the logic chain to lock in: the photon gives energy to the electron → the photon now has less energy → since E = hc/λ, less energy means longer wavelength. Energy down, wavelength up — they move in opposite directions. That single relationship is the heart of every Compton question, so make sure you can explain why losing energy stretches the wavelength.

The Compton formula

The increase in wavelength depends only on the scattering angle θ of the photon — not on the incident wavelength. This is the equation you’ll use for every calculation:

The Compton scattering formula Δλ = (h / m_ec)(1 − cosθ) Δλ = λ_f − λ_i (increase in wavelength) • θ = photon scattering angle

The constant h/m_ec is the Compton wavelength of the electron — a fixed number, about 2.43 × 10⁻¹² m. The (1 − cosθ) part controls how the shift grows with angle:

Scattering angle θ	(1 − cosθ)	Change in wavelength Δλ
0° (straight through)	0	Zero — no change
90° (sideways)	1	Equal to the Compton wavelength
180° (straight back)	2	Maximum — twice the Compton wavelength

WE 1

An X-ray photon is scattered by a stationary electron through an angle of 90°. Calculate the change in the photon’s wavelength. (h = 6.63 × 10⁻³⁴ J s, m_e = 9.11 × 10⁻³¹ kg, c = 3.00 × 10⁸ m s⁻¹)

Step 1 — the Compton formula at θ = 90° Δλ = (h/mₑc)(1 − cos90°) cos90° = 0, so (1 − cos90°) = 1 Step 2 — substitute the Compton wavelength Δλ = (6.63 × 10⁻³⁴) / (9.11 × 10⁻³¹ × 3.00 × 10⁴) × 1 Δλ = 2.43 × 10⁻¹² m At 90° the shift is exactly the Compton wavelength — a handy result worth remembering. Notice the answer doesn’t depend on the incident wavelength at all; only the angle matters.

WE 2

An X-ray photon of wavelength 0.0500 nm collides with a stationary electron and scatters through 60°. (a) Calculate the wavelength of the scattered photon. (b) Explain what happens to the photon’s energy. (h = 6.63 × 10⁻³⁴ J s, m_e = 9.11 × 10⁻³¹ kg, c = 3.00 × 10⁸ m s⁻¹)

(a) Step 1 — change in wavelength at θ = 60° cos60° = 0.5, so (1 − cos60°) = 0.5 Δλ = (2.43 × 10⁻¹²)(0.5) = 1.21 × 10⁻¹² m Step 2 — add to the incident wavelength λᶠ = λᵢ + Δλ = 0.0500 × 10⁻⁹ + 1.21 × 10⁻¹² λᶠ = 5.00 × 10⁻¹¹ + 0.121 × 10⁻¹¹ = 5.12 × 10⁻¹¹ m λᶠ = 0.0512 nm (b) the photon’s energy Wavelength increased, and E = hc/λ, so: the photon’s energy DECREASES (it gave energy to the electron) Convert everything to metres before adding — 0.0500 nm = 5.00 × 10⁻¹¹ m. The scattered wavelength is always LONGER than the incident one, because the photon always loses energy in the collision.

Finding the scattering angle

Questions often run the formula backwards: you’re told both wavelengths and asked for the angle. Just rearrange for cosθ.

Rearranged for the angle cosθ = 1 − (m_ec Δλ) / h

WE 3

A photon of wavelength 0.0600 nm scatters off a stationary electron and emerges with a wavelength of 0.0624 nm. Calculate the scattering angle of the photon. (h = 6.63 × 10⁻³⁴ J s, m_e = 9.11 × 10⁻³¹ kg, c = 3.00 × 10⁸ m s⁻¹)

Step 1 — change in wavelength Δλ = λᶠ − λᵢ = 0.0624 − 0.0600 = 0.0024 nm = 2.4 × 10⁻¹² m Step 2 — rearrange for cosθ cosθ = 1 − (mₑc Δλ)/h cosθ = 1 − (9.11 × 10⁻³¹ × 3.00 × 10⁴ × 2.4 × 10⁻¹²) / (6.63 × 10⁻³⁴) cosθ = 1 − 0.989 = 0.011 Step 3 — take the inverse cosine θ = cos⁻¹(0.011) = 89° (2 s.f.) A shift close to the full Compton wavelength (2.43 pm) tells you the angle is near 90° before you even finish — a nice sanity check. Keep Δλ in metres throughout.

Why it proves light is a particle

Classical wave theory can’t explain any of this. A wave scattering off electrons should keep the same wavelength — it would just make the electrons wobble and re-radiate at the same frequency. The fact that the wavelength increases, and does so in a way that depends on the collision angle, only makes sense if light is a stream of particles carrying energy and momentum that get shared in a collision.

Notice how neatly this closes the chapter. The photoelectric effect showed photons carry energy in packets. Compton scattering shows photons also carry momentum — and that both energy and momentum are conserved in a photon-electron collision, exactly as for two solid particles. Together they make the particle nature of light undeniable. And the same Planck’s constant h that appears in E = hf, in λ = h/p, and in this Compton formula ties the entire quantum story together.

⚛ Working a Compton question

Find Δλ from the angle: Δλ = (h/m_ec)(1 − cosθ).
Scattered wavelength? λ_f = λ_i + Δλ (it always gets longer).
Given both wavelengths, find the angle? Rearrange: cosθ = 1 − (m_ecΔλ)/h.
Photon energy change? Use E = hc/λ before and after; the difference goes to the electron.
Convert units: keep all wavelengths in metres (nm → ×10⁻⁹, pm → ×10⁻¹²).
Sanity check: Δλ can never exceed 2 × (Compton wavelength) ≈ 4.85 × 10⁻¹² m.

💡 Top tips

The photon always loses energy, so its wavelength always increases.
Δλ depends on the angle only, not the incident wavelength.
At θ = 0°, Δλ = 0; maximum shift is at θ = 180°.
At θ = 90°, the shift equals the Compton wavelength (2.43 pm) exactly.
Keep wavelengths in metres and remember the formula assumes the electron starts at rest.

⚠ Common mistakes

Thinking the wavelength decreases — it always increases (photon loses energy)
Forgetting to add Δλ to the incident wavelength for λ_f
Leaving wavelengths in nm or pm instead of metres
Using (1 + cosθ) instead of (1 − cosθ)
Forgetting the shift is zero at θ = 0°
Assuming Δλ depends on the photon’s starting wavelength — it doesn’t

Quick recap: Compton scattering is a high-energy photon colliding with a stationary electron like a particle, conserving energy and momentum. The photon hands over energy, so it leaves with less energy and a longer wavelength, while the electron recoils. The shift follows Δλ = (h/m_ec)(1 − cosθ), depending only on the scattering angle: zero at 0°, one Compton wavelength at 90°, maximum at 180°. Because a pure wave couldn’t change wavelength this way, Compton scattering is powerful evidence for the particle nature of light.

And that completes the Quantum Physics chapter — from the photoelectric effect and Einstein’s equation, through the photon model, de Broglie’s matter waves and wave-particle duality, to Compton scattering. You can now calculate photon energies, work functions, matter wavelengths and Compton shifts, explain every classic experiment, and argue confidently for both the wave and particle natures of light and matter. These ideas underpin everything from electron microscopes to medical imaging — brilliant work getting through the whole set.