9. Compton Scattering¶

Module

Particle Nature of Light · Experiment 2

Try it live

Explore this on the interactive simulator with real-time sliders and graphs: 🔬 Launch in Simulator

Aim¶

To verify the Compton scattering formula by studying the wavelength shift of X-ray photons scattered by a free electron as a function of scattering angle.

Theoretical Background¶

When a high-energy photon (typically an X-ray) collides with a loosely bound (effectively free) electron, the interaction can be treated as an elastic, relativistic billiard-ball collision between two particles, provided the photon is assigned momentum p = h/λ in addition to its energy E = hν = hc/λ.

Applying conservation of relativistic energy and momentum to this two-body collision shows that the scattered photon must have a longer wavelength (lower energy) than the incident photon, with the wavelength shift Δλ depending only on the scattering angle θ and fundamental constants — not on the target material or the incident intensity.

This angle-only dependence, and the fact that the shift is of the exact size predicted only if the photon is treated as a particle with definite momentum, was decisive evidence (alongside the photoelectric effect) for the particle (quantum) nature of light, distinct from its well-established wave behaviour.

Governing Formulas¶

Compton shift:

\[ \Delta\lambda = \lambda_C(1-\cos\theta) \]

Compton wavelength:

\[ \lambda_C = \dfrac{h}{m_e c} = 2.426\ \text{pm} \]

Scattered wavelength:

\[ \lambda' = \lambda + \Delta\lambda \]

Apparatus layout diagram for this experiment

Labelled apparatus layout — original illustration prepared for this guide (not a photograph). See the drag-to-rotate 3-D schematic below for a spatial view of the same setup.

3-D apparatus schematic · drag to rotate, scroll to zoom · X-ray source → target electron → angled detector

Simulator Controls¶

Control	Symbol	Range	Unit
Incident X-ray wavelength	λ	1 – 100	pm
Scattering angle	θ	0 – 180	°

Procedure¶

Fix the incident wavelength λ and vary the scattering angle θ from 0° to 180° in steps of 15°, recording the Compton shift Δλ and scattered wavelength λ' at each step.
Repeat the run for a second incident wavelength and confirm that Δλ at a given angle is unchanged (i.e. independent of λ).
Plot Δλ against (1 − cosθ) and verify a straight line through the origin of slope λC.

Observation Table¶

Table: Compton wavelength shift as a function of scattering angle (θ = 0° to 180°)

S. No.	θ (°)	1 − cosθ	λ (pm)	Δλ (pm) observed	λ' = λ+Δλ (pm)
1
2
3
4
5
6
7
8
9
10
11
12
13

Graph¶

Plot Compton shift Δλ (pm) (y-axis) against (1 − cosθ) (x-axis).

Expected trend: a straight line through the origin of slope λC = 2.426 pm.

Δλ vs (1 − cosθ), used to extract the Compton wavelength

15×15 grid — choose your own scale per square; plot from your observation table

Calculation¶

Plot Δλ against (1 − cosθ); the slope of the best-fit straight line gives the experimental Compton wavelength λC.
Compare the experimental λC with the accepted value h/(mec) = 2.426 pm and compute the percentage error.

Result¶

Table: Determination of the Compton wavelength

λC (accepted, pm)	λC (from slope, pm)	% error
1
2
3

Maximum Permissible Error¶

Δλ = λ_C(1−cosθ); differentiating with respect to the scattering angle gives this term. The Compton wavelength λ_C is a fundamental constant with negligible uncertainty.

Precautions¶

Note that Δλ depends only on θ and never on the incident wavelength or the scattering material — use this as a built-in consistency check.
Take closely spaced readings near θ = 90° and θ = 180° where the shift changes most/reaches its maximum.

Self-Check Quiz¶

1. The Compton shift Δλ depends on:

A) The scattering angle θ only (plus fundamental constants)
B) The target material only
C) The incident beam intensity only
D) Both material and intensity, but not angle

Answer: A — Δλ = λ_C(1 − cosθ) depends only on the scattering angle and fundamental constants — famously independent of the target material, which was strong evidence for the photon picture.

2. At θ = 0° (no scattering), the Compton shift Δλ is:

A) Maximum
B) Zero
C) Equal to λ_C
D) Negative

Answer: B — Δλ = λ_C(1 − cos0°) = λ_C(1 − 1) = 0 — no deflection means no wavelength shift.

3. At what scattering angle is the Compton shift maximum?

A) 0°
B) 90°
C) 180° (direct backscatter)
D) 45°

Answer: C — Δλ = λ_C(1 − cosθ) is maximised when cosθ = −1, i.e. θ = 180°, giving Δλ = 2λ_C.

4. Compton's experiment required treating the photon as having, in addition to energy E = hν, a definite:

A) Electric charge
B) Rest mass
C) Momentum, p = h/λ
D) Magnetic moment

Answer: C — Assigning the photon momentum p = h/λ was essential to apply conservation of momentum in the photon–electron collision, alongside conservation of energy.

Viva-Voce Questions¶

Why does Compton scattering require the photon to be assigned a definite momentum p = h/λ?
Why is the Compton shift independent of the scattering material?
Why is the Compton wavelength shift independent of the wavelength of the incident photon?
Why is the Compton effect readily observed with X-rays but not with visible light?
What does the Compton wavelength of the electron physically represent?
Why does the scattered photon lose energy while the recoiling electron gains kinetic energy in this process?
How does Compton scattering provide evidence for the particle (photon) nature of light?