Compton Effect

The Compton Effect is experimental proof that photons carry momentum. When X-rays scatter off electrons, the scattered photons have longer wavelengths than the incident photons — a result that classical wave theory cannot explain.

The Experiment (1923)

Arthur Compton fired X-rays at a target and measured the scattered radiation.

What He Observed

  1. Some X-rays scattered with the same wavelength
  2. Some X-rays scattered with longer wavelength
  3. The wavelength shift depended on the scattering angle

Classical wave theory predicted no wavelength change — only particle collision with momentum transfer explains this.

What Happens

Before Collision

  • Photon: moving with momentum $p = \frac{h}{\lambda}$
  • Electron: nearly at rest

During Collision

  • Photon strikes electron
  • Momentum is transferred

After Collision

  • Electron: recoils with kinetic energy
  • Photon: scatters with lower momentum → longer wavelength

The Compton Shift Formula

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

Where:

Symbol Meaning Value
$\Delta \lambda$ Wavelength shift —
$\lambda'$ Scattered photon wavelength —
$\lambda$ Incident photon wavelength —
$h$ Planck's constant $6.63 \times 10^{-34}$ J·s
$m_e$ Electron mass $9.11 \times 10^{-31}$ kg
$c$ Speed of light $3.0 \times 10^8$ m/s
$\theta$ Scattering angle —

Compton Wavelength of Electron

$$\frac{h}{m_e c} = 2.43 \times 10^{-12} \text{ m}$$

Key Insights

Maximum Shift

Maximum wavelength increase occurs at $\theta = 180°$ (backscattering): $$\Delta \lambda_{max} = \frac{2h}{m_e c} = 4.86 \times 10^{-12} \text{ m}$$

No Shift at Zero Angle

At $\theta = 0°$: $\cos 0° = 1$, so $\Delta \lambda = 0$

Why Classical Theory Fails

  • Classical wave theory: scattering doesn't change wavelength
  • Only particle collision with momentum conservation explains the shift

Example Calculation

Problem: X-rays with $\lambda = 1.0 \times 10^{-11}$ m scatter at $\theta = 90°$. Find $\lambda'$.

Solution: $$\Delta \lambda = \frac{h}{m_e c}(1 - \cos 90°) = 2.43 \times 10^{-12} \times (1 - 0) = 2.43 \times 10^{-12} \text{ m}$$

$$\lambda' = \lambda + \Delta \lambda = 1.0 \times 10^{-11} + 2.43 \times 10^{-12} = 1.243 \times 10^{-11} \text{ m}$$

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