21.3 Compton's Effect

Photon Momentum

Although a photon has zero rest mass, it carries momentum. According to Einstein's theory and Planck's relation, the momentum $p$ of a photon is:

$p=\frac{E}{c}=\frac{hf}{c}=\frac{h}{\lambda }$

where $h$ is Planck's constant, $f$ is the frequency, $c$ is the speed of light, and $\lambda$ is the wavelength. This shows that a photon's momentum is inversely proportional to its wavelength.

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The Compton Effect

In 1923, Arthur H. Compton directed a beam of X-rays at a graphite (carbon) target and analysed the scattered X-rays. He found that the scattered X-rays had a longer wavelength (lower energy) than the incident X-rays. This phenomenon is called the Compton Effect.

Explanation

Compton explained this by treating the X-ray photon as a particle that collides elastically with a free (loosely bound) electron in the target, similar to a billiard-ball collision:

1. The photon transfers some of its energy and momentum to the electron.
2. The electron recoils (the Compton electron).
3. The scattered photon has less energy → longer wavelength.

This experiment provided direct evidence that photons carry momentum and behave as particles.

Compton Shift Formula

Applying conservation of energy and conservation of momentum to the photon–electron collision, Compton derived:

$\Delta \lambda ={\lambda }^{\prime }-\lambda =\frac{h}{m_{0}c}(1-\cos \theta )$

Compton Wavelength

The constant $\frac{h}{m_{0}c}$ is called the Compton wavelength of the electron:

$\frac{h}{m_{0}c}=2.43\times 10^{-12}\text{ m}=0.00243\text{ nm}$

Special Cases

Why Compton Effect is Not Observed with Visible Light

The Compton shift $\Delta \lambda \approx 0.00243$ nm is fixed regardless of the incident wavelength. For visible light ($\lambda \approx 400$–$700$ nm), the fractional change:

$\frac{\Delta \lambda }{\lambda }\approx \frac{0.00243}{500}\approx 5\times 10^{-6}$

This is negligibly small and undetectable. For X-rays ($\lambda \approx 0.1$ nm), the fractional shift is significant (~2%), making it observable.

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Pair Production and Annihilation

Pair Production

Pair production is the process in which a high-energy gamma-ray photon ($\gamma$) is converted into an electron–positron pair in the vicinity of a nucleus:

$\gamma \to e^{-}+e^{+}$

• The photon must have energy at least equal to the rest-mass energy of both particles: $E_{\text{min}}=2m_0{c}^2=2\times (9.11\times 10^{-31})(3\times 10^8{)}^2\approx 1.02\text{ MeV}$
• A nearby nucleus is required to conserve momentum.
• This is a direct demonstration of mass–energy equivalence ($E=m{c}^2$): energy is converted into matter.

Pair Annihilation

Pair annihilation is the reverse of pair production. When an electron ($e^{-}$) meets a positron ($e^{+}$), they annihilate each other and produce two gamma-ray photons:

$e^{-}+e^{+}\to 2\gamma$

• Two photons are produced (not one) to conserve momentum — they travel in opposite directions.
• Each photon has energy equal to the rest-mass energy of one particle: $E_{\gamma }=m_0{c}^2=0.511\text{ MeV}$
• This demonstrates conversion of matter into energy.