Compton Wavelength

The Compton wavelength $ \lambda_c $ quantifies how much a photon's wavelength $ \Delta \lambda $ shifts when it scatters elastically off a massive particle such as an electron. $$ \Delta \lambda = \lambda_c (1 - \cos \theta) $$ Here, $ \theta $ is the scattering angle, and the Compton wavelength is defined as $$ \lambda_c = \frac{h}{m c} $$ where $ h $ is Planck’s constant, $ m $ is the particle’s mass (e.g., electron), and $ c $ is the speed of light.

In essence, the Compton wavelength tells us how much a photon’s wavelength increases when it collides with a massive particle in a perfectly elastic interaction.

This is a cornerstone concept in relativistic quantum mechanics.

Compton Scattering

The effect was first observed by Arthur H. Compton in 1923. In what became known as Compton scattering, he found that when X-rays strike a material, the scattered radiation has a longer wavelength (and thus lower energy) than the incident radiation.

This behavior couldn't be explained using classical wave theory.

However, when photons are treated as particles carrying momentum, the result becomes intuitively clear.

During the collision, the photon transfers part of its energy and momentum to the electron-just as in a classical elastic collision between two objects.

Note. Compton scattering provided one of the earliest and most direct experimental confirmations of the particle nature of light.

Compton Scattering Equation
A Worked Example

Compton Scattering Equation

The change in the photon’s wavelength due to scattering is described by the Compton formula:

$$ \Delta \lambda = \lambda' - \lambda = \lambda_c (1 - \cos \theta) $$

Where:

$ \lambda $ is the initial wavelength of the photon
$ \lambda' $ is the wavelength after scattering
$ \theta $ is the scattering angle (the photon's deflection angle)
$ \lambda_c $ is the Compton wavelength of the target particle (e.g., the electron)

The Compton wavelength is given by:

$$ \lambda_c = \frac{h}{m c} $$

where $ h $ is Planck’s constant, $ m $ is the particle’s mass, and $ c $ is the speed of light.

Example. For an electron, the Compton wavelength is: $$ \lambda_c \approx 2.426 \times 10^{-12} \text{ m} = 2.426 \, \text{pm} $$

The Compton wavelength is a fundamental length scale for massive particles. It sets the limit below which quantum and relativistic effects become non-negligible.

If one tries to localize a particle within a region smaller than its Compton wavelength, quantum fluctuations become strong enough to allow for the spontaneous creation of particle-antiparticle pairs.

Note. The Compton wavelength is closely tied to wave-particle duality: it assigns a characteristic wavelength to a massive particle, but it differs from the de Broglie wavelength (which depends on momentum). Specifically, the Compton wavelength is mass-dependent: $$ \lambda_c = \frac{h}{mc} $$ while the de Broglie wavelength is momentum-dependent: $$ \lambda = \frac{h}{p} $$

A Worked Example

Let’s walk through a concrete example of Compton scattering-calculating the increase in a photon’s wavelength after it collides with an electron.

Suppose we have an X-ray photon with an initial wavelength:

$$ \lambda = 0.030 \ \text{nm} = 3.00 \times 10^{-11} \ \text{m} $$

The photon strikes a free electron and is scattered at an angle of 90°:

$$ \theta = 90^\circ $$

The Compton wavelength is given by:

$$ \lambda_c = \frac{h}{m c} $$

With Planck’s constant $ h = 6.626 \times 10^{-34} \ \text{J·s} $ and the speed of light $ c = 3.00 \times 10^8 \ \text{m/s} $

$$ \lambda_c = \frac{6.626 \times 10^{-34} \ \text{J·s}}{m \times 3.00 \times 10^8 \ \text{m/s}} $$

The mass of an electron is $ m_e = 9.11 \times 10^{-31} \ \text{kg} $

$$ \lambda_c = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \cdot 3.00 \times 10^8} $$

$$ \lambda_c \approx 2.426 \times 10^{-12} \ \text{m} $$

Now apply the Compton formula:

$$ \Delta \lambda = \lambda_c (1 - \cos \theta) $$

For $ \theta = 90^\circ $, $ \cos \theta = 0 $

$$ \Delta \lambda = \lambda_c (1 - 0) = \lambda_c $$

So the shift in wavelength is:

$$ \Delta \lambda = 2.426 \times 10^{-12} \ \text{m} $$

The new (scattered) wavelength is:

$$ \lambda' = \lambda + \Delta \lambda $$

$$ \lambda' = (3.00 \times 10^{-11}) + (2.426 \times 10^{-12}) \ \text{m} $$

$$ \lambda' = 3.2426 \times 10^{-11} \ \text{m} $$

So, after the collision with the electron, the photon’s wavelength increases:

$$ \lambda' > \lambda = 3.00 \times 10^{-11} \ \text{m} $$

This means the photon’s energy has slightly decreased.

The energy loss is due to the photon transferring momentum to the electron during the collision.

Why does a longer wavelength mean lower energy? A photon's energy $ E $ is given by: $$ E = h \nu $$ where $ h $ is Planck’s constant and $ \nu $ is the frequency. Frequency and wavelength are inversely related: $$ \nu = \frac{c}{\lambda} $$ Substituting into the energy formula: $$ E = h \cdot \frac{c}{\lambda} $$ This shows that as $ \lambda $ increases, the energy $ E $ decreases-and vice versa. With light speed constant, a lower frequency (i.e., longer wavelength) means less energy per photon.

This experiment provides strong evidence that light and matter exchange momentum in elastic collisions, just as predicted by quantum-relativistic theory.

It remains one of the landmark confirmations of light’s particle-like behavior.

And so on.