How Do You Calculate the Wavelength Shift in Compton Scattering?

For IB Physics HL only: Calculate the wavelength shift in Compton scattering using

$$\Delta \lambda = \lambda_f-\lambda_i=\frac{h}{m_ec}(1-\cos\theta).$$

The scattered photon always has an equal or longer wavelength than the incident photon, so $\Delta\lambda\geq 0$.

The Reasoning and Method

Compton scattering occurs when a photon collides with an electron initially at rest. The photon transfers energy and momentum to the electron. Consequently, the scattered photon has lower energy, lower frequency, and a longer wavelength because $E=hf=hc/\lambda$.

In the equation:

• $h$ is the Planck constant, $6.63\times10^{-34}\ \mathrm{J\,s}$
• $m_e$ is the electron rest mass,

The quantity

$$\frac{h}{m_ec}=2.43\times10^{-12}\ \mathrm{m}$$

is the electron's Compton wavelength.

This relation follows from simultaneous conservation of energy and momentum for the photon-electron interaction. It predicts no shift at $\theta=0^\circ$, because the photon continues forward, and the largest shift at $\theta=180^\circ$, because the reversal produces the greatest momentum transfer. The factor $1-\cos\theta$ is dimensionless, so the result has units of metres.

Worked Example

A photon is scattered through $60^\circ$. Calculate its wavelength shift.

$$\Delta\lambda=(2.43\times10^{-12})(1-\cos60^\circ)$$

Since $\cos60^\circ=0.500$:

$$\Delta\lambda=(2.43\times10^{-12})(0.500)=1.22\times10^{-12}\ \mathrm{m}.$$

If $\lambda_i=5.00\times10^{-11}\ \mathrm{m}$, then

$$\lambda_f=\lambda_i+\Delta\lambda=5.12\times10^{-11}\ \mathrm{m}.$$

A useful check is to compare $\Delta\lambda$ with $\lambda_i$. In the example, the shift is about 2.4% of the incident wavelength, so $\lambda_f$ should be slightly larger. Keep unrounded calculator values until the final line, and ensure the calculator is in degree mode.

The maximum shift occurs when the photon reverses direction. A common misconception is to substitute the electron's recoil angle; the equation requires the scattered photon's angle. The shift is also independent of the incident wavelength and radiation intensity.

IB Exam Technique

This is E.2 Quantum physics, an Additional higher level (AHL) topic. For a Calculate question, show the equation, substitution, and final value with units. Add $\Delta\lambda$ to $\lambda_i$ only when the final wavelength is requested, and check that your result is non-negative.