Compton scattering at 90 degrees with X-rays

Problem

An X-ray photon with $\lambda = 50\;\text{pm}$ is scattered by an electron initially at rest and deflected by $\theta = 90^\circ$ (Compton effect). Calculate the change $\Delta\lambda$ in the wavelength and the new $\lambda'$. What fraction of energy did the photon give up to the electron?

Solution

Step 1 — Change in wavelength (Compton formula). With $\lambda_C = h/(m_e c) = 2{,}43\;\text{pm}$ and $\theta = 90^\circ$ ($\cos 90^\circ = 0$):

$\Delta\lambda = \lambda_C\,(1 - \cos\theta) = 2{,}43\;\text{pm}\cdot(1 - 0) = 2{,}43\;\text{pm}$

$\ev{\Delta\lambda \approx 2{,}43\cdot 10^{-12}\;\text{m}}$

Step 2 — New wavelength.

$\lambda' = \lambda + \Delta\lambda = 50 + 2{,}43 = 52{,}43\;\text{pm}$

Step 3 — Fraction of energy given up. Since $E = hc/\lambda$, the fraction lost by the photon is

$\frac{\Delta E}{E} = 1 - \frac{E'}{E} = 1 - \frac{\lambda}{\lambda'} = 1 - \frac{50}{52{,}43} \approx 0{,}046$

$\ev{\frac{\Delta E}{E} \approx 4{,}6\%}$

The photon gives up about 4.6% of its energy to the recoiling electron. The absolute shift $\Delta\lambda$ is fixed (it depends only on the angle), but it only becomes significant as a percentage for small $\lambda$, i.e. for X-rays and $\gamma$-rays: this is why the Compton effect is observed with very energetic light.

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Topics: Quantum physics Concepts: Compton effect · Photon