Worked exercise — Compton scattering with a 0.8 MeV photon

Problem

A photon of initial energy $E = 0{,}8\;\text{MeV}$ undergoes Compton scattering, losing a third of its energy. Calculate: (a) the scattering angle $\theta$; (b) the kinetic energy and speed of the electron; (c) check whether it is legitimate to use the classical formula for kinetic energy.

Solution

Step 1 — Initial and final wavelengths. The photon loses a third of its energy, so $E' = \tfrac{2}{3}E$. Since $\lambda = hc/E$, the wavelength grows by a factor of $3/2$:

$\lambda = \frac{hc}{E} = \frac{1240\;\text{eV·nm}}{0{,}8\cdot 10^6\;\text{eV}} \approx 1{,}55\cdot 10^{-3}\;\text{nm} = 1{,}55\;\text{pm}, \qquad \lambda' = \tfrac{3}{2}\lambda \approx 2{,}33\;\text{pm}$

Step 2 — Scattering angle (Compton formula). With $\lambda_C = h/(m_e c) = 2{,}43\;\text{pm}$:

$1 - \cos\theta = \frac{\lambda' - \lambda}{\lambda_C} = \frac{0{,}78\;\text{pm}}{2{,}43\;\text{pm}} \approx 0{,}32 \;\Rightarrow\; \cos\theta \approx 0{,}68$

$\ev{\theta \approx 47^\circ}$

Step 3 — Kinetic energy of the electron. By conservation of energy the electron acquires what the photon loses:

$K_e = E - E' = 0{,}8 - 0{,}53 \approx 0{,}27\;\text{MeV}$

Step 4 — Classical or relativistic? With the classical formula $K_e = \tfrac{1}{2}m_e v^2$ one would obtain

$v = \sqrt{\frac{2K_e}{m_e}} \approx 3{,}08\cdot 10^8\;\text{m/s} > c \quad \text{(impossible!)}$

One must use the relativistic formula $K_e = (\gamma - 1)m_e c^2$. With $m_e c^2 = 0{,}511\;\text{MeV}$:

$\gamma = 1 + \frac{K_e}{m_e c^2} = 1 + \frac{0{,}27}{0{,}511} \approx 1{,}53$

$\ev{v = c\sqrt{1 - 1/\gamma^2} \approx 2{,}3\cdot 10^8\;\text{m/s} \approx 0{,}77\,c}$

This is an excellent example of how classical kinematics fails when the kinetic energy becomes comparable with $m_e c^2 = 511\;\text{keV}$.

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Topics: Fisica quantistica Concepts: Effetto Compton · Fotone · Equivalenza massa-energia