1.3 The Compton Effect

According to special relativity, a particle with zero mass should have momentum-energy relation given by $$E=pc.$$

From previous two sections, we assumed photon energy is determined by $$E=hf.$$

Thus $$p=\frac{h}{\lambda}.$$

1.3.1 A Two-Particle Collision (Elastic)

Image from Wikipedia

• Before collision: A photon of wavelength \(\lambda\) approaches an electron at rest.
• Photon: \(E=h/\lambda\), \(P=hc/\lambda\);
• Electron: \(E=0\), \(m_ec^2\).
• After collision: The electron scatters at speed of \(u\), angle \(\phi\). A photon of wavelength \(\lambda'\) scatters at angle \(\theta\).
• Photon: \(E=h/\lambda'\), \(P=hc/\lambda\);
• Electron: \(E=\gamma_u m_e u\), \(P=\gamma_u m_e c^2\).

Here \(\lambda_u=1/\sqrt{1-u^2/c^2}\) is the Lorentz factor, as electron is moving fast and relativistic correction needs to be considered.

To solve for scattering angle, consider conservation of both energy and momentum.

• Momentum
• x-comp.: \(\frac{h}{\lambda}=\frac{h}{\lambda'}\cos(\theta)+\gamma_u m_e u\cos(\phi);\)
• y-comp.: \(0=\frac{h}{\lambda'}\sin(\theta)-\gamma_u m_e u\sin(\phi).\)
• Energy
• \(h\frac{c}{\lambda}+m_ec^2=h\frac{c}{\lambda'}+\gamma_u m_ec^2.\)

The result of Compton effect by solving above set of equation is $$\bbox[silver]{\lambda'-\lambda=\frac{h}{m_e c}(1-\cos(\theta)).}$$

The expression for \(\phi\) leaves as exercise.

Problems

1. (Harris) An X-ray of unknown wavelength is directed at a carbon sample (source of electron). An electron is scattered with a speed of \(4.5\times10^{7}\) m/s at an angle of \(60^\circ\). Determine the wavelength of the X-ray source.

2. (Harris) Obtain $$\lambda'-\lambda=\frac{h}{m_e c}(1-\cos(\theta)),$$ from $$\begin{cases}\frac{h}{\lambda}=\frac{h}{\lambda'}\cos(\theta)+\gamma_u m_e u\cos(\phi),\\0=\frac{h}{\lambda'}\sin(\theta)-\gamma_u m_e u\sin(\phi),\\h\frac{c}{\lambda}+m_ec^2=h\frac{c}{\lambda'}+\gamma_u m_ec^2.\end{cases}$$

   (Hint: use \(\cos^2(\phi)+\sin^2(\phi)=1\))

3. (Harris) Show the scattering angles of the photon and the electron following $$\cot(\frac{\theta}{2})=\left(1+\frac{h}{mc\lambda}\right)\tan(\phi).$$