 Solution

Write the incident and scattered photon energies as $\epsilon =hc/\lambda$ and ${\epsilon }^{\prime }=hc/{\lambda }^{\prime }$. Conservation of relativistic energy gives the recoiling electron energy $E_e=m{c}^2+\epsilon -{\epsilon }^{\prime }$. By momentum conservation, its momentum is the difference of the two photon momenta; henceInsert these into the relativistic energy-momentum relation for the electron:Expanding and cancelling terms givesSince $1/\epsilon =\lambda /(hc)$, the Compton scattering formula follows:The scale $h/(mc)$ is the electron Compton wavelength. At $\theta =0$, ${\lambda }^{\prime }=\lambda$: there is no energy or momentum transfer in this ideal forward limit, so the electron remains at rest. At $\theta =\pi$, the wavelength shift is maximal:The electron gains its largest kinetic energy, $\epsilon -{\epsilon }_{\min }^{\prime }$, and recoils forward while the photon reverses direction.