= Solution
For a <photon> scattering through angle $\theta$ from a free <electron> initially at rest, conserve <energy> and <momentum>. Eliminating the final <electron> <momentum> by its relativistic <energy> relation gives
$$
mc^2(E-E')=EE'(1-\cos\theta),\qquad
\boxed{\lambda'-\lambda=\frac h{mc}(1-\cos\theta)},
$$
or $E'=E/[1+(E/mc^2)(1-\cos\theta)]$. The recoil <electron> gains <kinetic energy> $E-E'$. The shift is zero forward, one <electron> <Compton wavelength> at $90^\circ$, and twice that <wavelength> backward.
Compton's X-ray scattering measurements compared scattered spectra at different angles and identified an angle-dependent shifted component. This supports <photon> <momentum> $p=E/c$ and inelastic recoil, beyond the unchanged-frequency prediction of classical elastic <Thomson scattering>. Bound <electrons> and coherent scattering can also produce an unshifted component; initial <momentum> and binding broaden or modify the ideal free-electron result. Treating an <electron> as initially free and stationary is therefore a kinematic approximation appropriate when the relevant <photon> transfer dominates binding scales.<A Quantum Theory of the Scattering of X-rays by Light Elements by Arthur Holly Compton>[Printed pp. 494–495, experimental angular-spectrum comparison; p. 496, recoil discussion]{ref}
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