= Solution
In <Compton scattering>, a <photon> transfers <energy> and <momentum> to an initially stationary, approximately free <electron>. With incident and scattered <wavelengths> $\lambda,\lambda'$ and <photon> deflection angle $\theta$, <momentum> conservation gives
$$
p_e^2=h^2\left(\lambda^{-2}+\lambda'^{-2}-\frac{2\cos\theta}{\lambda\lambda'}\right).
$$
Combine this with $hc/\lambda+mc^2=hc/\lambda'+\sqrt{m^2c^4+p_e^2c^2}$ to obtain
$$
\boxed{\lambda'-\lambda=\frac{h}{mc}(1-\cos\theta)}.
$$
The <photon> loses <energy> to <electron> <recoil>; the shift is independent of incident <wavelength> in this free-electron model and is largest at backscattering. At <energies> comparable with atomic binding, initial <electron> motion, binding and coherent scattering from the atom require additional treatment. Classical elastic scattering from a fixed charge alone does not produce this <recoil> <wavelength> shift.
In <Arthur Holly Compton>'s spectral test, molybdenum-target X-rays scattered from <graphite> at $90^\circ$ were compared with the primary spectrum using the same slits, crystal and tube <potential energy>. A <calcite> crystal resolved the shifted line by <Bragg's law>; lines on both sides established angular zero. The crystal glancing angle on the plotted axis differs from the <photon> scattering angle. The compared curves had different intensity scales, and the paper did not directly observe the <recoil> <electrons> in this <X-ray> test. These controls support the <wavelength> comparison without converting it into a direct recoil-track measurement.<A Quantum Theory of the Scattering of X-rays by Light Elements by Arthur Holly Compton>[pp. 494–496, Experimental Test, Fig. 4 and opening of Velocity of recoil of the scattering electrons]{ref}
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