= Solution
For complete photon backscatter $\theta=\pi$, the electron recoils forward and <momentum conservation> gives
$$
m_ev=h\left(\frac1\lambda+\frac1{\lambda'}\right).
$$
Combining this with <conservation of energy> and writing $\Delta\lambda=\lambda'-\lambda$ gives
$$
\frac{hc\Delta\lambda}{\lambda\lambda'}=\frac{h^2}{2m_e}\left(\frac1\lambda+\frac1{\lambda'}\right)^2,
\quad
\Delta\lambda=\frac{h}{2m_ec}\frac{(\lambda+\lambda')^2}{\lambda\lambda'}.
$$
For $\Delta\lambda\ll\lambda,\lambda'$, the final ratio is approximately four, so
$$
\boxed{\Delta\lambda\simeq\frac{2h}{m_ec}=2\lambda_C\simeq4.85\times10^{-12}\,\mathrm m.}
$$
Within the stated nonrelativistic equations the unsimplified relation is implicit and has higher-order recoil corrections; the displayed final approximation is all that the small-shift assumption supports. Exact relativistic <Compton recoil kinematics> gives $\Delta\lambda=(h/m_ec)(1-\cos\theta)$, and therefore exactly $2\lambda_C$ at backscatter. Agreement at leading order does not make the nonrelativistic derivation exact at arbitrary photon <energy>.
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