Suppose that a 3.64-nm photon moving in the $+x$ -direction collides head-on with a $2 \times 10^{5} \mathrm{~m} / \mathrm{s}$ electron moving in the $-x$ direction. If the collision is perfectly elastic, find the conditions after collision.
From the law of conservation of momentum,
Momentum before = Momentum after
$$
\frac{h}{\lambda_{0}}-m v_{0}=\frac{h}{\lambda}-m v
$$
But, from $\underline{\text { Problem } 42.9,} h / \lambda_{0}=m u$ in this case. Hence, $h / \lambda=m v$. Also, for a perfectly elastic collision,
$$
\begin{array}{l}
\text { KE before }=\text { KE after } \\
\frac{h c}{\lambda_{0}}+\frac{1}{2} m v_{0}^{2}=\frac{h c}{\lambda}+\frac{1}{2} m v^{2}
\end{array}
$$
Using the facts that $h / \lambda_{0}=m v_{0}$ and $h / \lambda=m v$, we find
$$
v_{0}\left(\mathrm{c}+\frac{1}{2} v_{0}\right)=v\left(\mathrm{c}+\frac{1}{2} v\right)
$$
Therefore, $v=\mathrm{u}_{0}$ and the electron moves in the $+\chi$ -direction with its original speed. Because $h / \lambda=m v=m u_{0}$, the photon also "rebounds," and with its original wavelength.