Problem 28.13

28.13.
The kinetic energy initially is given by E₀. The final kinetic energy of the two are each given by $\frac{E_{0}}{2}$. Since the energy of a photon is given by
$E = \frac{hc}{\lambda}$
We have
$\lambda' = \lambda_{0} + \lambda_{c}(1 - \cos \theta)$
Since the energy of the photon is half of its initial value, its wavelength is twice of the inital value $\lambda_{0}$.
$2\lambda_{0} = \lambda_{0} + \lambda_{c}(1 - \cos \theta)$
Solving for $\theta$, we have
$(1 - \cos \theta) = \frac{\lambda_{0}}{\lambda_{c}}$
$\theta = 70.1{}^{\circ}$