\begin{problem*}{28.15} % Compton effect
X-rays having an energy of $300\U{keV}$ undergo Compton scattering
from a target.  The scattered rays are detected at $37.0\dg$ relative
to the incident rays.  Find \Part{a} the Compton shift at this
angle, \Part{b} the energy of the scattered x-ray, and \Part{c} the
energy of the recoiling electron.
\end{problem*}

\begin{solution}
\Part{a}
From the Compton shift equation (Eq.~28.8)
\begin{align}
  \lambda' - \lambda_0 &= \frac{h}{m_e c} (1-\cos\theta) \\
  \Delta \lambda &= 2.43\U{pm}(1-\cos 37.0\dg)
    = \ans{489\U{fm}} \;.
\end{align}

\Part{b}
The wavelength of the incoming photon was
\begin{equation}
  \lambda_0 = \frac{hc}{E_0} = \frac{1240\U{eV$\cdot$nm}}{300\U{keV}}
    = 4.13\U{pm} \;.
\end{equation}
The scattered wavelength is thus
\begin{equation}
  \lambda' = \lambda_0 + \Delta \lambda = (4.13+0.489)\U{pm} = 4.62\U{pm} \;,
\end{equation}
and the energy of the scattered photon is
\begin{equation}
  E' = \frac{hc}{\lambda'} = \frac{1240\U{eV$\cdot$nm}}{4.62\U{pm}}
    = \ans{268\U{keV}} \;.
\end{equation}

\Part{c}
All the energy lost by the photon must go into the recoiling electron so
\begin{equation}
  E_e = E_0 - E' = (300-268)\U{keV} = \ans{31.7\U{keV}} \;.
\end{equation}
\end{solution}
