1 \begin{problem*}{28.15}
2 X-rays having an energy of $300\U{keV}$ undergo Compton scattering
3 from a target. The scattered rays are detected at $37.0\dg$ relative
4 to the incident rays. Find \Part{a} the Compton shift at this
5 angle, \Part{b} the energy of the scattered x-ray, and \Part{c} the
6 energy of the recoiling electron.
7 \end{problem*} % problem 28.15
11 From the Compton shift equation (Eq.~28.8)
13 \lambda' - \lambda_0 &= \frac{h}{m_e c} (1-\cos\theta) \\
14 \Delta \lambda &= 2.43\U{pm}(1-\cos 37.0\dg)
19 The wavelength of the incoming photon was
21 \lambda_0 = \frac{hc}{E_0} = \frac{1240\U{eV$\cdot$nm}}{300\U{keV}}
24 The scattered wavelength is thus
26 \lambda' = \lambda_0 + \Delta \lambda = (4.13+0.489)\U{pm} = 4.62\U{pm} \;,
28 and the energy of the scattered photon is
30 E' = \frac{hc}{\lambda'} = \frac{1240\U{eV$\cdot$nm}}{4.62\U{pm}}
31 = \ans{268\U{keV}} \;.
35 All the energy lost by the photon must go into the recoiling electron so
37 E_e = E_0 - E' = (300-268)\U{keV} = \ans{31.7\U{keV}} \;.