Answer
$\phi = 44.3^{\circ}$
Work Step by Step
We can find the original wavelength of the photon:
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(6.63\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(200\times 10^3~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda = 6.2156\times 10^{-12}~m$
We can find $\Delta \lambda$:
$\vert \Delta E \vert = \frac{\Delta \lambda}{\lambda+\Delta \lambda} = 0.10$
$\Delta \lambda = 0.10~(\lambda+\Delta \lambda)$
$0.90~\Delta \lambda = 0.10~\lambda$
$\Delta \lambda = \frac{\lambda}{9}$
$\Delta \lambda = \frac{6.2156\times 10^{-12}~m}{9}$
$\Delta \lambda = 6.906\times 10^{-13}~m$
We can find the required angle $\phi$:
$\Delta \lambda = \frac{h}{mc}~(1-cos~\phi)$
$\frac{\Delta \lambda mc}{h} = 1-cos~\phi$
$cos~\phi = 1 - \frac{\Delta \lambda mc}{h}$
$cos~\phi = 1 - \frac{(6.906\times 10^{-13}~m)(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)}{6.63\times 10^{-34}~J~s}$
$cos~\phi = 1 - 0.2846$
$cos~\phi = 0.7154$
$\phi = cos^{-1}~(0.7154)$
$\phi = 44.3^{\circ}$
