Answer
incident X-ray wavelength is $\lambda=0.353292nm$
Work Step by Step
Given that Scattering angle is $\theta=122^{0}$
Scattered X-ray has a momentum $p'=1.856\times10^{-24}kg.m/s$
suppose scattered X-ray has wavelength $\lambda'$
we know that momentum of photon is given by $p=\frac{h}{\lambda}$
so $\lambda=\frac{h}{p}$,
for scattered photon $\lambda'=\frac{h}{p'}$
putting the values $h=6.626\times10^{-34}J.s$ and $p'=1.856\times10^{-24}kg.m/s$
$\lambda'=\frac{6.626\times10^{-34}J.s}{1.856\times10^{-24}kg.m/s}=3.570043\times10^{-10}m=0.3570043\times10^{-9}m$
$\lambda'=0.3570043\times10^{-9}m=0.3570043nm$
Suppose wavelength of incident X-ray is $\lambda$
so from Compton's scattering equation 29.7
$\lambda'-\lambda=\frac{h}{mc}(1-cos\theta)$
putting the values of
$h=6.626\times10^{-34}J.s$,
$\lambda'=0.3570043\times10^{-9}m$,
$\theta=122^{0}$,
$m=9.109\times10^{-31}kg$
$c=2.998\times10^{8}m/s$
$0.3570043\times10^{-9}m-\lambda=\frac{6.626\times10^{-34}J.s}{9.109\times10^{-31}kg\times2.998\times10^{8}m/s}(1-cos122^{0})$
$0.3570043\times10^{-9}m-\lambda=0.242632\times10^{-11}m[1-(-0.529919)]$
$0.3570043\times10^{-9}m-\lambda=0.371207\times10^{-11}m$
$0.3570043\times10^{-9}m-\lambda=0.00371207\times10^{-9}m$
$\lambda=0.3570043\times10^{-9}m-0.00371207\times10^{-9}m$
$\lambda=0.353292\times10^{-9}m$
or incident X-ray wavelength is $\lambda=0.353292nm$
