Answer
Please see the work below.
Work Step by Step
(a) We know that
$\lambda_f\lambda_i=\frac{6.63\times 10^{-34}J.s}{(9.11\times 10^{-31})(3\times 10^8m/s)}[1-cos175^{\circ}]$
$\lambda_f-\lambda_i=0.0048\times 10^{-9}m=0.0048nm$
Now $\lambda_i=0.320nm-0.0048nm=0.315nm$
(b) As $E_i=\frac{hc}{\lambda_i}$
$\implies E_i=\frac{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{0.315\times 10^{-9}m}=6.31\times 10^{-16}J$
$E_i=3.94KeV$
and $E_f=\frac{hc}{\lambda_f}$
$E_f=\frac{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{0.315\times 10^{-9}m}$
$E_f=6.22\times 10^{-16}J=3.88KeV$
(c) We can find the kinetic energy of the recoil electron as follows:
$K=E_i-E_f$
We plug in the known values to obtain:
$K=(3.94IKeV)-(3.88KeV)$
$K=0.06eV=60eV$
