Answer
$K_e = 1.13 \space keV$
Work Step by Step
when X-ray and stationary electron collides, it produces energy, $$E= \frac{hc}{\lambda}$$
From conservation of energy equation we get : $$\frac{hc}{\lambda} = \frac{hc}{\lambda '} + K_e$$
- $\lambda$ , $ \lambda '$ are the wavelengths of the incident and the scattered photons
Solving for $K_e$,
$K_e = hc [ \frac{1}{\lambda} - \frac{1}{\lambda '}]$
$K_e = (6.63 \times 10^{-34} J.s)(3.00 \times 10^8 m/s) [ \frac{1}{0.100 \times 10^{-9} m} - \frac{1}{0.110 \times 10^{-9} m}] $
$K_e = 1.81 \times 10^{-16} J $
Convert to eV
$K_e = 1.13 \space keV$
