Given :-
Wavelength of incident X-rays, lamda = 1 A⁰ = 1 x 10^-10m
Scattering angle, theta = 90⁰

We know that, Compton shift = [h / (m₀ c)] [1 - cos (theta)]
i.e., Compton shift = [(6.6 x 10^-34 J.s)] / [(9.1 x 10^-31 kg) (3 x 10⁸ m/s)] {1 - cos (90⁰)}
i.e., Compton shift = [(6.6 x 10^-34 J.s)] / [(9.1 x 10^-31 kg) (3 x 10⁸ m/s)]
i.e., Compton shift = 0.0242 A⁰

Compton shift = (lamda)' - (lamda) = 0.0242 A⁰
So, wavelength of scattered X-ray = (lamda)' = (lamda) + 0.0242 A⁰
i.e., wavelength of scattered X-ray = (lamda)' = 1 A⁰ + 0.0242 A⁰
i.e., wavelength of scattered X-ray = (lamda)' = 1.0242 A⁰

Kinetic energy imparted to recoil electron =
Energy of incident X-ray - Energy of scattered X-ray
= (h c) / (lamda) - (h c) / (lamda)'
= h c [{1 / (lamda)} - {1 / (lamda)'}]
= (6.6 x 10^-34 J.s) (3 x 10⁸ m/s) [(1 / 10^-10 m) - (1 / 1.0242 x 10^-10 m)]
= 1.986 x 10^-15 (1 - 0.976) J.
= 4.7 x 10^-17 J.

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By :-   Dr. A. W. Pangantiwar