Answer
$39.23\;keV$
Work Step by Step
Let $E$ is the energy of the incident photon, $E^\prime$ is the energy of the scattered photon, and $K$ is the kinetic energy of the recoiling electron, which is imparted to the electron by photon
According to conservation of energy,
$E=E^\prime+K$
or, $K=E-E^\prime$ ...................$(1)$
According to quantum theory of light,
$E=hf=\frac{ch}{\lambda}$
and $E^\prime=hf^\prime=\frac{ch}{\lambda^\prime}$
where the symbols have their usual meanings.
Now, we need to find the fractional energy loss for photons that scatter from the electrons:
$\text{frac}=\frac{\text{energy loss}}{\text{initial energy}}=\frac{E-E^\prime}{E}=\frac{K}{E}$
$\frac{K}{E}=\frac{\frac{ch}{\lambda}-\frac{ch}{\lambda^\prime}}{\frac{ch}{\lambda}}$
or, $\frac{K}{E}=\frac{\frac{1}{\lambda}-\frac{1}{\lambda^\prime}}{\frac{1}{\lambda}}$
or, $\frac{K}{E}=\frac{\lambda^\prime-\lambda}{\lambda^\prime}$
or, $\frac{K}{E}=\frac{\Delta\lambda}{\lambda+\Delta\lambda}$
where, $\Delta\lambda=\lambda^\prime-\lambda=\frac{h}{mc}(1-\cos\phi)$ is called the Compton wavelength. $\phi$ is the angle at which it is scattered from its initial direction of motion.
$\therefore \frac{K}{E}=\frac{\frac{h}{mc}(1-\cos\phi)}{\lambda+\frac{h}{mc}(1-\cos\phi)}$
or, $K=\frac{E(1-\cos\phi)}{\frac{\lambda mc}{h}+(1-\cos\phi)}$
For head-on collision, $\phi=180^{\circ}$, hence
$K=\frac{2E}{\frac{\lambda mc}{h}+2}$
or, $K=\frac{E}{\frac{\lambda mc}{2h}+1}$
Substitute the given values,
or, $K=\frac{E}{\frac{\lambda mc}{2h}+1}$
or, $K=\frac{1.2\times 10^{5}\;eV}{\frac{10\times 10^{-12}\times 9.1\times 10^{-31}\times 3\times 10^{8}}{2\times6.63\times 10^{-34}}+1}$
or, $K=3.923\times 10^{4}\;eV=39.23\;keV$
$\therefore$ The photon imparts $39.23\;keV$ energy to an electron in a head-on collision.
