Answer
The electron does not acquire any fraction of the light's energy.
Work Step by Step
According to the conservation of energy,
$hf=hf^\prime+K$,
in which $hf$ is the energy of the incident x-ray photon, $hf^\prime$ is the energy of the scattered x-ray photon, and $K$ is the kinetic energy of the recoiling electron.
$\therefore$ $K=hf-hf^\prime$
or, $K=\frac{ch}{\lambda}-\frac{ch}{\lambda^\prime}$ ...........$(1)$
Now the Compton shift $\Delta\lambda(=\lambda^\prime-\lambda)$ is given by the formula:
$\Delta\lambda=\frac{h}{mc}(1-\cos\phi)$
Given, $\phi=0^{\circ}$
$\therefore$ $\Delta\lambda=\frac{h}{mc}(1-\cos0^{\circ})$
or, $\Delta\lambda=0$
or, $\lambda^\prime=\lambda$
Thus, the relation $(1)$ becomes,
$K=0$
$\therefore$ The kinetic energy of the recoiling energy is zero. This implies that the electron does not acquire any fraction of the light's energy.
