A photon strikes a free electron at rest and is scattered straight backward. If the speed of electron after collision is $\alpha c$, where $\alpha \ll 1$, then,

Given, $v=\alpha c$ where $\alpha \ll 1$ the electron is moving at non-relativistic speed. 

By Law of Conservation of Momentum.

$\frac{h}{\lambda }=\left(-\frac{h}{{\lambda }^{\text{'}}}\right)+mv$

$\Rightarrow \frac{hc}{\lambda }+\frac{hc}{{\lambda }^{\text{'}}}=mcv=\frac{m{v}^2}{\alpha } \cdots \left(i\right)$

By Law of Conservation of Energy

$\frac{hc}{\lambda }-\frac{hc}{{\lambda }^{\text{'}}}=\frac{1}{2}m{v}^2 \cdots \left(ii\right)$

Adding (i) and (ii), we get

$2\frac{hc}{\lambda }=\frac{1}{2}m{v}^2+\frac{1}{\alpha }\left(m{v}^2\right)$

$\Rightarrow \frac{hc}{\lambda }=E_p=\frac{m{v}^2}{4}+\frac{m{v}^2}{2\alpha }$

Since, $\alpha \ll 1$

$$\begin{gathered} \Rightarrow \frac{1}{\alpha }\gg 1 \\ \Rightarrow \frac{m{v}^2}{4}\ll \frac{m{v}^2}{2\alpha } \\ \Rightarrow E_p\cong \frac{1}{\alpha }\left(\frac{m{v}^2}{2}\right) \\ \Rightarrow \frac{1}{2}m{v}^2=\alpha E_p \end{gathered}$$