Answer
See explanation. With such a large kinetic energy, the electrons are relativistic.
Work Step by Step
a. $\gamma=\frac{1}{\sqrt{1-v^{2}/c^{2}}}$
The relativistic kinetic energy is $(\gamma-1)mc^2$.
We are told $(\gamma-1)mc^2=(\gamma-1)0.511\times10^6 eV=7.50\times10^5 eV $
Solve for v when $\gamma-1=\frac{7.50\times10^5 eV }{0.511\times10^6 eV }=1.46$
$$\gamma=\frac{1}{\sqrt{1-v^{2}/c^{2}}}=2.46$$
$$v=c\sqrt{1-1/\gamma^2}=0.914c$$
The ratio of v/c is 0.914.
b. Using $\frac{1}{2}mv^2=7.50\times10^5 eV $, we obtain $v = 5.13\times10^8m/s$, which is greater than c.
