Answer
The required work is $~~1.75\times 10^{-12}~J$
Work Step by Step
We can find $\gamma$:
$\gamma = \frac{1}{\sqrt{1-\beta^2}}$
$\gamma = \frac{1}{\sqrt{1-0.9990^2}}$
$\gamma = 22.37$
We can find the kinetic energy of the electron at this speed:
$K = mc^2(\gamma-1)$
$K = (9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2(22.37-1)$
$K = 1.75\times 10^{-12}~J$
The required work is $~~1.75\times 10^{-12}~J$
