Answer
The required work is $~~189~MeV$
Work Step by Step
We can find $\gamma_i$:
$\gamma_i = \frac{1}{\sqrt{1-\beta_i^2}}$
$\gamma_i = \frac{1}{\sqrt{1-0.9850^2}}$
$\gamma_i = 5.79528$
We can find an expression for the initial kinetic energy of the electron:
$K_i = mc^2(5.79528-1)$
$K_i = 4.79528~mc^2$
We can find $\gamma_f$:
$\gamma_f = \frac{1}{\sqrt{1-\beta_f^2}}$
$\gamma_f = \frac{1}{\sqrt{1-0.9860^2}}$
$\gamma_f = 5.99717$
We can find an expression for the final kinetic energy of the electron:
$K_f = mc^2(5.99717-1)$
$K_f = 4.99717~mc^2$
We can find the required work:
$Work = \Delta K$
$Work = K_f-K_i$
$Work = (4.99717~mc^2)-(4.79528~mc^2)$
$Work = 0.20189~mc^2$
$Work = (0.20189)~(938\times 10^6~eV)$
$Work = 189\times 10^6~eV$
$Work = 189~MeV$
The required work is $~~189~MeV$
