Answer
$v = 0.822c$
Work Step by Step
The electron’s KE comes from the decrease in potential energy, so it has a kinetic energy of $6.20\times10^{-14}J$. Use equation 26–5b.
$$KE=6.20\times10^{-14}J =(\gamma-1)mc^2=$$
$$\frac{1}{\sqrt{1-v^2/c^2}}-1=\frac{6.20\times10^{-14}J }{(9.11\times10^{-31}kg)(3.00\times10^8m/s)^2}$$
$$v=c\sqrt{1-\frac{1}{(\frac{6.20\times10^{-14}J }{(9.11\times10^{-31}kg)(3.00\times10^8m/s)^2}+1)^2}}$$
$$v=0.822c$$