From, $q V=\frac{1}{2} m v^2$
where, $V$ is stopping potential.
$V=\frac{m v^2}{2 q}$
$=\frac{9 \times 10^{-31} \times\left(4 \times 10^6\right)^2}{2 \times 16 \times 10^{-19}}$
$\approx 45 \,V$
So, electron must move across a potential difference of $45 \,V$ from higher to lower potential.


