Answer
$v=1.11\times 10^7\frac{m}{s}$
Work Step by Step
From law of conservation of energy,
K.E=P.E
$\frac{1}{2}mv^2=qV$
Rearranging this formula to solve for $v$:
$v^2=\frac{2qV}{m}$
Now, Taking square root on both sides, we obtain
$v=\sqrt{\frac{2qV}{m}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{2(1.6\times 10^{-19}\times 350)}{9.109\times 10^{-31}}}$
$v=1.11\times 10^7\frac{m}{s}$
