Explanation

(c) When electron is accelerated through a potential difference of $V$ volts, then
Kinetic energy –$eV$
$ie.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}mv^{2} = eV$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \sqrt{\left( \frac{2eV}{m} \right)}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v =$ $\sqrt{\left( \frac{2 \times 1.6 \times 10^{- 19} \times 45.5}{9.1 \times 10^{- 31}} \right)}$
=$\ 4 \times 10^{6}\ {ms}^{- 1}$