Explanation

(d) Here, mass of $\alpha -$ particle $m_{a} = 6.4 \times 10^{- 27}kg$
Charge of $\alpha -$ particle $q_{a} = 3.2 \times 10^{- 19}C$
Electric field $E = 1.6 \times 10^{5}\ Vm^{- 1}$
Force on $\alpha -$ particle in a uniform electric field $E$ is
$F = q_{\alpha}E = 3.2 \times 10^{- 19} \times 1.6 \times 10^{5}N$
$= 3.2 \times 1.6 \times 10^{- 14}N$
Acceleration of the $\alpha -$ particle is
$a = \frac{F}{m_{\alpha}} = \frac{3.2 \times 1.6 \times 10^{- 14}}{6.4 \times 10^{- 27}}ms^{- 2}$
Given $:u = 0$
$\therefore v^{2} = 2aS = \frac{2 \times 3.2 \times 1.6 \times 10^{- 14} \times 2 \times 10^{- 2}}{6.4 \times 10^{- 27}}$
[As $S = 2 \times 10^{- 2}m$]
$= 3.2 \times 10^{11} = 32 \times 10^{10}$
or $v = 4\sqrt{2} \times 10^{5}\ {ms}^{- 1}$