Explanation

(d) Given, $m_{\alpha} = 6.4 \times 10^{- 27}kg$
$q_{\alpha} = 3.2 \times 10^{- 19}$ C, $E = 1.6 \times 10^{5}\ {Vm}^{- 1}$
Force on $\alpha$-particle
$F = q_{\alpha}F = 3.2 \times 10^{- 19} \times 1.6 \times 10^{5}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 51.2 \times 10^{- 15}$N
Now, acceleration of the particle
$\alpha = \frac{F}{m_{\alpha}} = \frac{51.2 \times 10^{- 15}}{6.4 \times 10^{- 27}} = 0.8 \times 10^{13}\ {ms}^{- 2}$
$\because Initial\ velocity,\ \ \ \ \ \ \ \ \ \ \ \ u = 0$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v^{2} = 2\alpha s$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2 \times 8 \times 10^{12} \times 2 \times 10^{- 2} = 32 \times 10^{10}\ $
or $v = 4\sqrt{2} \times 10^{5}{ms}^{- 1}$