An $\alpha$-particle of mass  $6.4\times {10}^{-27}$Kg and charge  $3.2\times {10}^{-19}$C is situated in a uniform electric field of $1.6 \times {10}^5$v/m. The velocity of the particle at the end of  $2\times {10}^{-2}$ m path when it starts from rest is 

Here mass of  $\alpha$-particle $M_{\alpha }=6.4\times {10}^{-27}$

Force on charge of $\alpha$-particle $q_{\alpha }=3.2\times {10}^{-19}\times 1.6\times {10}^5$

                                                     $=3.2\times 1.6\times {10}^{-14}$

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$

Given u=0 $\Rightarrow$$v^2=2as$

$v^2=\frac{2\times 3.2\times 1.6\times {10}^{-14}\times 2\times {10}^{-2}}{6.4\times {10}^{-27}}$ as $s=2\times {10}^{-2}$

=$3.2\times {10}^{11}$=$32\times {10}^{10}$

$\therefore v=4\sqrt{2}\times {10}^{5}m{s}^{-1}$