Question

An $\alpha$-partic!e of mass $6.4\times 10^{-27}$ kg and charge $3.2\times 10^{-19}C$ is situated in a uniform electric field of $1.6\times 10^{5}V{m}^{-1}$. The velocity of the particle at the end of $2\times 10^{-2}m$ path when it starts from rest is

• A. $8\times 10^{5}m{s}^{-1}$
• B. $16\times 10^{5}m{s}^{-1}$
• C. $4\sqrt{2}\times 10^{5}m{s}^{-1}$
• D. $2\sqrt{3}\times 10^{5}m{s}^{-1}$

Answer 1

Answer: (C)

Given, $m_{\alpha }=6.4\times 10^{-27}kg$,
$q_{\alpha }=3.2\times 10^{-19}C,E=1.6\times 10^{5}V{m}^{-1}$
Force on $\alpha$ -particle
$F=q_{\alpha }E=3.2\times 10^{-19}\times 1.6\times 10^5$
$=51.2\times 10^{-15}N$
Now, acceleration of the particle
$\alpha =\frac{F}{m_{\alpha }}=\frac{51.2\times 10^{-15}}{6.4\times 10^{-27}}$
$=0.8\times 10^{13}m{s}^{-2}$
$\because$ 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 $v=4\sqrt{2}\times 10^{5}m{s}^{-1}$