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 $1.6 \times10^{5} V m^{-1}$. The velocity of the particle at the end of $2 \times10^{-2}m$ path when it starts from rest is

• A. $ 8\times10^{5} ms^{-1}$
• B. $ 16\times10^{5} ms^{-1}$
• C. $ 4\sqrt{2}\times 10^{5} ms^{-1} $
• D. $ 2\sqrt{3}\times 10^{5} ms^{-1} $

Answer 1

Answer: (C)

Given, $m_{\alpha}=6.4 \times 10^{-27}\, kg$,
$q_{\alpha}=3.2 \times 10^{-19} C , E=1.6 \times 10^{5}\, Vm ^{-1}$
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}} $
$=0.8 \times 10^{13} \,ms ^{-2}$
$\because$ 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 $\,\,\,\,v =4 \sqrt{2} \times 10^{5}\, ms ^{-1}$