Q168 / 180 · KCET 2009 · Physics
An $\alpha$-particle of mass $6.4 \times 10^{-27} \mathrm{~kg}$ and charge $3.2 \times 10^{-19} \mathrm{C}$ is situated in a uniform electric field of $1.6 \times 10^{5} \mathrm{Vm}^{-1}$. The velocity of the particle at the end of $2 \times 10^{-2} \mathrm{~m}$ path when it starts from rest is
- A$2 \sqrt{3} \times 10^{5} \mathrm{~ms}^{-1}$
- B$8 \times 10^{5} \mathrm{~ms}^{-1}$
- C$16 \times 10^{5} \mathrm{~ms}^{-1}$
- Correct answerD$4 \sqrt{2} \times 10^{5} \mathrm{~ms}^{-1}$
Solution
Given, $m_{\alpha}=6.4 \times 10^{-27} \mathrm{~kg}$,
$$
\mathrm{q}_{\alpha}=3.2 \times 10^{-19} \mathrm{C}, \mathrm{E}=1.6 \times 10^{5} \mathrm{Vm}^{-1}
$$
Force on $\alpha$-particle
$$
\begin{gathered}
\mathrm{F}=\mathrm{q}_{\alpha} \mathrm{E}=3.2 \times 10^{-19} \times 1.6 \times 10^{5} \\
=51.2 \times 10^{-15} \mathrm{~N}
\end{gathered}
$$
Now, acceleration of the particle
$\alpha=\frac{\mathrm{F}}{\mathrm{m}_{\alpha}}=\frac{51.2 \times 10^{-15}}{6.4 \times 10^{-27}}$
$=0.8 \times 10^{13} \mathrm{~ms}^{-2}$
$\because$ Initial velocity, $\mathrm{u}=0$
$\mathrm{v}^{2}=2 \alpha \mathrm{S}$
$=2 \times 8 \times 10^{12} \times 2 \times 10^{-2}$
$=32 \times 10^{10}$
or
$\mathrm{v}=4 \sqrt{2} \times 10^{5} \mathrm{~ms}^{-1}$
Hence, the correct answer is option: 'D'.