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

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
\(\mathrm{v}=4 \sqrt{2} \times 10^{5} \mathrm{~ms}^{-1}\)