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(D) The acceleration $a$ of a particle of mass $m$ and charge $q$ in a uniform electric field $E$ is given by $a = \frac{qE}{m}$.For a proton $(p)$: $

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$1: \sqrt{2}$

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(D) The acceleration $a$ of a particle of mass $m$ and charge $q$ in a uniform electric field $E$ is given by $a = \frac{qE}{m}$.For a proton $(p)$: $q_p = e$, $m_p = m$. Thus, $a_p = \frac{eE}{m}$.For an $\alpha$-particle $(\alpha)$: $q_{\alpha} = 2e$, $m_{\alpha} = 4m$. Thus, $a_{\alpha} = \frac{2eE}{4m} = \frac{eE}{2m}$.Since they start from rest, the distance $s$ traveled in time $t$ is $s = \frac{1}{2}at^2$. Given $s$ is the same for both, $\frac{1}{2}a_p t_p^2 = \frac{1}{2}a_{\alpha} t_{\alpha}^2$.$\frac{t_p^2}{t_{\alpha}^2} = \frac{a_{\alpha}}{a_p} = \frac{eE/2m}{eE/m} = \frac{1}{2}$.Therefore, the ratio of times taken by the proton to the $\alpha$-particle is $\frac{t_p}{t_{\alpha}} = \frac{1}{\sqrt{2}}$.

$A$ proton and an $\alpha$-particle start from rest in a uniform electric field. The ratio of times taken by them to travel the same distance in the field is

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