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An electron (mass = $m_{e}$ ) and a proton (mass= $m_{p}$ ) initially at rest move through a certain distance in a uniform electric field in times $t_{1}$ and $t_{2}$ . Neglect the effect of gravity. then

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The acceleration of electron is much greater than that of proton

The acceleration of proton is much greater than that of electron

$\frac{t_{1}}{t_{2}} = {(\frac{m_{e}}{m_{p}})}^{1 / 2}$

$\frac{t_{1}}{t_{2}} = {(\frac{m_{p}}{m_{e}})}^{1 / 2}$

answer is M.

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Detailed Solution

$a_{e} = \frac{F}{m_{e}}, a_{p} = \frac{F}{m_{p}}$

Since $m_{e} < < m_{p}$ , so

$a_{e} > > a_{p}$

Now, $d = \frac{1}{2} a_{e} t_{1}^{2}$ and $d = \frac{1}{2} a_{p} t_{2}^{2}$

$\frac{t_{1}}{t_{2}} = {(\frac{m_{e}}{m_{p}})}^{1 / 2}$

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