An electron of mass $m_{e}$ , initially at rest, moves through a certain distance in a uniform electric field in time $t_{1}$ . A proton of mass $m_{p}$ , which is also initially at rest, takes time $t_{2}$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio $t_{2} / t_{1}$ is nearly equal to

1

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

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

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(\frac{m_{e}}{m_{p}})

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Solution

The correct option is B ${(\frac{m_{p}}{m_{e}})}^{1 / 2}$
Electrostatic force for both the particles,
$F = e E$
where, charge $e$ is the magnitude of the charge on both the particles.

But, acceleration of electron, $a_{e} = \frac{e E}{m_{e}}$

and acceleration of proton, $a_{e} = \frac{e E}{m_{p}}$

Applying equation of motion,

$s = u t + \frac{1}{2} a t^{2}$

Since, both the particles are at rest initially, so $u = 0$ and we get,

$s = \frac{1}{2} a t^{2}$

Now, substituting the value of acceleration for both the particles, we get

For electron: $s_{e} = \frac{1}{2} (\frac{q E}{m_{e}}) t_{1}^{2} . . . (1)$

For proton: $s_{p} = \frac{1}{2} (\frac{q E}{m_{p}}) t_{2}^{2} . . . (2)$

According to the problem, both the particles move the equal distance, so $s_{e} = s_{p}$

Therefore, from $(1)$ and $(2)$ , we have

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

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

Hence, option (b) is the correct answer.

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Similar questions

Q. An electron of mass

m_{e}

, initially at rest, moves through a certain distance in a uniform electric field in time

t_{1}

. A proton of mass

m_{p}

, which is also initially at rest, takes time

t_{2}

to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio

t_{2} / t_{1}

is nearly equal to

An electron of mass $m_{e}$ , initially at rest, moves through a certain distance in a uniform electric field in time $t_{1}$ . A proton of mass $m_{p}$ , also initially at rest, takes time $t_{2}$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio $\frac{t_{2}}{t_{1}}$ is nearly equal to,