Question

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

• A. $1$
• B. ${\left(\frac{m_p}{m_e}\right)}^{1/2}$
• C. ${\left(\frac{m_e}{m_p}\right)}^{1/2}$
• D. $\left(\frac{m_e}{m_p}\right)$

Answer 1

The correct option is B ${\left(\frac{m_p}{m_e}\right)}^{1/2}$

Electrostatic force for both the particles,
$F=eE$
where, charge $e$ is the magnitude of the charge on both the particles.

But, acceleration of electron, $a_e=\frac{eE}{m_e}$
​
and acceleration of proton, $a_e=\frac{eE}{m_p}$

Applying equation of motion,

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

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: $s_e=\frac{1}{2}\left(\frac{qE}{m_e}\right)t_1^2...\left(1\right)$

For proton: $s_p=\frac{1}{2}\left(\frac{qE}{m_p}\right)t_2^2...\left(2\right)$

According to the problem, both the particles move the equal distance, so $s_e=s_p$

Therefore, from $\left(1\right)$ and $\left(2\right)$, we have

$\frac{t_1^2}{m_e}=\frac{t_2^2}{m_p}$

$\frac{t_2}{t_1}={\left(\frac{m_p}{m_e}\right)}^{1/2}$

Hence, option (b) is the correct answer.