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 \( t_2/t_1 \) is nearly equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \left(m_p/m_e\right)^{1/2} \)
The electron and the proton carry charges of the same magnitude \( e \). So in the same field both feel a force of the same size \( eE \). The acceleration \( a=eE/m \) is smaller for the heavier particle. Starting from rest, the distance covered is \( h=\frac{1}{2}at^{2} \).
The time to cover the distance \( h \) is:
\( \begin{aligned} t &= \sqrt{\frac{2h}{a}} \\ &= \sqrt{\frac{2hm}{eE}} \end{aligned} \)
For the same \( h \) and \( E \), \( t\propto\sqrt{m} \). So
\( \begin{aligned} \frac{t_2}{t_1} &= \sqrt{\frac{m_p}{m_e}} \\ &= \left(m_p/m_e\right)^{1/2} \end{aligned} \)
Since \( m_p\approx1836\,m_e \), the proton takes about 43 times as long as the electron.
NCERT (2026-27):
Example 1.7 – “… Starting from rest, the time required by the electron to fall through a distance h is given by
\( \displaystyle t_e=\sqrt{\frac{2h}{a_e}}= \)\( \displaystyle \sqrt{\frac{2hm_e}{eE}} \)
… The time of fall for the proton is
\( \displaystyle t_p=\sqrt{\frac{2h}{a_p}}= \)\( \displaystyle \sqrt{\frac{2hm_p}{eE}} \)\( \displaystyle =1.3\times10^{-7}\,\mathrm{s} \)
Thus, the heavier particle (proton) takes a greater time to fall through the same distance.”