An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is ?

An electron falls from rest through a vertical  distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed,  keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is ?

(a) smaller

(b) 5 times greater

(c) equal

(d) 10 times greater

Answer :  (a) smaller

$$ F = qE $$

 $$ a_e = \frac{eE}{m_e} $$

 $$ a_p = \frac{eE}{m_p} $$

\ Since \( m_p \gg m_e \), we have \( a_p \ll a_e \) Thus, the electron accelerates faster and falls quicker. Using the kinematic equation: 

 $$ s = ut + \frac{1}{2} a t^2 \quad \text{(with } u = 0 \text{)} $$

 $$ h = \frac{1}{2} a t^2 $$

 Solving for \( t \): $$ t = \sqrt{ \frac{2h}{a} } $$

 Time for electron: $$ t_e = \sqrt{ \frac{2 m_e h}{eE} } $$

 Time for proton: $$ t_p = \sqrt{ \frac{2 m_p h}{eE} } $$

 Comparing times: $$ \frac{t_p}{t_e} = \sqrt{ \frac{m_p}{m_e} } \approx \sqrt{1800} $$

 $$ t_p \gg t_e \Rightarrow t_e < t_p $$