Question

A proton and an alpha particle enter a uniform electric field perpendicular to the direction of the field. The ratio of distances travelled by the proton and the alpha particle in the direction of the field after a time of $t$ seconds is:

• A. 1:2
• B. 1:4
• C. 2:1
• D. 4:1

Hint:

Acceleration in uniform field: $a = qE/m$.
Distance under constant acceleration: $s = \frac12 a t^2$.
Compare proton and alpha particle: q and m must be used.

Answer 1

The Correct Option is B

• The acceleration of proton: $a_p = \dfrac{eE}{m}$,
acceleration of alpha particle: $a_\alpha = \dfrac{eE}{2m}$.
• Distance travelled under uniform acceleration: $s = \frac{1}{2} a t^2$,
with the same time $t$ for both particles.
• Therefore, the ratio of distances: \[ s_p : s_\alpha = a_p : a_\alpha = \frac{eE/m}{eE/(2m)} = 2 : 1. \] • Hence, the ratio of distances travelled by proton and alpha particle in the field is 2:1.