Question

A proton and an $\alpha $- particle start from rest in a uniform electric field, then the ratio of times of flight to travel same distance in the field is
(A) $\sqrt 5 :\sqrt 2 $
(B) $\sqrt 3 :1$
(C) $2:1$
(D) $1:\sqrt 2 $

Answer 1

Hint: We need to first calculate the acceleration of each of the particles in the electric field. Then by substituting this value of acceleration in the equation of motion and equating the distances travelled, we get the ratio of the times of flight.
Formula used: In this solution we will be using the following formula,
$\Rightarrow F = ma$
where $F$ is the force applied to the mass $m$ causing the acceleration $a$
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
where $s$ is the distance travelled, $u$ is the initial velocity and $t$is the time.

Complete step by step solution:
 In the question we are told that the charged particles enter an electric field $E$.
Let us first consider the proton. Let the mass of the proton is $m$ and the charge on the proton be $q$. On entering the electric field, the proton experiences an acceleration of $a$.
Therefore, we can write $F = ma$, where the force on the proton will be the product of its charge and the value of the electric field. So, $F = qE$. Substituting we get,
$\Rightarrow qE = ma$…..(1)
Now for the alpha particle, the mass is $4m$ and the charge is $2q$. So the force on the alpha particle in the electric field is $F = 2qE$. Let this force cause an acceleration $b$. So we can write, $F = mb$. Substituting the value of the force we get,
$\Rightarrow 2qE = mb$……(2)
Taking the ratio of (1) and (2) we get,
$\Rightarrow \dfrac{{qE}}{{2qE}} = \dfrac{{ma}}{{mb}}$
Cancelling the common terms we have,
$\Rightarrow \dfrac{a}{b} = \dfrac{1}{2}$
Now the distance travelled by the proton is given by the formula, $s = ut + \dfrac{1}{2}a{t^2}$. The initial velocity of the proton is zero. So we have,
$\Rightarrow s = \dfrac{1}{2}a{t_a}^2$
Similarly for the alpha particle the distance travelled will be,
$\Rightarrow s = \dfrac{1}{2}b{t_b}^2$
According to the problem this distance travelled is equal. So we have,
$\Rightarrow \dfrac{1}{2}a{t_a}^2 = \dfrac{1}{2}b{t_b}^2$
On cancelling the common terms and taking the same terms on one side we get,
$\Rightarrow \dfrac{{{t_b}^2}}{{{t_a}^2}} = \dfrac{a}{b}$
Therefore we can substitute $\dfrac{a}{b} = \dfrac{1}{2}$ and get,
$\Rightarrow \dfrac{{{t_b}^2}}{{{t_a}^2}} = \dfrac{1}{2}$
On taking square root,
$\Rightarrow \dfrac{{{t_a}}}{{{t_b}}} = \dfrac{1}{{\sqrt 2 }}$
So the correct answer is option D.

Note:
In this problem we have taken the mass of the alpha particle as four times the mass of the proton. This is because the alpha particle is a doubly ionised helium nucleus. So it contains 2 protons and 2 neutrons. Since the mass of neutrons is comparable to the mass of protons, we can write the mass of an alpha particle as 4 times the mass of a proton.