Question

An electron of mass ${m_e}$ . initially at rest takes time ${t_1}$ to move a distance in a uniform electric field in the same field environment, a proton of mass ${m_p}$ initially at rest takes time ${t_2}$ to move the same distance (in the opposite direction). Ignoring gravity, the ratio $\dfrac{{{t_2}}}{{{t_1}}}$ is
A. $1$
B. ${\left( {\dfrac{{{m_e}}}{{{m_p}}}} \right)^{\dfrac{1}{2}}}$
C. $\dfrac{{{m_p}}}{{{m_e}}}$
D. ${\left( {\dfrac{{{m_p}}}{{{m_e}}}} \right)^{\dfrac{1}{2}}}$

Answer 1

Here, we are given an electron and a proton that are initially at rest. We will use the formula of the distance in case of electrons and protons to calculate the time taken to move a distance by both electrons and protons. Also, we will use Newton’s second law of motion to calculate the acceleration of protons and electrons. Then we will calculate the ratio of both the accelerations to get the required answer.

Complete step by step answer:
As we know that the acceleration of the particle is $a$ , then the distance $s$ travelled by the particle in time $t$ is given by
$s = \dfrac{1}{2}a{t^2}$
Now, it is given in the question that the electron of mass ${m_e}$ takes time ${t_1}$ to move some distance, therefore, the distance travelled by an electron is given by
$s = \dfrac{1}{2}{a_e}t_1^2$
Also, the time taken by the proton of mass ${m_p}$ to move some distance is ${t_p}$ . therefore, the distance travelled by a proton is given by
$s = \dfrac{1}{2}{a_p}t_2^2$
Now, dividing the distance travelled by a proton by the distance travelled by an electron, we get
$\dfrac{{t_2^2}}{{t_1^2}} = \dfrac{{{a_p}}}{{{a_e}}}$
$\Rightarrow \,\dfrac{{{t_2}}}{{{t_1}}} = \sqrt {\dfrac{{{a_p}}}{{{a_e}}}}$
Now, the force acting on an electron in uniform electric field is given by
${F_e} = eE$
Now, according to Newton’s second law of motion, the force acting on the body is equal to the product of mass and acceleration of the body and is given below
$F = ma$
$\Rightarrow \,a = \dfrac{F}{m}$
Now, the acceleration of the electron having mass ${m_e}$ is given by
${a_e} = \dfrac{{{F_e}}}{{{m_e}}}$
$\Rightarrow \,{a_e} = \dfrac{{eE}}{{{m_e}}}$
Also, the acceleration of the proton having mass ${m_p}$ is given by
${a_p} = \dfrac{{{F_p}}}{{{m_p}}}$
$\Rightarrow \,{a_p} = \dfrac{{eE}}{{{m_p}}}$
Now, dividing the acceleration of the electron by acceleration of the proton, we get
$\dfrac{{{a_e}}}{{{a_p}}} = \dfrac{{eE}}{{{m_e}}} \times \dfrac{{{m_p}}}{{eE}}$
$\Rightarrow \,\dfrac{{{a_e}}}{{{a_p}}} = \dfrac{{{m_p}}}{{{m_e}}}$
Now, putting the above value in the ratio of the time, we get
$\therefore\dfrac{{{t_2}}}{{{t_1}}} = \sqrt {\dfrac{{{m_p}}}{{{m_e}}}}$
Therefore, the ratio of $\dfrac{{{t_2}}}{{{t_1}}}$ is $\sqrt {\dfrac{{{m_p}}}{{{m_e}}}}$ .

Hence, option D is the correct option.

Note: The force on the charge of magnitude $q$ at any point in space is equal to the product of the charge and the electric field and is given by $F = qE$. Here, in the above example, the charge is an electron that is of magnitude $e$ , that is why the force on the electron is given by $F = eE$.