Question

A deuteron and an $\alpha $-particle are placed in an electrical field. The forces acting on them are ${F_1}$ and ${F_2}$ and their accelerations are ${a_1}$ and ${a_2}$ respectively.(This is the multiple choice question)
A. ${F_1} = {F_2}$
B. ${F_1} \ne {F_2}$
C. ${a_1} = {a_2}$
D. ${a_1} \ne {a_2}$

Answer 1

Hint:
Force acting on a particle in an electric field is directly proportional to the charge. And the acceleration is also directly proportional to the force.

Complete step by step answer:
(i)Deuteron is an isotope of hydrogen. It is represented as ${}_1{H^2}$. Therefore, it has one electron in it which means $q = 1$. And it has mass $m = 2$.
(ii)An $\alpha $-particle is simply a helium molecule. It is represented as${}_2H{e^4}$. Therefore, it has two electrons in it which means$q = 2$ and it has mass, $m = 4$.
(iii)Taking the forces of a deuteron and an $\alpha $-particle are ${F_1}$ and ${F_2}$ respectively. And also the mass of the deuteron and $\alpha $-particle is represented as ${m_1}$ and ${m_2}$ respectively. Similarly accelerations of the deuteron and an $\alpha $-particle are represented as ${a_1}$ and ${a_2}$ respectively.
(iii)We know the force is directly proportional to the charge from coulomb’s law which states that coulomb’s force is directly proportional to the charges in contact and inversely proportional to the distance between those charges.
$ \Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{q_1}}}{{{q_2}}}$
Now let us substitute the values in the given formula:
$\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{1}{2}$
${F_2} = 2{F_1}$,I t leads to ${F_1} \ne {F_2}$
(iv)According to Newton's second law of motion, $F = ma$. Newton’s second law of motion states that the force is directly proportional to the mass of the object and its acceleration in the definite direction.
$ \Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{m_1}{a_1}}}{{{m_2}{a_2}}}$
We can substitute the values of mass value in the formula:
$\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{2{a_1}}}{{4{a_2}}}$
We know that $\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{1}{2}$ substitute this value in the above equation. And cancelling the values in the right-hand side of the equation.
$\dfrac{1}{2} = \dfrac{{{a_1}}}{{2{a_2}}}$
Now just cross multiply the equation. Then we have,
$ \Rightarrow {a_1} = {a_2}$

(v)From these processes, we found that ${F_1} = {F_2}$ and ${a_1} = {a_2}$. As this is the multiple-choice question,So, the correct answers are “Option B & C”.


Additional Information:
Coulomb’s law applied to the particle which is smaller as an atom which equals ${10^{ - 10}}m$.

Note:
Compared the values of mass, charge and the accelerations of the deuteron and an $\alpha $-particle to know the relations by using Coulomb's law and Newton's second law of motion. There are three laws of motion framed by Newton. The remaining laws of motion are given.
Newton’s first law:
Newton’s first law states that an object remains in rest or of a uniform motion unless it is compelled to change its state by an external applied force.
Newton’s third law:
Newton’s third law states that there is an equal and opposite reaction for every action.