Question

When an $\alpha $ particle of mass $m$ moving with velocity $v$ bombarded on a heavy nucleus of charge $Ze$, its distance of closest approach from the nucleus depends in $m$ as:
(A) $\dfrac{1}{m}$
(B) $\dfrac{1}{{\sqrt m }}$
(C) $\dfrac{1}{{{m^2}}}$
(D) $m$

Answer 1

Hint the energy can neither be created nor destroyed, the energy can be transferred from one form of the energy to the other form of the energy. This can be given by the law of conservation of energy. Here, the kinetic energy is converted to the potential energy.
Useful formula:
The kinetic energy can be given by,
$KE = \dfrac{1}{2}m{v^2}$
Where, $KE$ is the kinetic energy of the particle, $m$ is the mass of the particle and $v$ is the velocity of the particle.
The potential energy can be given by,
$PE = k\dfrac{{Qq}}{d}$
Where, $PE$ is the potential energy of the particle, $k$ is the constant, $Q$ is the charge of the one particle, $q$ is the charge of the other particle and $d$ is the distance between the two charges.

Complete step by step answer
Given that,
The mass of the particle is given as, $m$,
The velocity of the particle is given as, $v$,
Now,
The kinetic energy of the particle can be given by,
$KE = \dfrac{1}{2}m{v^2}\,.....................\left( 1 \right)$
Now,
The potential energy of the particle can be given by,
$PE = k\dfrac{{Qq}}{d}\,...................\left( 2 \right)$
By the law of the conversation of the energy, then the equation (1) is equated with the equation (2), then
$\dfrac{1}{2}m{v^2} = k\dfrac{{Qq}}{d}$
In the question, the relation between the distance and the mass is asked, so assume the remaining terms as the constant, then
$m = \dfrac{1}{d}$
By rearranging the terms in the above equation, then the above equation is written as,
$d = \dfrac{1}{m}$

Hence, the option (A) is the correct answer.

Note The kinetic energy of the particle is directly proportional to the mass of the particle and the square of the velocity of the particle. As the mass of the particle and the square of the velocity of the particle increases then the kinetic energy of the particle also increases.