Question

A heavy nucleus at rest breaks into two fragments which fly off with velocities 8:1. The ratio of radii of fragments is:
(A) 1:2
(B) 1:4
(C) 4:1
(D) 2:1

Answer 1

In any reaction, such as in the question, the momentum is always conserved i.e. total momentum before reaction is equal to the total momentum after reaction. The volume of the nucleus is directly proportional to the cube of its radius.

Formula used: In this solution we will be using the following formula;
$\Rightarrow p = mv$ where $p$ is the linear momentum of an object, $m$ is mass of the object and $v$ is the velocity of the object.

Complete step by step answer
In the question, a particle is at rest hence velocity and momentum are zero. This particle breaks off into two other particles.
According to the principle of conservation of momentum, the total momentum before the break off (decay) must be equal to the total momentum after
Hence
$\Rightarrow 0 = {p_1} + {p_2}$
$\Rightarrow {p_1} = - {p_2}$. The negative simply shows the direction of the second particle. It signifies they are in the opposite direction. We are interested in their magnitudes, thus, it can be dropped.
Hence,
$\Rightarrow {p_1} = {p_2}$
Now momentum is given as
$\Rightarrow p = mv$ where, $m$ is mass of the object and $v$ is the velocity of the object,
Thus,
$\Rightarrow {m_1}{v_1} = {m_2}{v_2}$
$\Rightarrow \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{{v_2}}}{{{v_1}}}$
The ratio of the velocities is say ${v_1}:{v_2} = 8:1$
Thus,
$\Rightarrow \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{8}$
But the mass is directly proportional to the volume, which in turn is directly proportional to the cube of the radius hence, we can write
$\Rightarrow \dfrac{{r_1^3}}{{r_2^3}} = \dfrac{1}{8}$
$\Rightarrow {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^3} = \dfrac{1}{8}$
Then by finding the cube root of both sides, we have
$\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{1}{2}$
$\Rightarrow {r_1}:{r_2} = 1:2$

Hence, the correct option is A.

Note
The mass of a nucleus is proportional to the cube of the radius because, the mass can be written as
$m = \rho V$ where $\rho$ is called the nuclear density and $V$ is the volume of the nucleus. In physics, the nucleus is considered a sphere and thus, its volume can be given as
$V = \dfrac{4}{3}\pi {R^3}$ where $R$ is the radius of the particular nucleus.
Hence, $m = \rho \dfrac{4}{3}\pi {R^3}$
$\Rightarrow m \propto {R^3}$