Question

A nucleus ruptures into two nuclear parts which have their velocity ratio equal to $1:2$ . The ratio of their nuclear sizes (nuclear radius) is
A) ${2^{\dfrac{1}{3}}}:1$
B) $2:{1^{\dfrac{1}{3}}}$
C) $3:{1^{\dfrac{1}{2}}}$
D) $1:{3^{\dfrac{1}{2}}}$

Answer 1

Hint:In this problem, we are given that a nucleus ruptures into two nuclear parts and their velocities are given in some ratio. There are no unbalanced external forces acting on the system hence, the momentum of the two nuclear parts must be balanced. Related mass in terms of radius to find the required ratio.

Complete step-by-step solution: As the nucleus breaks into two parts, the given process is nuclear fission. Nuclear fission is a process in which the nucleus of an atom splits into two or more smaller nuclei as fission products, and usually some by-product particles are obtained. Fission is a form of elemental transmutation.
We are given with the velocities of the parts after the fission. We know that as no external force was acting on the nucleus, the momentum of the system must be conserved.
Now, initially the nucleus was at rest. Therefore, the initial momentum must be zero as the initial velocity is zero. The final momentum is the sum of momentum of each part. Let ${m_1}$ be the mass of the first part and ${m_2}$ be the mass of the second part. Thus, the momentum can be given as:

$0 = {m_1}{v_1} + {m_2}{v_2}$

As initial momentum was zero. Here, ${v_1},{v_2}$ are the velocities of the parts respectively. We are given that the ratio of velocities is ${v_1}:{v_2} = 1:2$ .
Considering magnitude only, we can have the mass in the ratio as:

$\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{{v_2}}}{{{v_1}}}$

Substituting ${v_1}:{v_2} = 1:2$ , we get

$\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{2}{1}$ --equation $1$

The ratio is ${m_1}:{m_2} = 2:1$ ,

We know that, density of the two parts will be equal as they are made from the same nucleus hence the ratio of their mass to volume must also be equal. This implies, mass is directly proportional to volume. Considering the nucleus parts are in the shape of a sphere, the volume $V$ in terms of radius $r$ is given as $V = \dfrac{4}{3}\pi {r^3}$ .
We have achieved that mass is proportional to volume, therefore the ratio of mass in terms of radius is given as:

$\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{{r_1}^3}}{{{r_2}^3}}$

Here, ${r_1},{r_2}$ are the radius of the first and second part respectively.

Using equation $1$ we have;

$\dfrac{{{r_1}^3}}{{{r_2}^3}} = \dfrac{2}{1}$
$ \Rightarrow {r_1}:{r_2} = {2^{\dfrac{1}{3}}}:1$

Thus, option A is the correct option.

Note:Remember that mass can be expressed in terms of radius. Also, note that the density of the two parts will be equal. As no force was acting in the given system, the momentum of the system was conserved. Unlike fission, fusion is the process in which two smaller atoms combine to make a new atom.