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a. ${2^{\dfrac{1}{3}}}:1$

b. $1:{3^{\dfrac{1}{2}}}$

c. ${3^{\dfrac{1}{2}}}:1$

d. ${2^{\dfrac{1}{3}}}:1$

Answer

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Total momentum of the system:

$ \Rightarrow M = {m_1}{v_1} + {m_2}{v_2}$

Where, ${m_1},{v_1}$ is the mass and velocity of the first nuclear part and ${m_2},{v_2}$ is the mass and velocity of the second nuclear part.

In the question, it is given that the nucleus disintegrates into two nuclear parts which have their velocities in the ratio $2:1$.

Consider the diagram. A nucleus disintegrates into two nuclear parts. The first nuclear part has the momentum ${m_1}$ and moves with the velocity of ${v_1}$. The second nuclear part has the momentum ${m_2}$ and moves with the velocity of ${v_2}$.

The masses of the system are constant. Thus, it is conserved in the total momentum of the system. That is the initial and the final momentum is conserved. We have the formula to calculate the total momentum.

Total momentum of the system:

$ \Rightarrow M = {m_1}{v_1} + {m_2}{v_2}$

Where, ${m_1},{v_1}$ is the mass and velocity of the first nuclear part and ${m_2},{v_2}$ is the mass and velocity of the second nuclear part.

The total momentum is conserved and hence the value of ${m_1}{v_1} + {m_2}{v_2}$ is zero. We can represent as,

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

We can bring the term ${m_2},{v_2}$ to the right hand side. The sign of the term will be changed into the opposite sign.

$ \Rightarrow {m_1}{v_1} = - {m_2}{v_2}$

We can rearrange the common terms. That is, we can bring the mass to the right hand side and velocity terms into the left hand side. We get,

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

We have the value for $\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{ - {v_2}}}{{{v_1}}}$ as $\dfrac{1}{2}$ that is,

$ \Rightarrow \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{ - {v_2}}}{{{v_1}}} = \dfrac{1}{2}$

As discussed before the masses of the nuclear particles are the same. So, the densities are also the same. We know the value for the mass, that is,

$ \Rightarrow m = \dfrac{4}{3}\pi {r^3}\rho $

Where $\rho $ is the density, $m$ is the mass and $r$ is the radius

Consider,

$ \Rightarrow {m_1} = \dfrac{4}{3}\pi {r_1}^3\rho $

$ \Rightarrow {m_2} = \dfrac{4}{3}\pi {r_2}^3\rho $

We can divide the masses. We get,

$ \Rightarrow \dfrac{{\dfrac{4}{3}\pi {r_1}^3\rho }}{{\dfrac{4}{3}\pi {r_2}^3\rho }}$

To simplify the given equation, we can cancel out the common terms we get,

$ \Rightarrow \dfrac{{{r_1}^3}}{{{r_2}^3}}$

We can take the cube as the whole term cube. That is,

$ \Rightarrow {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^3}$

The value of ${\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^3}$ is equal to $\dfrac{1}{2}$ as the value of $\dfrac{{{m_1}}}{{{m_2}}}$ is equal to $\dfrac{1}{2}$.

$ \Rightarrow {\left( {\dfrac{{{r_1}}}{{{r_2}}}} \right)^3} = \dfrac{1}{2}$

We can take the cube root for the left-hand side part to remove the cube power in the right-hand side. That is,

$ \Rightarrow \left( {\dfrac{{{r_1}}}{{{r_2}}}} \right) = {\left( {\dfrac{1}{2}} \right)^{\dfrac{1}{3}}}$

$ \Rightarrow \left( {\dfrac{{{r_1}}}{{{r_2}}}} \right) = \left( {\dfrac{1}{{{2^{\dfrac{1}{3}}}}}} \right)$

The ratio is ${1 : 2^{\dfrac{1}{3}}}$.