A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2:1. The ratio of their nuclear sizes will be
a) ${{2}^{1/3}}:1$
b)$1:{{3}^{1/2}}$
c) $3:1$
d) $1:{{2}^{1/3}}$
Answer
605.1k+ views
Hint: Initially the nucleus just disintegrates without having any momentum and the nuclear particle move apart from each other having some velocities. The density of both the fragments is also the same as they are from the same parent nucleus. Hence using the principle of conservation of momentum, we can obtain the ratio of the sizes of the nuclear particles.
Formula used:
${{m}_{1}}{{v}_{1}}+{{m}_{2}}(-{{v}_{2}})=0$
Complete answer:
In the question it is given that the nucleus disintegrates into two nuclear parts whose ratio of velocities is 2:1. Let us say that the mass of the nuclear part 1 is ${{m}_{1}}$ and after disintegration moves with velocity ${{v}_{1}}$ . Similarly, the mass of nuclear part 2 be ${{m}_{2}}$ and after disintegration moves with velocity ${{v}_{2}}$. The parent nucleus is at rest and the nuclear parts after disintegration move way from each other. Hence using the principle of conservation of momentum we get,
$\begin{align}
& {{m}_{1}}{{v}_{1}}+{{m}_{2}}(-{{v}_{2}})=0 \\
& \Rightarrow {{m}_{1}}{{v}_{1}}={{m}_{2}}{{v}_{2}} \\
& \therefore \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{v}_{2}}}{{{v}_{1}}}=\dfrac{1}{2}.....(1) \\
\end{align}$
Therefore the ratio of ${{m}_{1}}:{{m}_{2}}$ is 1:2. The two nuclear parts have the same density ‘d’. Let us say the radius of nuclear part 1 be ${{r}_{1}}$ and that of part 2 be ${{r}_{2}}$. Therefore the ratio of their sizes from equation 1 assuming that the two nuclear parts are small spheres is,
$\begin{align}
& \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{1}{2} \\
& \Rightarrow \dfrac{d\dfrac{4}{3}\pi {{r}_{1}}^{3}}{d\dfrac{4}{3}\pi {{r}_{2}}^{3}}=\dfrac{1}{2} \\
& \Rightarrow {{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}=\dfrac{1}{2} \\
& \therefore \dfrac{{{r}_{1}}}{{{r}_{2}}}={{\left( \dfrac{1}{2} \right)}^{1/3}} \\
\end{align}$
Therefore the ratio of the sizes of the two nuclear parts is $1:{{2}^{1/3}}$ .
Hence the correct answer of the above question is option d.
Note:
The mass of a body can be written as the product of its density times the volume. The principle of conservation of momentum states that the momentum of a system is always to be conserved i.e. momentum of a body cannot be lost but can be transferred. This is a consequence of the law of conservation of energy.
Formula used:
${{m}_{1}}{{v}_{1}}+{{m}_{2}}(-{{v}_{2}})=0$
Complete answer:
In the question it is given that the nucleus disintegrates into two nuclear parts whose ratio of velocities is 2:1. Let us say that the mass of the nuclear part 1 is ${{m}_{1}}$ and after disintegration moves with velocity ${{v}_{1}}$ . Similarly, the mass of nuclear part 2 be ${{m}_{2}}$ and after disintegration moves with velocity ${{v}_{2}}$. The parent nucleus is at rest and the nuclear parts after disintegration move way from each other. Hence using the principle of conservation of momentum we get,
$\begin{align}
& {{m}_{1}}{{v}_{1}}+{{m}_{2}}(-{{v}_{2}})=0 \\
& \Rightarrow {{m}_{1}}{{v}_{1}}={{m}_{2}}{{v}_{2}} \\
& \therefore \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{v}_{2}}}{{{v}_{1}}}=\dfrac{1}{2}.....(1) \\
\end{align}$
Therefore the ratio of ${{m}_{1}}:{{m}_{2}}$ is 1:2. The two nuclear parts have the same density ‘d’. Let us say the radius of nuclear part 1 be ${{r}_{1}}$ and that of part 2 be ${{r}_{2}}$. Therefore the ratio of their sizes from equation 1 assuming that the two nuclear parts are small spheres is,
$\begin{align}
& \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{1}{2} \\
& \Rightarrow \dfrac{d\dfrac{4}{3}\pi {{r}_{1}}^{3}}{d\dfrac{4}{3}\pi {{r}_{2}}^{3}}=\dfrac{1}{2} \\
& \Rightarrow {{\left( \dfrac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3}}=\dfrac{1}{2} \\
& \therefore \dfrac{{{r}_{1}}}{{{r}_{2}}}={{\left( \dfrac{1}{2} \right)}^{1/3}} \\
\end{align}$
Therefore the ratio of the sizes of the two nuclear parts is $1:{{2}^{1/3}}$ .
Hence the correct answer of the above question is option d.
Note:
The mass of a body can be written as the product of its density times the volume. The principle of conservation of momentum states that the momentum of a system is always to be conserved i.e. momentum of a body cannot be lost but can be transferred. This is a consequence of the law of conservation of energy.
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