A nucleus at rest splits into two nuclear par | vedclass

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(A) According to the law of conservation of linear momentum,for a nucleus initially at rest,the total momentum after splitting must be zero.$m_1 v_1 =

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$8$

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(A) According to the law of conservation of linear momentum,for a nucleus initially at rest,the total momentum after splitting must be zero.$m_1 v_1 = m_2 v_2 \implies \frac{v_1}{v_2} = \frac{m_2}{m_1}$Since the density of nuclear matter is constant,the mass $m$ of a nucleus is proportional to its volume,which is proportional to the cube of its radius $R$.$m \propto R^3 \implies \frac{m_2}{m_1} = \left(\frac{R_2}{R_1}\right)^3$Given the ratio of radii $\frac{R_1}{R_2} = \frac{1}{2}$,we have $\frac{R_2}{R_1} = \frac{2}{1}$.Substituting this into the velocity ratio equation:$\frac{v_1}{v_2} = \left(\frac{2}{1}\right)^3 = \frac{8}{1}$.Thus,the ratio of their velocities is $8:1$.

$A$ nucleus at rest splits into two nuclear parts having radii in the ratio $1:2$. Their velocities are in the ratio: (in $:1$)

Similar Questions

$A$ nucleus has mass number $A_1$ and volume $V_1$. Another nucleus has mass number $A_2$ and volume $V_2$. If the relation between mass numbers is $A_2 = 4 A_1$,then $\frac{V_2}{V_1} = $ . . . . . . .

The graph of $\ln \left(\frac{R}{R_0}\right)$ versus $\ln A$ is,where $R$ is the radius of a nucleus,$A$ is its mass number,and $R_0$ is a constant.

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Consider the following two statements, $A$ and $B$ and identify the correct answer given below:
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The radius of the nucleus of ${ }_{13}^{27} Al$ is $\qquad$ . (Given: $R_{0}=1.2 \text{ fm}$)

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