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 $A_2 = 4A_1$, then $\frac{V_2}{V_1} =$ _______.

Question

The average energy released per fission for the nucleus of $^{235_{92}U$ is 190 MeV. When all the atoms of 47 g pure $^{235}_{92}U$ undergo fission process, the energy released is $\alpha \times 10^{23}$ MeV. The value of $\alpha$ is ___. (Avogadro Number $= 6 \times 10^{23}$ per mole)}

Answer 1

Nucleus Volume Calculation:

Volume Formula: \[ V = \frac{4}{3} \pi R^3 \]

Radius Dependence on Mass Number (A):

\[ R = R_0 \left( A \right)^{1/3} \]

Volume Equation with Radius Substitution:

\[ V = \frac{4}{3} \pi \left( R_0 (A)^{1/3} \right)^3 \]

Volume Ratio Simplification:

\[ \Rightarrow \frac{V_2}{V_1} = \frac{A_2}{A_1} = 4 \]

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 \(A_2 = 4A_1\), then \(\frac{V_2}{V_1} =\) _______.

Correct Answer: 4

Solution and Explanation

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