Question:

The radius of nucleus is expressed as $R = R_0 A^{1/3}$, where $A$ is mass number and $R_0 = 1.2 \times 10^{-15} \, \text{m}$. Prove that the density of nucleus does not depend upon the mass number $A$.

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Nuclear density is constant ($\sim 2.3 \times 10^{17} \, \text{kg/m}^3$) for all nuclei, showing tightly packed nucleons.
Updated On: Oct 8, 2025
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Solution and Explanation

Step 1: Write expression for volume of nucleus.
\[ V = \frac{4}{3} \pi R^3 \] \[ R = R_0 A^{1/3} \quad \Rightarrow \quad V = \frac{4}{3} \pi (R_0 A^{1/3})^3. \] \[ V = \frac{4}{3} \pi R_0^3 A. \]
Step 2: Write expression for mass of nucleus.
Mass of nucleus $\approx A m_p$ (where $m_p$ is mass of one nucleon, nearly proton/neutron mass).
Step 3: Density of nucleus.
\[ \rho = \frac{\text{Mass}}{\text{Volume}} = \frac{A m_p}{\frac{4}{3} \pi R_0^3 A}. \] \[ \rho = \frac{m_p}{\frac{4}{3} \pi R_0^3}. \]
Step 4: Conclusion.
Since $A$ cancels out, $\rho$ is independent of mass number $A$.
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