The ratio of nuclear densities and nuclear volumes of ${}_{26}^{56}\mathrm{Fe}$ and ${}_{2}^{4}\mathrm{He}$ ar

Question

The ratio of nuclear densities and nuclear volumes of ${}_{26}^{56}\mathrm{Fe}$ and ${}_{2}^{4}\mathrm{He}$ are, respectively:

Accepted Answer

1. **Nuclear Density:** The nuclear radius $R$ is related to the mass number $A$ by $R = R_0 A^{1/3}$ . The nuclear volume $V$ is $\frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 A$. Since the nuclear mass $M$ is proportional to $A$ ($M \approx m_n A$), the nuclear density $\rho = \frac{M}{V}$ is independent of $A$ (constant for all nuclei). Therefore, the ratio of densities is **1:1**.
2. **Nuclear Volume:** As derived above, nuclear volume is directly proportional to the mass number ($V \propto A$).
   - For ${}_{26}^{56}\mathrm{Fe}$, $A = 56$.
   - For ${}_{2}^{4}\mathrm{He}$, $A = 4$.
   - Ratio of volumes $\frac{V_{\mathrm{Fe}}}{V_{\mathrm{He}}} = \frac{56}{4} = \frac{14}{1}$.

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