Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

13:1 and 14:1

14:1 and 1:1

1:1 and 14:1

1:1 and 13:1

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.
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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