Step 1: Conceptual Understanding:
This question evaluates comprehension of fundamental atomic nucleus properties: size (radius), volume, and density, and their relationship with mass number A.
Step 2: Detailed Analysis:
(A) Nuclear Radius Formula:
A nucleus with mass number A has a radius R described by \(R = R_0 A^{1/3}\). Here, \(R_0\) is a constant, approximately 1.2 fm. This empirical formula is derived from experimental data and is a cornerstone of nuclear physics. This statement is correct.
(B) Volume-Mass Number Proportionality:
Assuming a spherical nucleus, its volume \(V\) is \(V = \frac{4}{3}\pi R^3\). Substituting the radius expression from (A):\[ V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A \]Since \(\frac{4}{3}\pi R_0^3\) is constant, the volume \(V\) is directly proportional to the mass number A (\(V \propto A\)). This statement is correct.
(C) Density vs. Radius:
Nuclear density (\(\rho\)) is mass per unit volume (\(\rho = \text{Mass}/\text{Volume}\)).Nuclear mass is approximately \(A \times m_p\), where \(m_p\) is the nucleon mass.Nuclear volume is \(V = \left(\frac{4}{3}\pi R_0^3\right) A\).Therefore, density is:\[ \rho = \frac{A \cdot m_p}{\left(\frac{4}{3}\pi R_0^3\right) A} = \frac{m_p}{\frac{4}{3}\pi R_0^3} \]The mass number \(A\) cancels out. Nuclear density \(\rho\) is approximately constant for all nuclei, irrespective of their radius or mass number. Thus, the statement that density increases with radius is incorrect.
(D) Density Independence from Mass Number:
As demonstrated above, the nuclear density formula is independent of the mass number A. This significant characteristic implies that all atomic nuclei possess nearly uniform, extremely high densities. This statement is correct.
Step 3: Conclusion:
Statements (A), (B), and (D) are accurate. Statement (C) is inaccurate. The correct option is (C).