Step 1: Recall the nuclear radius formula.
The empirical formula for the radius of a nucleus with mass number $A$ is: \[ R = R_0 A^{1/3} \] where $R_0 \approx 1.2 \times 10^{-15}\,\text{m} = 1.2\,\text{fm}$ is the nuclear radius constant.
Step 2: Analyze option (1) -- nuclear density.
Volume of nucleus: $V = \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi R_0^3 A$. Since $A$ nucleons each have mass approximately $m_p$: $M \approx A m_p$. So density $\rho = \dfrac{M}{V} = \dfrac{A m_p}{\frac{4}{3}\pi R_0^3 A} = \dfrac{3m_p}{4\pi R_0^3}$. This is constant -- independent of $A$. Option (1) is CORRECT.
Step 3: Analyze option (2) -- radius proportional to A.
From $R = R_0 A^{1/3}$, radius is proportional to $A^{1/3}$, NOT to $A$. So option (2) is INCORRECT.
Step 4: Analyze option (3) -- binding energy and mass defect.
Binding energy = (mass defect) $\times c^2$, so binding energy is directly proportional to mass defect, not inversely. Option (3) is INCORRECT.
Step 5: Analyze option (4) -- fission of heavy nuclei.
When heavy nuclei split into lighter nuclei (nuclear fission), energy is released (not absorbed) because the products have higher binding energy per nucleon. Option (4) is INCORRECT.
Step 6: State the correct answer.
The nuclear density is approximately the same for all nuclei regardless of their mass number $A$. \[ \boxed{\text{Option (1): Nuclear density is independent of mass number } A} \]