Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The nuclear density of nuclides $$^{10}_5$$B, $$^{6}_3$$Li, $$^{56}_{26}$$Fe, $$^{20}_{10}$$Ne and $$^{209}_{83}$$Bi can be arranged as
$$\rho^N_{Bi} > \rho^N_{Fe} > \rho^N_{Ne} > \rho^N_{B} > \rho^N_{Li}$$
Reason R: The radius $$R$$ of nucleus is related to its mass number $$A$$ as $$R = R_0 A^{\frac{1}{3}}$$, where $$R_0$$ is a constant.
In the light of the above statement, choose the correct answer from the options given below:

Nuclear density is given by $$\rho = \frac{\text{mass}}{\text{volume}} = \frac{A \cdot m_u}{\frac{4}{3}\pi R^3}$$, where $$R = R_0 A^{1/3}$$. Substituting, $$\rho = \frac{A \cdot m_u}{\frac{4}{3}\pi R_0^3 A} = \frac{m_u}{\frac{4}{3}\pi R_0^3}$$, which is independent of the mass number $$A$$. This means all nuclei have approximately the same nuclear density, regardless of their size.

Therefore, the ordering $$\rho^N_{Bi} > \rho^N_{Fe} > \rho^N_{Ne} > \rho^N_{B} > \rho^N_{Li}$$ given in Assertion A is false — all these nuclear densities are essentially equal. Reason R, which states $$R = R_0 A^{1/3}$$, is a well-established empirical relation and is true. In fact, R being true is precisely what leads to the conclusion that nuclear density is constant, making A false.

The correct answer is Option B: A is false but R is true.