Assume that protons and neutrons have equal masses. Mass of a nucleon is $$1.6 \times 10^{-27}$$ kg and radius of nucleus is $$1.5 \times 10^{-15} A^{1/3}$$ m. The approximate ratio of the nuclear density and water density is $$n \times 10^{13}$$. The value of $$n$$ is _____.

Find the ratio of nuclear density to water density. To calculate the nuclear density, consider the mass of a nucleon $$m = 1.6 \times 10^{-27}$$ kg and the nuclear radius $$r = 1.5 \times 10^{-15} A^{1/3}$$ m. The volume of the nucleus is $$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (1.5 \times 10^{-15})^3 A$$ m^3 while the mass of the nucleus is $$M = Am$$. Since the nuclear density is independent of A, we have $$ \rho_n = \frac{Am}{\frac{4}{3}\pi(1.5)^3 \times 10^{-45} \times A} = \frac{m}{\frac{4}{3}\pi(3.375) \times 10^{-45}} = \frac{1.6 \times 10^{-27}}{14.14 \times 10^{-45}} $$ which yields $$ = 1.13 \times 10^{17}\;\text{kg/m}^3 $$.

To find the ratio of nuclear density to water density, we compute $$ \frac{\rho_n}{\rho_w} = \frac{1.13 \times 10^{17}}{10^3} \approx 11 \times 10^{13} $$. Therefore $$n = 11$$ and the answer is 11.