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 _____.

The nuclear density is calculated using the mass of the nucleus and its volume. The mass of a nucleon is given as $$ m = 1.6 \times 10^{-27} $$ kg, and since protons and neutrons have equal masses, the mass of a nucleus with mass number $$ A $$ is $$ M = A \times m = A \times 1.6 \times 10^{-27} $$ kg.

The radius of the nucleus is given by $$ R = 1.5 \times 10^{-15} \times A^{1/3} $$ m. The volume of the nucleus is:

$$ V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \left(1.5 \times 10^{-15} \times A^{1/3}\right)^3 $$

First, compute $$ R^3 $$:

$$ R^3 = \left(1.5 \times 10^{-15}\right)^3 \times \left(A^{1/3}\right)^3 = (1.5)^3 \times 10^{-45} \times A = 3.375 \times 10^{-45} \times A \text{ m}^3 $$

Now substitute into the volume formula:

$$ V = \frac{4}{3} \pi \times 3.375 \times 10^{-45} \times A = \frac{4}{3} \times 3.375 \times \pi \times 10^{-45} \times A $$

Compute $$ \frac{4}{3} \times 3.375 $$:

$$ \frac{4}{3} \times 3.375 = \frac{4}{3} \times \frac{27}{8} = \frac{4 \times 27}{3 \times 8} = \frac{108}{24} = 4.5 $$

So,

$$ V = 4.5 \times \pi \times 10^{-45} \times A \text{ m}^3 $$

The nuclear density $$ \rho_{\text{nucleus}} $$ is:

$$ \rho_{\text{nucleus}} = \frac{M}{V} = \frac{A \times 1.6 \times 10^{-27}}{4.5 \times \pi \times 10^{-45} \times A} = \frac{1.6 \times 10^{-27}}{4.5 \times \pi \times 10^{-45}} $$

Simplify the expression:

$$ \rho_{\text{nucleus}} = \frac{1.6}{4.5 \pi} \times 10^{-27 - (-45)} = \frac{1.6}{4.5 \pi} \times 10^{18} \text{ kg/m}^3 $$

Reduce $$ \frac{1.6}{4.5} = \frac{16}{45} $$, so:

$$ \rho_{\text{nucleus}} = \frac{16}{45 \pi} \times 10^{18} \text{ kg/m}^3 $$

The density of water is $$ \rho_{\text{water}} = 1000 \text{ kg/m}^3 = 10^3 \text{ kg/m}^3 $$. The ratio of nuclear density to water density is:

$$ \text{Ratio} = \frac{\rho_{\text{nucleus}}}{\rho_{\text{water}}} = \frac{\frac{16}{45 \pi} \times 10^{18}}{10^3} = \frac{16}{45 \pi} \times 10^{15} $$

The problem states that this ratio is approximately $$ n \times 10^{13} $$. Therefore:

$$ \frac{16}{45 \pi} \times 10^{15} = n \times 10^{13} $$

Solve for $$ n $$:

$$ n = \frac{16}{45 \pi} \times 10^{15} \times \frac{1}{10^{13}} = \frac{16}{45 \pi} \times 10^{2} = \frac{1600}{45 \pi} $$

Simplify $$ \frac{1600}{45} $$ by dividing numerator and denominator by 5:

$$ \frac{1600 \div 5}{45 \div 5} = \frac{320}{9} $$

So,

$$ n = \frac{320}{9 \pi} $$

Now compute the numerical value. Using $$ \pi \approx 3.1416 $$:

$$ 9 \pi = 9 \times 3.1416 = 28.2744 $$

$$ n = \frac{320}{28.2744} \approx 11.32 $$

Since the ratio is approximate and the problem specifies that it is $$ n \times 10^{13} $$ with $$ n $$ to be found, and given that the calculated value is approximately 11.32, which rounds to 11 as the nearest integer for the approximation, we have $$ n = 11 $$.

Thus, the value of $$ n $$ is 11.