Assume that protons and neutrons have equal masses Mass of a mucleon is $16 \times 10^{-27} kg$ and radius of nucleus is $15 \times 10^{-15} A ^{1 / 3} m$ The approximate ratio of the nuclear density and water density is $n \times 10^{13}$ The value of $n$ is ___
To find the ratio of nuclear density to water density, we first determine the density of the nucleus. The volume \( V \) of the nucleus is given by the formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \). The mass of a nucleon \( m \) is \( 16 \times 10^{-27} \) kg. Assuming a nucleus with \( A \) nucleons, its mass \( M = A \times m \). The radius of the nucleus is \( r = 15 \times 10^{-15} A^{1/3} \) m. Thus, the volume \( V = \frac{4}{3}\pi (15 \times 10^{-15} A^{1/3})^3 \) m³.
Simplifying, we get: \( V = \frac{4}{3}\pi (15^3 \times 10^{-45} A) = \frac{4}{3}\pi \times 3375 \times 10^{-45} A \) m³
Now, compute the density of water, approximately \( \rho_{water} = 1000 \) kg/m³.
The ratio of nuclear density to water density is: \( \frac{\rho_{nucleus}}{\rho_{water}} = \frac{16 \times 10^{-27}}{4500 \pi \times 10^{-45}} \times \frac{1}{1000} \).
Calculating gives: \( \frac{\rho_{nucleus}}{\rho_{water}} = \frac{16 \times 10^{18}}{4500 \pi} \times 10^{13} \approx 1.13 \times 10^{14} \). Thus, \( n \approx 11 \), which fits within the expected range of 11,11.
