Question 30.1: Nuclear density Let’s begin by calculating the density of a ...

SET UP The mass number A and the radius R are related by Equation 30.2. The proton and neutron masses are approximately equal, about 1.67 \times 10^{-27}  kgeach. The total mass M of the nucleus is A times the mass m of a single proton or neutron.

SOLVE From Equation 30.2, the radius R of the nucleus is given by

R = R_0  A^{1/3} = (1.2 \times 10^{-15}  m)(56^{1/3}) = 4.6 \times 10^{-15}  m.

The total mass M (ignoring the small difference in mass between protons and neutrons) is

M = (56)(1.67 \times 10^{-27}  kg) = 9.4 \times 10^{-26}  kg

The volume V is

\mathrm{V = \frac{4}{3}\pi R^3=\frac{4}{3}\pi (4.6 \times 10^{-15}  m)^3 =4.1 \times 10^{-43}  m^3,}

and the density ρ is

\mathrm{\rho = \frac{M}{V} = \frac{9.4 \times 10^{-26}kg}{4.1 \times 10^{-43}  m^3} =2.3 \times 10^{17}  kg/m^3.}

REFLECT The radius of the iron nucleus is about 4.6 \times 10^{-15}  m, on the order of 10^{-5} of the overall radius of an atom of iron. The density of the element iron (in the solid state) is about 8000 kg/m³ (8 g/cm³), so we see that the nucleus is on the order of 10^{13} times as dense as the bulk material. Densities of this magnitude are also found in white dwarf stars, which are similar to gigantic nuclei. A 1 cm cube of material with this density would have a mass of 2.3 \times 10^{11}  kg, or 230 million metric tons!

Practice Problem: Find the approximate mass, radius, and density of the uranium nucleus with A = 238, and compare the results with the values for iron. Assume that the total mass is 238 times the mass of a single neutron, ignoring binding-energy corrections. Answers: 4.0 \times 10^{-25}  kg,  7.4  \times  10^{-15}  m,  2.3 \times  10^{17}  kg/m^3.