Nuclear density Let’s begin by calculating the density of a nucleus. The most common variety of iron nucleus has 26 protons and 30 neutrons, so A = 56. Find the radius, approximate mass, and approximate density of this nucleus.

Question

Nuclear density

Let’s begin by calculating the density of a nucleus. The most common variety of iron nucleus has 26 protons and 30 neutrons, so A = 56. Find the radius, approximate mass, and approximate density of this nucleus.

Accepted Answer

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 [latex]1.67 imes 10^{-27} kg[/latex]each. 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

[latex]R = R_0 A^{1/3} = (1.2 imes 10^{-15} m)(56^{1/3}) = 4.6 imes 10^{-15} m.[/latex]

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

[latex]M = (56)(1.67 imes 10^{-27} kg) = 9.4 imes 10^{-26} kg[/latex]

The volume V is

[latex]\mathrm{V = \frac{4}{3}\pi R^3=\frac{4}{3}\pi (4.6 imes 10^{-15} m)^3 =4.1 imes 10^{-43} m^3,} [/latex]

and the density ρ is

[latex]\mathrm{ ho = \frac{M}{V} = \frac{9.4 imes 10^{-26}kg}{4.1 imes 10^{-43} m^3} =2.3 imes 10^{17} kg/m^3.} [/latex]

REFLECT The radius of the iron nucleus is about [latex]4.6 imes 10^{-15} m,[/latex] on the order of [latex]10^{-5}[/latex] 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 [latex]10^{13}[/latex] 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 [latex]2.3 imes 10^{11} kg,[/latex] 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: [latex]4.0 imes 10^{-25} kg, 7.4 imes 10^{-15} m, 2.3 imes 10^{17} kg/m^3.[/latex]

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

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