Actually, the condition gives average mass for 1 proton and 1 neutron (in reality they have different masses).
(a) Calculate density of a nucleon representing average particle between a proton and a neutron considering it as a ball:
"\\rho=\\frac{M}{V}=\\frac{M}{\\frac{4}{3}\\pi\\cdot r^3}=\\frac{1.673\\cdot 10^{-27}}{\\frac{4}{3}\\pi\\cdot (10^{-15})^3}=3.994\\cdot 10^{17} \\text{ kg\/m}^3."Awesome.
A nucleus of iron has 56 nucleons that form a ball - a nucleus - with a radius of approximately
"4\\cdot 10^{-15} \\text{ m}."Thus density of the nucleus:
"\\rho_{\\text{Fe}}=\\frac{nM}{\\frac{4}{3}\\pi\\cdot r^3}=\\frac{56\\cdot 1.673\\cdot 10^{-27}}{\\frac{4}{3}\\pi\\cdot (4\\cdot 10^{-15})^3}=3.495\\cdot 10^{17} \\text{ kg\/m}^3."(b) And common density of iron is
"7874 \\text{ kg\/m}^3."So the nucleons are extremely dense. Nuclei are also very dense. Density of the material in common scale is very low.
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