Question #84573

5. The nucleus of an atom can be modeled as several protons and neutrons closely packed together.
Each particle has a mass of 1.67 3 10227 kg and radius
on the order of 10215 m. (a) Use this model and the
data provided to estimate the density of the nucleus of
an atom. (b) Compare your result with the density of
a material such as iron. What do your result and comparison suggest about the structure of matter?

Expert's answer

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:

ρ=MV=M43π⋅r3=1.673⋅10−2743π⋅(10−15)3=3.994⋅1017 kg/m3.\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⋅10−15 m.4\cdot 10^{-15} \text{ m}.

Thus density of the nucleus:

ρFe=nM43π⋅r3=56⋅1.673⋅10−2743π⋅(4⋅10−15)3=3.495⋅1017 kg/m3.\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 kg/m3.7874 \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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