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The nucleus of an atom can be modeled as several protons and neutrons closely packed together. Each particle has a mass of 1.67 10-27 kg and radius on the order of 10-15 m. (a) Use this model and the data provided to estimate the density of the nucleus of an atom. 3.9868*10^17kg/m^3 Incorrect: Your answer is incorrect. Check the syntax of your response. (b) Compare your result with the density of a material such as iron (rho = 7874 kg/m3). What do your result and comparison suggest about the structure of matter?

Answer :

Explanation:

The nucleus of an atom can be modeled as several protons and neutrons closely packed together.

Mass of the particle, [tex]m=1.67\times 10^{-27}\ kg[/tex]

Radius of the particle, [tex]R=10^{-15}\ m[/tex]

(a) The density of the nucleus of an atom is given by mass per unit area of the particle. Mathematically, it is given by :

[tex]d=\dfrac{m}{V}[/tex], V is the volume of the particle

[tex]d=\dfrac{m}{(4/3)\pi r^3}[/tex]

[tex]d=\dfrac{1.67\times 10^{-27}}{(4/3)\pi (10^{-15})^3}[/tex]

[tex]d=3.98\times 10^{17}\ kg/m^3[/tex]

So, the density of the nucleus of an atom is [tex]3.98\times 10^{17}\ kg/m^3[/tex].

(b) Density of iron, [tex]d'=7874\ kg/m^3[/tex]

Taking ratio of the density of nucleus of an atom and the density of iron as :

[tex]\dfrac{d}{d'}=\dfrac{3.98\times 10^{17}}{7874}[/tex]

[tex]\dfrac{d}{d'}=5.05\times 10^{13}[/tex]

[tex]d=5.05\times 10^{13}\ d'[/tex]

So, the density of the nucleus of an atom is [tex]5.05\times 10^{13}[/tex] times greater than the density of iron. Hence, this is the required solution.

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