Question

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 1

Explanation:

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

Mass of the particle, $m=1.67\times 10^{-27}\ kg$

Radius of the particle, $R=10^{-15}\ m$

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

$d=\dfrac{m}{V}$, V is the volume of the particle

$d=\dfrac{m}{(4/3)\pi r^3}$

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

$d=3.98\times 10^{17}\ kg/m^3$

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

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

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

$\dfrac{d}{d'}=\dfrac{3.98\times 10^{17}}{7874}$

$\dfrac{d}{d'}=5.05\times 10^{13}$

$d=5.05\times 10^{13}\ d'$

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