Question

given the following data, calculate the densities of a carbon - 14 nucleus and a carbon - 14 atom.
particle\tmass
electron\t9.11×10^(-31) kg
proton\t1.67×10^(-27) kg
neutron\t1.67×10^(-27) kg
the diameter of the nucleus is about 10^(-13) cm and electrons move about the nucleus at an average distance of about 10^(-8) cm away from it.
(a) calculate the density of a carbon - 14 nucleus.
density =
(b) calculate the density of a carbon - 14 atom.
density =

Explanation:

Step1: Calculate the mass of carbon - 14 nucleus

Carbon - 14 has 6 protons and 8 neutrons.
$m_{nucleus}=6\times1.67\times 10^{-27}\text{ kg}+8\times1.67\times 10^{-27}\text{ kg}=(6 + 8)\times1.67\times 10^{-27}\text{ kg}=14\times1.67\times 10^{-27}\text{ kg}=2.338\times 10^{-26}\text{ kg}$

Step2: Calculate the volume of carbon - 14 nucleus

Assume the nucleus is spherical, $V_{nucleus}=\frac{4}{3}\pi r^{3}$, with diameter $d = 10^{-13}\text{ cm}=10^{-15}\text{ m}$, so $r=\frac{d}{2}=5\times 10^{-16}\text{ m}$
$V_{nucleus}=\frac{4}{3}\pi(5\times 10^{-16})^{3}\text{ m}^{3}=\frac{4}{3}\pi\times125\times 10^{-48}\text{ m}^{3}\approx5.24\times 10^{-46}\text{ m}^{3}$

Step3: Calculate the density of carbon - 14 nucleus

$
ho_{nucleus}=\frac{m_{nucleus}}{V_{nucleus}}=\frac{2.338\times 10^{-26}\text{ kg}}{5.24\times 10^{-46}\text{ m}^{3}}\approx4.46\times 10^{19}\text{ kg/m}^{3}$
To convert to g/cm³, $1\text{ kg/m}^{3}=10^{-3}\text{ g/cm}^{3}$
$
ho_{nucleus}=4.46\times 10^{16}\text{ g/cm}^{3}$

Step4: Calculate the mass of carbon - 14 atom

The mass of electrons is negligible compared to the mass of the nucleus, so $m_{atom}\approx m_{nucleus}=2.338\times 10^{-26}\text{ kg}$

Step5: Calculate the volume of carbon - 14 atom

Assume the atom is spherical with radius $r_{atom}=10^{-8}\text{ cm}=10^{-10}\text{ m}$
$V_{atom}=\frac{4}{3}\pi r_{atom}^{3}=\frac{4}{3}\pi(10^{-10})^{3}\text{ m}^{3}=\frac{4}{3}\pi\times 10^{-30}\text{ m}^{3}\approx4.19\times 10^{-30}\text{ m}^{3}$

Step6: Calculate the density of carbon - 14 atom

$
ho_{atom}=\frac{m_{atom}}{V_{atom}}=\frac{2.338\times 10^{-26}\text{ kg}}{4.19\times 10^{-30}\text{ m}^{3}}\approx5.58\times 10^{3}\text{ kg/m}^{3}=5.58\text{ g/cm}^{3}$

Answer:

(a) Density of carbon - 14 nucleus = $4.46\times 10^{16}$ g/cm³
(b) Density of carbon - 14 atom = $5.58$ g/cm³