Question

Assume the radius of a neutron to be approximately 1.0×10−13 cm, and calculate the density of a neutron. [hint: for a sphere v=(4/3)πr3.]

Answer 1

The density of the neutron,

$\rho = \frac{m}{V}$

Here, m is the mass and V is the volume.

Now to calculate the volume of a neutron, we use the formula

$V=\frac{4}{3} \times \pi\times r^3$

Here r is the radius of the neutron and its value of $1.0\times10^{-13} cm$

So,

$V=\frac{4}{3} \times 3.14\times (1.0\times 10^{-13})^3 \\ V =4.186\times 10^{-39} cm^{3}$

We know the mass of neutron, $m= 1.674929 \times 10^{-27} kg = 1.674929 \times10^{-24} g$

Thus,  

$\rho =\frac{1.674929 \times10^{-24} \ g}{4.186\times 10^{-39} cm^{3}} \\\\\ \rho=4.00\times10^{14} \ g/cm^3$