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 a neutron is calculated by finding the volume of the neutron (treated as a sphere) using its radius, and then dividing the neutron's mass by this volume. The density of a neutron star, however, is much higher, reaching levels that are much greater than the density of individual nucleons or any regular atomic nucleus like that of hydrogen.

Step-by-step explanation:

To calculate the density of a neutron, one can use the formula for the volume of a sphere V = (4/3)πr³ and then use the formula for density ρ = m/V. Here, 'r' is the radius of the neutron and 'm' is the mass of the neutron. Considering a neutron to be a sphere with a radius of approximately 1.0×10⁻¹³ cm, we use the given volume formula for a sphere to find its volume. To find density, we would then divide the mass of a neutron by this volume.

Comparatively speaking, the density of a neutron star is astronomically high. A neutron star with a mass of 1.97 solar masses and a diameter of 13 km would have a tremendously greater density compared to a single neutron or any typical matter we normally encounter, such as the density of a hydrogen nucleus. It's important to notice that while neutrons and protons (collectively known as nucleons) are themselves very dense, the density of a neutron star surpasses even that of a nucleus by several orders of magnitude.

Answer 2

The density of the neutron,

`\rho = (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=(4)/(3) * \pi* r^3`

Here r is the radius of the neutron and its value of
`1.0*10^(-13) cm`

So,

`V=(4)/(3) * 3.14* (1.0* 10^(-13))^3 \\ V =4.186* 10^(-39) cm^(3)`

We know the mass of neutron,
`m= 1.674929 * 10^(-27) kg = 1.674929 *10^(-24) g`

Thus,

`\rho =(1.674929 *10^(-24) \ g)/(4.186* 10^(-39) cm^(3)) \\\\\ \rho=4.00*10^(14) \ g/cm^3`