Question

Assume that the mass of a nucleus is approximately given by M = Amp where A is the mass number. Estimate the density of matter in kgm−3 inside a nucleus. What is the specific gravity of nuclear matter?

Answer 1

The nuclear matter’s specific gravity is = $3 \times 10^{14} \mathrm{kg} / \mathrm{m}^{3}$

Explanation:

It is given that:

Nucleus mass, M = Amp (A is the mass number) Nucleus volume, V = $43 \pi \mathrm{RO} 3 \mathrm{A}$

Matter’s density inside the nucleus, d = $A m p 43 \pi R 03 A=M V$

= $=3 \mathrm{mp} 4 \times \pi \mathrm{RO} 3=3 \times 1017 \mathrm{kg} / \mathrm{m} 3$

The nuclear matter’s specific gravity = Matter density and water’s density  

Therefore, specific gravity = $3 \times 1017103=3 \times 10^{14} \mathrm{kg} / \mathrm{m}^{3}$