Question

From the relation $ R = {R_0}{A^{\dfrac{1}{3}}} $ , where $ {R_0} $ is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e., independent of A).

Answer 1

Hint :In order to solve this question, you must be aware about the concept of nuclear density. Nuclear density is the density of the nucleus of an atom, averaging about 2.3×1017 kg/m3. The descriptive term nuclear density is also applied to situations where similarly high densities occur, such as within neutron stars.

Complete Step By Step Answer:
 $ R = {R_0}{A^{\dfrac{1}{3}}} $ Radius of nucleus of mass number A is given by:
Where $ {R_0} $ is a constant
For a stable nucleus, the nuclear radius is approximately proportional to the cube root of the mass number (A) of the nucleus and the nucleons arranged in more spherical configurations.
Volume of nucleus = $ \dfrac{4}{3}\pi {R^3} $ , where R is the radius of the nucleus
= $ \dfrac{4}{3}\pi {({R_0}{A^{\dfrac{1}{3}}})^3} $
= $ \dfrac{4}{3}\pi {R_0}^3A $
\[Density{\text{ }}of{\text{ }}nuclear{\text{ }}matter = \dfrac{{Mass\;of\;nucleus}}{{Volume\;of\;nucleus}}\]
Let m be the average mass of the nucleus
Hence, mass of the nucleus = mA
Density of nuclear matter $ = \dfrac{A}{{\dfrac{4}{3}\pi {R^3}}} $
 $ = \dfrac{{mA}}{{\dfrac{4}{3}\pi {R_0}^3A}} $
 $ = \dfrac{{3m}}{{4\pi {R_0}^3}} $
This equation shows that the density of the nuclear matter is independent of A.
Therefore, the density of the nuclear matter is nearly constant.
Hence proved.

Note :
Two particles, both called nucleons, are found inside nuclei. The two types of nucleons are protons and neutrons; they are very similar, except that the proton is positively charged while the neutron is neutral. The protons, inside the nucleus, being positive, exert tremendous repulsive forces on one another. These forces are called the weak and strong nuclear forces.