Question

(i) What characteristic property of nuclear force explains the constancy of binding energy per nucleon $\left(BE/A\right)$ in the range of mass number ${}^{\prime }{A}^{\prime }$ lying $30<A<170$?
(ii) Show that the density of nucleus over a wide range of nuclei is constant-independent of mass number $A$.

Answer 1

(i)

The approximate constancy of $\frac{BE}{A}$ over most of the range is saturation property of nuclear force.

In heavy nuclei, nuclear size > range of the nuclear force.

So a nuclear sense approximately a constant number of neighbors and thus the nuclear $\frac{BE}{A}$ levels off at high A. This saturation of the nuclear force.

(ii)

The mass of the nucleus of the atom of the mass number A

A= A a.m.u.

= A $\times 1.660565\times {10}^{-27}kg$

The volume of the nucleus is $\frac{4\pi r^3}{3}=\frac{4\pi }{3}(R_0{A}^{-3}{)}^3=\frac{4\pi R_0^{3}A}{3}$

$\therefore Volume=\frac{4\pi }{3}(1.1\times {10}^{-15}{)}^3\times A{m}^3$

Now the density of the nucleus= $\frac{m}{V}$; where m is mass and V is the volume of the nucleus.

$\rho =\frac{A\times 1.660565\times {10}^{-27}}{\frac{4\pi }{3}(1.1\times {10}^{-15}{)}^3\times A}=2.97\times {10}^{17}Kg{m}^{-3}$

Thus, we can see the density of the nuclei is independent of mass number and it is constant for all nuclei .