Given the mass of the iron nucleus as 55.85u and A = 56. Find the nuclear density.
                                                          OR
Draw a plot of potential energy between a pair of nucleons as a function of their separation.
Mark the Regions, where potential energy is (i) positive and (ii) negative.

$$\begin{gathered} \mathrm{Given}, \mathrm{mass}, \mathrm{m}=55.85 \mathrm{u} \\ = 55.85\times 1.67\times {10}^{-27}\mathrm{kg} \\ \mathrm{Volume}, \mathrm{V}=\frac{4}{3}{\mathrm{\pi R}}^3=\frac{4}{3}\mathrm{\pi }{\left({\mathrm{R}}_0{\mathrm{A}}^{1/3}\right)}^3=\frac{4}{3}{\mathrm{\pi R}}_0^3\times \mathrm{A} \\ \therefore \mathrm{Nuclear} \mathrm{density}, \mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{V}} \\ = \frac{3\times 55.85\times 1.67\times {10}^{-27}}{4\times {\displaystyle \frac{22}{7}}\times {(1.2\times {10}^{-15})}^3\times 56} \\ =2.29\times {10}^{17}\mathrm{kg}/{\mathrm{m}}^3 \end{gathered}$$

OR

he graph between the potential energy of a pair of nucleons as a function of their separation is given below

i. For distance less than 0.8 fm, negative PE decreases to zero and then becomes positive. 

ii. For distance greater than 0.8 fm, negative PE decreases to zero and then becomes positive.