Group the following six nuclides into three pairs of (i) isotones, (ii) isotopes and (iii) isobars:
${}_6{}^{12} \mathrm{C}, {}_2{}^3 \mathrm{He}, {}_{80}{}^{198} \mathrm{Hg}, {}_1{}^3 \mathrm{H}, {}_{79}{}^{197} \mathrm{Au}, {}_6{}^{14} \mathrm{C}$
How does the size of a nucleus depend on its mass number? Hence explain why the density of nuclear matter should be independent of the size of the nucleus.  

Classification of nuclides is as follows: 

Isotopes: ${}_{ 6}{}^{12}\mathrm{C}, {}_6{}^{14}\mathrm{C}$ ; has the same atomic number but, different mass number. 

Isobars $: {}_2{}^3\mathrm{He}, {}_1{}^3\mathrm{H}$ ; has the same mass number but, different atomic number.

Isotones: ${}_{80}{}^{198} \mathrm{Hg}, {}_{79}{}^{197} \mathrm{Au}$ ; has same neutron number but, different proton number. 

Now, 

Radius of nucleus, $\boxed{\mathrm{R} = {\mathrm{R}}_0 {\mathrm{A}}^{1/3}}$                          ...(i) 
where,  ${\mathrm{R}}_0$is the range of nuclear force (or Nuclear Unit Radius) 

Density is given by, $\mathrm{\rho } = \frac{\mathrm{M}}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}$                        ...(ii) 

But,                         $\mathrm{M}\propto \mathrm{A}$                               ...(iii) 

From (i) and (iii), we get

$\mathrm{R}\propto {\mathrm{M}}^{1/3} \Rightarrow {\mathrm{R}}^3 \propto \mathrm{M} \Rightarrow {\mathrm{R}}^3 = \mathrm{kM}$ where,

$\mathrm{k}\to \mathrm{proportionality} \mathrm{constant}$                                 ...(iv) 

From (ii) and (iv), we have
                                   $\mathrm{\rho } = \frac{\mathrm{M}}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}= \frac{3}{4\mathrm{\pi }}\frac{\mathrm{M}}{\mathrm{KM}} = \mathrm{constant}.$

Thus, we can see that nuclear density is clearly independent of mass number A.