# (a) Classify the following six nuclides into (i) isotones, (ii) isotopes, and (iii) isobars :        $_{6}^{12}\textrm{C},\; _{2}^{3}\textrm{He},\; _{80}^{198}\textrm{Hg},\; _{1}^{3}\textrm{H},\; _{79}^{197}\textrm{Au},\; _{6}^{14}\textrm{C}$(b) 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.

(a) Classification of nuclides are as follows;-

(i) $isotopes\; ;\; _{6}^{12}\textrm{C}\; and\; _{6}^{14}\textrm{C}$

(ii) $isobars\; ;\; _{2}^{3}\textrm{He}\; and\; _{1}^{3}\textrm{H}$

(iii) $isotones\; ;\; _{80}^{198}\textrm{Hg}\; and\; _{79}^{197}\textrm{Au}$

(c) The radius of the nucleus having mass number A is given as ;

$R=r_{0}(A)^{\frac{1}{3}}$

Where, $A=$ mass no.

$R=$ radius

$r_{0}=$ constant value

Such that, the volume of the nucleus;-

$=\frac{4}{3}\pi R^{3}$

$=\frac{4}{3}\pi \times r_{0}^{3}\times A^{\frac{1}{3}\times 3}$

$=\frac{4}{3}\pi r_{0}^{3} A$

If m be the average mass of a nucleon, then the mass of nucleus =mA

$\therefore$ the nuclear density

$=\frac{mass }{volume }=\frac{mA}{\frac{4}{3}\pi (r_{0}^{3})\times A}=\frac{3m}{4\pi (r_{0})^{3}}$

Hence, nuclear density is independent of the size of the nucleus.

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