question_answer

Group the following six nuclides into three pairs of (i) isotones, (ii) isotopes and (iii) isobars; \[_{6}^{12}C,_{2}^{3}He,_{80}^{198}Hg,_{1}^{3}H,_{79}^{197}Au,_{6}^{14}C\] 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.                

Answer:

                (i) Isotones: \[_{80}^{198}Hg\] and \[_{79}^{197}Au\]    [Same (A-Z)] (ii) Isotopes: \[_{6}^{12}C\] and \[_{6}^{14}C\]   [Same Z but different A] (iii) Isobars: \[_{2}^{3}\] \[He\] and \[_{1}^{3}H\].    [Same A but different Z] The effective radius of a nucleus is related to its mass number A as \[r={{r}_{0}}{{A}^{1/3}},{{r}_{0}}\] is a constant. Volume of the nucleus \[=\frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{({{r}_{0}}{{A}^{1/3}})}^{3}}=\frac{4}{3}\pi r_{0}^{3}A\] If \[m\] is the average mass of a nucleon, then mass of the nucleus \[=mA\] Nuclear density \[\text{=}\frac{\text{Mass}}{\text{Volume}}=\frac{mA}{\frac{4}{3}\pi r_{0}^{3}A}=\frac{3m}{4\pi r_{0}^{3}}\] Clearly, nuclear density is independent of the size of the nucleus.