Chapter 31 - Nuclear Physics and Radioactivity - Problems - Page 899: 8

$ ^{120}_{50}Sn$, so unknown nucleus is nucleus of tin.

$ ^{60}_{28}Ni$ Comparing with $ ^{A}_{Z}X$ we get $A=60$ $Z=28$ number of neutrons $N=A-Z$ $N=60-28=2$ no of neutron in nickel is $N=2$ Suppose radius of nickel nucleus is $r_{Ni}$ so from equation number 31.2 $r=(1.2\times10^{-15}m)A^{\frac{1}{3}}$ we can write for nickel $r_{Ni}=(1.2\times10^{-15}m)A_{Ni}^{\frac{1}{3}}$ $r_{Ni}=(1.2\times10^{-15}m)\times(60)^{\frac{1}{3}}$ considering nucleus to be spherical volume of nickel nucleus $V_{Ni}=\frac{4}{3} \pi r_{Ni}^3$ $V_{Ni}=\frac{4}{3} \pi [(1.2\times10^{-15}m)\times(60)^{\frac{1}{3}}]^3$ $V_{Ni}=\frac{4}{3} \pi (1.2\times10^{-15}m)^3\times(60)$...........equation(1) now suppose unknown nucleus $ ^{A}_{Z}X$ has radius $r$ since it has number of nucleon as $A$ $r=(1.2\times10^{-15}m)A^{\frac{1}{3}}$ and volume of unknow nucleus is $V_{U}=\frac{4}{3} \pi r^3$ $V_{U}=\frac{4}{3} \pi [(1.2\times10^{-15}m)A^{\frac{1}{3}}]^3$ $V_{U}=\frac{4}{3} \pi (1.2\times10^{-15}m)^3A$.......equation(2) given that unknown nucleus has twice the volume of nickel nuclei. so $V_{U}=2\times V_{Ni}$ from equation(1) and (2) putting the value we will get $\frac{4}{3} \pi (1.2\times10^{-15}m)^3A=2\times \frac{4}{3} \pi (1.2\times10^{-15}m)^3\times(60)$ so $A=120$ so unknown nucleus will have $A=120$ nucleons given that number of neutron $N=70$ so $Z=A-N=120-70=50$ $ ^{120}_{50}X$=$ ^{120}_{50}Sn$