Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1152: 9

a) 3.7 fm b) 240 c) 240

(a) We know that $r=(1.2\times 10^{-15})A^{\frac{1}{3}}$ $\implies r=(1.2fm)(30)^{\frac{1}{3}}$ $r=3.7fm$ (b) We can find the required mass number as follows: $(2r)^3=8r^3$ Now $A=8(30)=240$ (c) As $r=(1.2fm)A^{\frac{1}{3}}$ $r=2r_p$ $\implies r=2(3.7fm)=7.4fm$ Now $A^{\frac{1}{3}}=\frac{7.4fm}{1.2fm}=6.2$ $A=(6.2)^3=240$