Answer
Ratio(largest/smallest) of the surface areas of sphere is $35.2179$
Work Step by Step
Given largest stable nucleus has Nucleon number $A_{l}=209$
suppose largest stable nucleus has radius $r_{l}$
so from equation number 31.2
$r=(1.2\times10^{-15}m)A^{\frac{1}{3}}$
$r_{l}=(1.2\times10^{-15}m)A_{i}^{\frac{1}{3}}$
assuming nucleus to be spherical surface area of nucleus will be
$(SA)_{l}=4\pi r_{i}^2$
$(SA)_{l}=4\pi [(1.2\times10^{-15}m)A_{i}^{\frac{1}{3}}]^2$
$(SA)_{l}=4\pi (1.2\times10^{-15}m)^2A_{i}^{\frac{2}{3}}$ .....equation(1)
Similarly for smallest stable nucleon
Given smallest stable nucleus has Nucleon number $A_{s}=1$
suppose smallest stable nucleus has radius $r_{s}$
so from equation number 31.2
$r=(1.2\times10^{-15}m)A^{\frac{1}{3}}$
$r_{s}=(1.2\times10^{-15}m)A_{s}^{\frac{1}{3}}$
assuming nucleus to be spherical surface area of nucleus will be
$(SA)_{s}=4\pi r_{s}^2$
$(SA)_{s}=4\pi [(1.2\times10^{-15}m)A_{s}^{\frac{1}{3}}]^2$
$(SA)_{s}=4\pi (1.2\times10^{-15}m)^2A_{s}^{\frac{2}{3}}$ ........equation (2)
on dividing equation(1) by equation(2) we will get
$\frac{(SA)_{l}}{(SA)_{s}}=\frac{4\pi (1.2\times10^{-15}m)^2A_{i}^{\frac{2}{3}}}{4\pi (1.2\times10^{-15}m)^2A_{s}^{\frac{2}{3}}}$
$\frac{(SA)_{l}}{(SA)_{s}}=(\frac{A_{l}}{A_{s}})^{\frac{2}{3}}$
in this equation putting
$A_{l}=209$
$A_{s}=1$
will give $\frac{(SA)_{l}}{(SA)_{s}}=(\frac{209}{1})^{\frac{2}{3}}$
$\frac{(SA)_{l}}{(SA)_{s}}=(209)^{\frac{2}{3}}=35.2179$
Ratio(largest/smallest) of the surface areas of sphere is $35.2179$
