Answer
(a) $1.2\times 10^{-15}m \space r\le 7.1\times 10^{-15}m$
(b) $35.2$
(c) $209$
Work Step by Step
(a) We can find the required range in nuclear radii as follows:
We know that
$r=(1.2\times 10^{-15}m)A^{\frac{1}{3}}$
The minimum radius is $r_{min}=(1.2\times 10^{-15m})(1)^{\frac{1}{3}}=1.2\times 10^{-15}m$
and the maximum radius is $r_{max}=(1.2\times 10^{-15}m)(209)^{\frac{1}{3}}=7.1\times 10^{-15}m$
Thus the radii of all the stable nuclei range is $1.2\times 10^{-15}m\le 7.1\times 10^{-15}m$.
(b) We can find the required ratio as follows:
$\frac{S_{max}}{S_{min}}=\frac{4\pi((1.2\times 10^{-15}m)(209)^{\frac{1}{3}})^2}{4\pi((1.2\times 10^{-15}m)(1)^{\frac{1}{3}})^2}=35.2$
(c) The ratio of volumes can be determined as
$\frac{V_{max}}{V_{min}}=\frac{{\frac{4}{3}\pi r_{max}^3}}{{\frac{4}{3}\pi r_{min}^3}}$
$\frac{V_{max}}{V_{min}}=\frac{r_{max}^3}{r_{min}^3}$
We plug in the known values to obtain:
$\frac{V_{max}}{V_{min}}=\frac{[(1.2\times 10^{-15}m)(209)^{\frac{1}{3}}]^3}{[(1.2\times 10^{-15}m)(1)^{\frac{1}{3}}]^3}=209$
