Answer
(a) $m = 1.16\times 10^{-26}~kg$
$R = 2.3~fm$
$\rho = 2.3\times 10^{17}~kg/m^3$
(b) $m = 3.44\times 10^{-25}~kg$
$R = 7.1~fm$
$\rho = 2.3\times 10^{17}~kg/m^3$
Work Step by Step
We can write a general equation for the radius $R$ of an atomic nucleus:
$R = r_0A^{1/3}$
$r_0 = 1.2~fm$
$A$ is the atomic mass number (number of nucleons)
(a) We can find the mass of $^7Li$:
$m = A~(1.66\times 10^{-27}~kg)$
$m = (7)~(1.66\times 10^{-27}~kg)$
$m = 1.16\times 10^{-26}~kg$
We can find the radius of $^7Li$:
$R = r_0A^{1/3}$
$R = (1.2~fm)(7)^{1/3}$
$R = 2.3~fm$
We can find the density of $^7Li$:
$\rho = \frac{m}{V}$
$\rho = \frac{m}{(4/3)\pi r^3}$
$\rho = \frac{3m}{4\pi r^3}$
$\rho = \frac{(3)(1.16\times 10^{-26}~kg)}{(4\pi) (2.3\times 10^{-15}~m)^3}$
$\rho = 2.3\times 10^{17}~kg/m^3$
(b) We can find the mass of $^{207}Pb$:
$m = A~(1.66\times 10^{-27}~kg)$
$m = (207)~(1.66\times 10^{-27}~kg)$
$m = 3.44\times 10^{-25}~kg$
We can find the radius of $^{207}Pb$:
$R = r_0A^{1/3}$
$R = (1.2~fm)(207)^{1/3}$
$R = 7.1~fm$
We can find the density of $^{207}Pb$:
$\rho = \frac{m}{V}$
$\rho = \frac{m}{(4/3)\pi r^3}$
$\rho = \frac{3m}{4\pi r^3}$
$\rho = \frac{(3)(3.44\times 10^{-25}~kg)}{(4\pi) (7.1\times 10^{-15}~m)^3}$
$\rho = 2.3\times 10^{17}~kg/m^3$
