Answer
(a) $2.3\times 10^{17}Kg/m^3$
(b) the same as
(c) $2.3\times 10^{17}Kg/m^3$
Work Step by Step
(a) We can calculate the nuclear density of $Th^{228}_{90}$ as follows:
$Mass \space of\space the\space nucleus\space M=Am$
$\implies M=(228)(1.67\times 10^{-27})Kg$
Now the volume of the nucleus is given as
$V=\frac{4}{3}\pi r^3$
$\implies V=\frac{4}{3}\pi[(1.2\times 10^{-15}m)(228)^{\frac{1}{3}}]^{3}$
$V=\frac{4}{3}\pi (1.2\times 10^{-15}m)^3(228)$
The nuclear density is given as
$\rho=\frac{M}{V}$
We plug in the known values to obtain:
$\rho=\frac{(228)(1.67\times 10^{-27}Kg)}{\frac{4}{3}\pi(1.2\times 10^{-15}m)^3(228)}$
$\rho=2.3\times 10^{17}Kg/m^3$
(b) We can find the required density as follows:
$\rho=\frac{M}{V}$
We plug in the known values to obtain:
$\rho=\frac{Am}{\frac{4}{3}\pi[(1.2\times 10^{-15}m)A^{\frac{1}{3}}]^3}$
This simplifies to:
$\rho=\frac{3m}{4\pi(1.7\times 10^{-45}m^3)}$
$\rho=2.3\times 10^{17}Kg/m^3$
Thus, the nuclear density is independent of the number of nucleons and the nuclear density of the alpha particle is same as that of $Th^{228}_{90}$.
(c) The nuclear density of alpha particle is calculated in part (b), which is $2.3\times 10^{17}Kg/m^3$
