#### Answer

(a) 79 protons, 118 neutrons, 79 electrons
(b) $\rho = 2.29\times 10^{17}~kg/m^3$
(c) The density of the gold nucleus is $~~2.0\times 10^{13}~~$ times the density of lead.

#### Work Step by Step

(a) $^{197}Au$
A gold atom has 79 protons.
Since this atom has 197 nucleons, the atom must have 118 neutrons.
Since this atom is neutral, it must have 79 electrons.
(b) We can assume that the nucleus is a sphere.
We can find the density of the nucleus:
$\rho = \frac{m}{V}$
$\rho = \frac{(197)(1.67\times 10^{-27}~kg)}{\frac{4}{3}\pi~r^3}$
$\rho = \frac{(3)(197)(1.67\times 10^{-27}~kg)}{(4\pi)~(7.0\times 10^{-15}~m)^3}$
$\rho = 2.29\times 10^{17}~kg/m^3$
(c) $\frac{2.29\times 10^{17}~kg/m^3}{11,400~kg/m^3} = 2.0\times 10^{13}$
The density of the gold nucleus is $~~2.0\times 10^{13}~~$ times the density of lead.