## College Physics (4th Edition)

$\rho = 2.3 \times 10^{17} kg/m^3$
The radius of nucleus is given by $r = r_0 A^{1/3}$ To find the density, $\rho$ $\rho = \frac{M}{V}$ $\rho = \frac{M}{\frac{4}{3} \pi r^3 }$ substitute $r = r_0 A^{1/3}$ and $M = Au$ into the $r$ in density equation $\rho = \frac{Au}{\frac{4}{3} \pi (r_0 A^{1/3})^3 }$ simplify the equation $\rho = \frac{1u}{\frac{4}{3} \pi (r_0^3) }$ Now solve for $\rho$ $\rho = \frac{1.66\times 10^{-27} kg}{\frac{4}{3} \pi (1.2 \times 10^{-15} m)^3 }$ $\rho = 2.3 \times 10^{17} kg/m^3$