## Essential University Physics: Volume 1 (3rd Edition)

We know that $n=\frac{\frac{4}{3}\pi(\frac{D_{cell}}{2})^3}{\frac{4}{3}\pi(\frac{D_{atom}}{2})^3}$ $n=(\frac{D_{cell}}{D_{atom}})^3$ We plug in the known values to obtain: $n=(\frac{10\times 10^{-6}}{0.1\times 10^{-9}})^3$ $n=(10^5)^3=10^{15}$ Now the number of atoms in the body can be determined as $n^{\prime}=(10^{15})(10^{15})\approx10^{30}atoms$