Question

A single human nerve cell may extend from 1 m from the sprial column to the foot and be as much as $\frac{1}{{10}^6}$ m in average diameter. If we assume that each of the constituent atoms of this structure occupies a volume of $3\times \frac{1}{{10}^{29}}{m}^3$, what is the order of the number of atoms building the nerve cell.

Answer 1

The volume of a single human nerve cell is $1m\times (2\times \pi \times \frac{1}{{10}^6}\times (0.5{)}^2)m^2=1.57\times {10}^{-12}{m}^3$
if one atom occupies $3\times {10}^{-29}{m}^3$
approx $15\times {10}^{11}$ atoms can fit in.