Question

Calculate approximately the diameter of an atom of mercury, assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom. The density of mercury is 13.6 g/cc.

Answer 1

Suppose the side of cube = x cm = diameter of mercury atom

∴ Volume of 1 Hg atom = $x^3$ and

Mass of 1 Hg atom

= $density\times volume$

= $13.6\times x^3$

Mass of 1 Hg atom

= $\frac{Atomicmass}{Avogadroconstant}$

= $\frac{200}{6.022\times {10}^{23}}$

$13.6\times x^3=\frac{200}{6.022\times {10}^{23}}$

=> $x^3=\frac{200}{13.6\times 6.022\times {10}^{23}}=2.44\times {10}^{-23}$

=> x = $(2.44\times {10}^{-23}{)}^{1/3}=2.9\times {10}^{-8}$ cm.