Question

Given mass number of gold = 197,
Density of gold = 19.7 g/cm${}^3$
Avogadro's number = $6\times {10}^{23}$. The radius of the gold atom is approximately

• A. $1.5\times {10}^{-8}m$
• B. $1.7\times {10}^{-9}m$
• C. $1.5\times {10}^{-10}m$
• D. $1.5\times {10}^{-12}m$

Answer 1

The correct option is C $1.5\times {10}^{-10}m$

Volume occupied by one gram atom of gold $=\frac{m}{\rho }=\frac{197g}{19.7g/c{m}^3}=10c{m}^3$

Volume of one atom $=\frac{10}{N_A}=\frac{10}{6\times {10}^{-23}}$ $=\frac{5}{3}\times {10}^{23}c{m}^3$
Let $r$ be the radius of the atom.
$\therefore \frac{4}{3}\pi r^3=\frac{5}{3}\times {10}^{23}$
$r^3=\frac{50\times {10}^{-24}}{4\times 3.14}$
$r=1.5\times {10}^{-10}m$