Question

Given mass number of gold $=197$,
Density of gold $=19.7g/c{m}^3$
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

Answer: (C)

Volume occupied by one gram atom of gold
$=\frac{197g}{19.7g/m^3}=10c{m}^3$
Volume of one atom $=\frac{10}{6\times 10^{-23}}$
$=\frac{5}{3}\times 10^{23}c{m}^3$
Let $r$ be the radius of the atom
$\therefore \frac{4}{3}\pi r^3=\frac{5}{3}\times 10^{23}$
$r^3=\frac{50\times 10^{-24}}{4\times 3.14}$
$r=1.5\times 10^{-10}m$