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

Answer: (C)

Volume occupied by one gram atom of gold
$=\frac{197\, g }{19.7\, g / m ^{3}}=10 \,cm ^{3}$
Volume of one atom $=\frac{10}{6 \times 10^{-23}}$
$=\frac{5}{3} \times 10^{23} \,cm ^{3}$
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} $
$r =1.5 \times 10^{-10}\, m$