Question

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

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

Answer 1

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

Given: Density of gold, $d=19.7gc{m}^{-3}=19.7\times 1000kg{m}^{-3}$

Mas of a gold atom, $m=197$ a.m.u (as $A=197$)

Let radius of the gold atom be $R$ m.

$\therefore d=\frac{m}{\frac{4}{3}\pi R^3}⟹R^3=\frac{3m}{4\pi d}$

Putting the values, $R^3=\frac{3\times 197\times 1.66\times {10}^{-27}}{4\pi \times 19.7\times 1000}=3.96\times {10}^{-30}{m}^3$

$⟹R\approx 1.5\times {10}^{-10}m$