Question

The mass number of gold$ = 197$, density of gold$ = 19.7g/c{m^3}$. Avogadro’s number$ = 6 \times {10^{23}}$. The radius of 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

Hint: we can find the density of one atom of gold from the information given in question. Then, equate this volume of one atom with $\dfrac{4}{3}\pi {r^3}$ to find the radius of the gold atom.

Complete step by step answer:
Following information is given in the question,
Density of gold,$\rho = 19.7g/c{m^3}$.
the mass number of gold$ = 197$,
We know that, density is the ratio of mass to volume of the given object,
$\rho = \dfrac{m}{V}$
$V = \dfrac{m}{\rho } = \dfrac{{197}}{{19.7}}$
$V = 10c{m^3}$
This is the volume of one mole of gold atoms. We know that there are Avogadro number of atoms in one.
So,
Volume of one atom$ = \dfrac{{10}}{{6 \times {{10}^{23}}}} = \dfrac{5}{3} \times {10^{ - 23}}c{m^3}$
We assume the atom to be perfectly sphere with perfectly rigid surface,
The volume of sphere$ = \dfrac{4}{3}\pi {r^3}$
Where $r = $radius of gold atom.
$\dfrac{4}{3}\pi {r^3} = \dfrac{5}{3} \times {10^{ - 23}}$
$r = {(\dfrac{5}{{4\pi }} \times {10^{ - 23}})^{ - 3}}$
$r = 1.5 \times {10^{ - 10}}m$
C) is correct.

Additional information:
An atom is the smallest unit of ordinary matter that forms everything in this universe. Every solid, liquid and gas is composed of neutral or ionized atoms. Atoms are so small that accurately predicting their behaviour using classical physics is not possible due to quantum effects.
Every atom is composed of a nucleus and one or more electrons bound to the nucleus. The nucleus is made of one or more protons and a number of neutrons.

Note: It is worth noting that, mass number is just the number of protons and neutrons in the nucleus of an atom while the same mass number in grams is the weight of one mole of that atom.