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

Hint: We know that the density of any object is the ratio of its mass upon its volume. Given are the value of density of gold, the mass number of gold, and Avogadro’s number. By using the mass number we get the mass of a gold atom. And as the volume of an atom is also known in its mathematical formula, equating all these together we get the value for the radius of a gold atom.

Formula Used:
$\eqalign{
  & {\text{Density}},d = \dfrac{{{\text{Mass, }}m}}{{{\text{Volume}},V}} \cr
  & {\text{Volume,}}V = \dfrac{4}{3}\pi {R^3} \cr
  & {\text{where }}R{\text{ is the radius of the atom}} \cr} $

Complete step by step answer:
Given:
The mass number of gold =$197$
Density of gold = $19.7g/c{m^3}$
Now, the density of any atom is given mathematically as:
${\text{Volume,}}V = \dfrac{4}{3}\pi {R^3} \cdots \cdots \cdots \cdots \left( 1 \right)$
But additionally, we also know that the density the atom will be given by:
$\eqalign{
  & {\text{Density}},d = \dfrac{{{\text{Mass, }}m}}{{{\text{Volume}},V}} \cr
  & \Rightarrow d = \dfrac{m}{V} \cdots \cdots \cdots \left( 2 \right) \cr} $
Combining equation (1) and (2) we get:
$\eqalign{
  & d = \dfrac{m}{{\dfrac{4}{3}\pi {R^3}}} \cr
  & \Rightarrow {R^3} = \dfrac{m}{{\dfrac{4}{3}\pi d}} \cr
  & \Rightarrow {R^3} = \dfrac{{3 \times 197 \times 1.66 \times {{10}^{ - 27}}}}{{4\pi \times 19.7 \times 1000}} + \cr
  & \Rightarrow {R^3} = 3.96 \times {10^{ - 30}}{m^3} \cr
  & \therefore R \approx 1.5 \times {10^{ - 10}}m \cr} $
Therefore, the correct option is C. i.e., the radius of the gold atom is $1.5 \times {10^{ - 10}}m$

Additional Information:
An atom is though very small in size, it still constitutes two basic parts i.e., the negatively charged electrons which revolve around it and the nucleus which constitutes of neutrons and positively charged protons.
The size of a nucleus in an atom is determined using the concept of the distance of the closest approach which was a consequence of Rutherford’s experiment on the scattering of alpha particles.
The nuclear radius is about ten thousand times smaller than the atomic radius.

Note: One atomic mass unit for an atom is defined as the $\dfrac{1}{2}th$ of the actual mass of carbon 12 atom. Additionally, it was found experimentally that the volume of a nucleus is also directly proportional to its mass number. So, the radius of a nucleus is also proportional to the cube root of its mass number.