Question

The density of iron is $7.87\,g\,c{m^{ - 3}}$. If the atoms are spherical and closely packed. The mass of an iron atom is $9.27 \times {10^{ - 26}}\,Kg$. What is the volume of an iron atom$?$
(i) $1.18 \times {10^{ - 29}}\,{m^3}$
(ii) $2.63 \times {10^{ - 29}}\,{m^3}$
(iii) $1.73 \times {10^{ - 29}}\,{m^3}$
(iv) $0.53 \times {10^{ - 29}}\,{m^3}$

Answer 1

Hint: Density is the degree of compactness of a substance and is given by mass per unit volume. The density of iron and the mass of the iron atom is given. Manipulate the formula relating them all a bit, put the values and convert the values to suitable units to get the volume of the iron atom.

Complete step-by-step answer:The density of a substance is its mass per unit volume. The symbol most often used to represent density is $\rho $. Mathematically, density of a substance is defined as mass divided by volume as: $\rho = \dfrac{m}{V}$ , where $\rho $ is the density of the substance, $m$ is the mass of the substance, and $V$ is the volume of the same substance.
Therefore the equation can also be written as: $V = \dfrac{m}{\rho }......\left( 1 \right)$.
In the question, it is given that the density of iron is $7.87\,g\,c{m^{ - 3}}$and the mass of iron atom is $9.27 \times {10^{ - 26}}\,Kg$.
But the values are not in the same unit and we need to convert them all either to SI units or CGS units. Since in the given options $metre$ is given therefore we convert them all to SI units.
Therefore, density of iron \[ = \,7.87 \times {10^{ - 3}}\,Kg\, \times {10^6}{m^{ - 3}}\, = \,7.87 \times {10^3}\,Kg\,{m^{ - 3}}\]in SI units.
Mass of iron is already in SI units.
Therefore putting these values in equation $\left( 1 \right)$ we get, \[V = \dfrac{{9.27 \times {{10}^{ - 26}}\,Kg}}{{7.87 \times {{10}^3}\,Kg\,{m^{ - 3}}}}\, = \,1.18 \times {10^{ - 29}}\,{m^3}\].

Hence the correct answer is (i) $1.18 \times {10^{ - 29}}\,{m^3}$.

Note:The most important part of this question is unit conversions. By looking at the question you might feel the question is easy and therefore in a hurry you can mix up SI and CGS units and do error in calculations. In order to avoid this proceed in a stepwise manner and convert to one unit before doing the calculations.