Question

Iron has a mass of 7.87 g per cubic centimeter of volume, and the mass of an iron atom is 9.27 × 10-26 kg. If you simplify and treat each atom as a cube, (a) what is the average volume (in cubic meters) required for each iron atom and (b) what is the distance (in meters) between the centers of adjacent atoms?

Answer 1

Concepts and reason

The main concept used to solve this problem is density of substance.

Initially, calculate the average volume for each iron atom by using the relation between density and volume. Finally, calculate the distance between the centers of adjacent atoms by using the volume of the atom.

Fundamentals

The density of a substance is defined as mass per unit volume of substance. The expression of density is given as follows:

$\rho = \frac{m}{V}$

Here, m is the mass and V is the volume.

The volume of the atomic cube is given as follows:

$V = {a^3}$

Here, a is the side of cube.

(a)

Re-arrange expression $\rho = \frac{m}{V}$ for V.

$V = \frac{m}{\rho }$

Convert the value of $\rho$from ${\rm{g/c}}{{\rm{m}}^{\rm{3}}}$ to ${\rm{kg/}}{{\rm{m}}^3}$.

$\begin{array}{c}\\\rho = 7.87{\rm{ g/c}}{{\rm{m}}^3}\left( {\frac{{{{10}^3}{\rm{ kg/}}{{\rm{m}}^3}}}{{1.0{\rm{ g/c}}{{\rm{m}}^3}}}} \right)\\\\ = 7.87 \times {10^3}{\rm{ kg/}}{{\rm{m}}^3}\\\end{array}$

Substitute $9.27 \times {10^{ - 26}}{\rm{ kg}}$for m and $7.87 \times {10^3}{\rm{ kg/}}{{\rm{m}}^3}$for $\rho$ in equation $V = \frac{m}{\rho }$.

$\begin{array}{c}\\V = \frac{{9.27 \times {{10}^{ - 26}}{\rm{ kg}}}}{{7.87 \times {{10}^3}{\rm{ kg/}}{{\rm{m}}^3}}}\\\\ = 1.178 \times {10^{ - 29}}{\rm{ }}{{\rm{m}}^3}\\\end{array}$

(b)

The volume of the atomic cube is given as follows:

$V = {a^3}$

Substitute $1.178 \times {10^{ - 29}}{\rm{ }}{{\rm{m}}^3}$ for V in the above expression.

$\begin{array}{c}\\{a^3} = 1.178 \times {10^{ - 29}}{\rm{ }}{{\rm{m}}^3}\\\\a = 2.28 \times {10^{ - 10}}{\rm{ m}}\\\end{array}$

Ans: Part a

The volume of each iron atom is $1.178 \times {10^{ - 29}}{\rm{ }}{{\rm{m}}^3}$.

Part b

The distance between the centers of adjacent atoms is $2.28 \times {10^{ - 10}}{\rm{ m}}$.