Q27P - Fundamentals Of Physics

Question

Iron has a density of $\mathbf{7}\mathbf{.}\mathbf{87}\mathbf{ }\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$ , and the mass of an iron atom is   $\mathbf{9}\mathbf{.}\mathbf{27}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{26}}\mathbf{ }\mathbf{kg}$ . If the atoms are spherical and tightly packed, (a) what is the volume of an iron atom, and (b) what is the distance between the centers of adjacent atoms?

Step-by-Step Solution

Answer

(a). The volume of an iron atom is $1.18\times {10}^{‐29} {\mathrm{m}}^3$.

(b). The distance between the centers of adjacent atoms is $2.82\times {10}^{‐10} \mathrm{m}$ .

Step 1: Given data

The density of iron, $\mathrm{p} = 7.87 \mathrm{g}/{\mathrm{cm}}^3$

The mass of iron atom, $\mathrm{m} = 9.27\times {10}^{‐26} \mathrm{kg}$

Step 2: Understanding the density of material

The density of a material (iron) is defined as the mass per unit volume. The distance between centers of adjacent atoms is equal to twice the radius of an atom.

The expression for density is given as: 

$\mathrm{p}=\frac{\mathrm{m}}{\mathrm{v}}$                                                                                         … (i)

Here, $\mathrm{p}$  is the density, $\mathrm{m}$  is the mass and $\mathrm{v}$ is the volume.

Step 3: (a) Determination of the volume of iron atom

Convert the density of iron into .

$\mathrm{p}=\left(7.87\frac{\mathrm{g}}{{\mathrm{cm}}^3}\right)\times \frac{1\mathrm{kg}}{1000 \mathrm{g}}\times {\left(\frac{100 \mathrm{cm}}{1 \mathrm{m}}\right)}^3$

Using equation (i), the volume of an iron atom is,

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{m}}{\mathrm{p}} \\ =\frac{9.27\times {10}^{‐26} \mathrm{kg}}{7870 \mathrm{kg}/{\mathrm{m}}^3} \\ =1.18\times {10}^{‐29} {\mathrm{m}}^3 \end{gathered}$$

Thus, the volume of iron atom is $1.18\times {10}^{‐29} {\mathrm{m}}^3$.

Step 4: (b) Determination of the distance between the centers of adjacent atoms

The volume of atom is given as follows: 

$\mathrm{v}=\frac{4{\mathrm{\pi R}}^3}{3}$

Here, $\mathrm{R}$  is the radius of the atom. 

Solving for R and substituting the values, 

$$\begin{gathered} \mathrm{R}=\sqrt{\left(\frac{3\mathrm{V}}{4\mathrm{\pi }}\right)} \\ =\sqrt{\left(\frac{3\times 1.18\times {10}^{‐29} {\mathrm{m}}^3}{4 \mathrm{m}}\right)} \\ =1.14\times {10}^{10} \mathrm{m} \end{gathered}$$

Now, the distance between the centers of two adjacent atoms is,

$$\begin{gathered} \mathrm{d}=2\mathrm{R} \\ =21.14\times {10}^{‐10} \mathrm{m} \\ =2.82\times {10}^{‐10} \mathrm{m} \end{gathered}$$

Thus, the distance between the centers of two adjacent atoms is, $2.82\times {10}^{‐10} \mathrm{m}$ .