Chapter 2, Problem 2.8P

To determine

(a)

The numbers of atoms in Silicon FCC unit cell.

Answer to Problem 2.8P

The numbers of atoms in Silicon FCC unit cell is $8$.

Explanation of Solution

Given:

Silicon is atom at $0,0,0$ and an atom at $\frac{1}{4},\frac{1}{4},\frac{1}{4}$.

Lattice parameter of silicon is $0.543\text{nm}$.

Density of the silicon atoms is $2.33\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$.

Calculation:

Draw the crystal structure of silicon atoms.

Figure (1)

FCC (Face-Centered structure) of Silicon has $4$ atoms. Other atoms are placed at $\frac{1}{4},\frac{1}{4},\frac{1}{4}$ as shown in figure (1). It can be concluded that after placing four new atoms in existing FCC structure, the total atoms in this FCC unit cell is $8\text{atoms}$.

Conclusion:

Therefore, the numbers of atoms in Silicon FCC unit cell is $8$.

To determine

(b)

The numbers of atoms per unit volume for the unit cell of Silicon.

Answer to Problem 2.8P

The numbers of atoms per unit volume for the unit cell of Silicon is $0.50\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}$.

Explanation of Solution

Formula used:

Write the expression to find the number of atoms per unit volume.

$n_{\text{a}}=\frac{\text{Number}\text{of}\text{atoms}}{a^3}$      ......... (I)

Here, $n_{\text{a}}$ is the number of atoms per unit volume, $a$ is the lattice parameter and $\text{Number}\text{of}\text{atoms}$ is the numbers of atoms in a unit cell.

Calculation:

Silicon has Face Centered Cubic (FCC) structure with $8$ number of atoms.

Substitute $8\text{atoms}$ for $\text{Number}\text{of}\text{atoms}$ and $0.543\text{nm}$ for $a$ in equation (I) to find $n_{\text{a}}$.

$\begin{array}{c}{n}_{\text{a}}=\frac{8\text{atoms}}{{\left(0.543\right)}^3}\\ =\frac{8\text{atoms}}{{\left(0.543\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}^3}\\ =\frac{8\text{atoms}}{0.16\times {10}^{-27}{\text{m}}^3}\\ =0.50\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the numbers of atoms per unit volume for the unit cell of Silicon is $0.50\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}$.

To determine

(c)

The number of atoms per unit volume by using the density of silicon.

Answer to Problem 2.8P

The number of atoms per unit volume is $0.50\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}$

Explanation of Solution

Formula used:

Write the formula to find the atoms per unit volume.

$N_{\text{V}}=\frac{{\rho }_{\gamma \text{Si}}\times N_{\text{A}}}{M_{\text{Si}}}$      ......... (III)

Here, $N_{\text{V}}$ is the atom per unit volume, ${\rho }_{\gamma \text{Si}}$ is thedensity of Silicon, $M_{\text{Si}}$ is the molecular weight of Silicon and $N_{\text{A}}$ is the Avogadro’s number.

Calculation:

The molecular weight of Silicon is $28.09\times {10}^{-3}\text{kg}/\text{mole}$.

The density of Silicon is $2.33\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$.

Avogadro’s number is $6.02\times {10}^{23}\text{atoms}/\text{mole}$.

Substitute $2.33\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\gamma \text{Si}}$, $28.09\times {10}^{-3}\text{kg}/\text{mole}$ for $M_{\text{Si}}$ and $6.02\times {10}^{23}\text{atoms}/\text{mole}$ for $N_{\text{A}}$ in equation (II) to find $N_{\text{V}}$.

$\begin{array}{c}{N}_{\text{V}}=\frac{2.33\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\times 6.02\times {10}^{23}\text{atoms}/\text{mole}}{28.09\times {10}^{-3}\text{kg}/\text{mole}}\\ =0.50\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the number of atoms per unit volume is $0.50\times {10}^{29}\text{atoms}/{\text{m}}^{\text{3}}$.