Chapter 2, Problem 2.12P

To determine

The planar atom density of the $\left\{100\right\}$ -type planes and $\left\{111\right\}$ -type planes and to compare their atom density.

Answer to Problem 2.12P

The planar density of $\left\{100\right\}$ -plane is $15.35\times {10}^{18}\text{atoms}/{\text{m}}^2$ and the planar density of $\left\{111\right\}$ -plane is $17.746\times {10}^{18}\text{atoms}/{\text{m}}^2$. $\left\{111\right\}$ -plane is more closely packed.

Explanation of Solution

Given:

Lattice parameter of FCC structure of copper is $0.361\text{nm}$.

Formula used:

The planar density of a plane is given as,

$PD=\frac{\text{Number of atoms contributed to plane}}{\text{Area of plane}}$   ...... (I)

Here, $PD$ is planar density.

The area of $\left\{100\right\}$ -plane is given by

$\text{Area of }\left\{100\right\}\text{-plane}={\left(a_0\right)}^2$   ...... (II)

Here, $a_0$ is lattice parameter.

Calculation:

The area of $\left\{100\right\}$ -plane in FCC structure has two atoms.

Substitute $0.361\text{nm}$ for $a_0$ in equation (II) to find the area of $\left\{100\right\}$ -plane,

$\begin{array}{c}\text{Area of }\left\{100\right\}\text{-plane}={\left(0.361\text{nm}\right)}^2\\ ={\left(0.361\text{ nm}\times \frac{{10}^{-9}\text{m}}{1\text{ nm}}\right)}^2\\ =0.1303\times {10}^{-18}{\text{ m}}^2\end{array}$

Substitute $0.1303\times {10}^{-18}{\text{ m}}^2$ for Area of plane and $2\text{ atoms}$ for Number of atoms contributed to plane in equation (I) to find the planar density of $\left\{100\right\}$ -plane,

$\begin{array}{c}PD=\frac{2\text{ atoms}}{0.1303\times {10}^{-18}{\text{ m}}^2}\\ =15.35\times {10}^{18}\text{atoms}/{\text{m}}^2\end{array}$

Draw $\left\{111\right\}$ -plane of FCC crystal.

Figure (1)

Calculate the plane area.

$\text{Area of plane}=\frac{1}{2}bh$

Here, $b$ is breadth and $h$ is height of the triangle.

From figure (1),

Substitute $\sqrt{2a_0}$ for $b$ and $\sqrt{\frac{3}{2}{a}_0}$ for $h$ in the above expression to find the expression for the area of plane.

$\text{Area of plane}=\frac{1}{2}\left(\sqrt{2a_0}\right)\left(\sqrt{\frac{3}{2}{a}_0}\right)$

Substitute $0.361\text{nm}$ for $a_0$ in the above equation to find area of the plane.

$\begin{array}{c}\text{Area of plane}=\frac{1}{2}\left[\sqrt{2}\left(0.361\text{nm}\right)\right]\left[\sqrt{\frac{3}{2}}\left(0.361\text{nm}\right)\right]\\ =\frac{1}{2}\left(0.51\text{nm}\right)\left(0.4421\text{nm}\right)\\ =\left(0.1127{\text{nm}}^2\right){\left(\frac{{10}^{-9}\text{ m}}{1\text{ nm}}\right)}^2\\ =0.1127\times {10}^{-18}{\text{ m}}^2\end{array}$

The area of $\left\{111\right\}$ -plane in FCC structure has two atoms.

Substitute $0.1127\times {10}^{-18}{\text{ m}}^2$ for Area of plane and $2\text{ atoms}$ for Number of atoms contributed to plane in equation (I) to find the planar density of $\left\{111\right\}$ -plane.

$\begin{array}{c}PD=\frac{2\text{ atoms}}{0.1127\times {10}^{-18}{\text{ m}}^2}\\ =17.746\times {10}^{18}\text{atoms}/{\text{m}}^2\end{array}$

Compare the planar density of $\left\{100\right\}$ -plane and the planar density of $\left\{111\right\}$ -plane, it can be observed that $\left\{111\right\}$ -plane is more closely packed.

Conclusion:

Therefore, the planar density of $\left\{100\right\}$ -plane is $15.35\times {10}^{18}\text{atoms}/{\text{m}}^2$ and the planar density of $\left\{111\right\}$ -plane is $17.746\times {10}^{18}\text{atoms}/{\text{m}}^2$. $\left\{111\right\}$ -plane is more closely packed.