Chapter 3.15, Problem 61AAP

(a)

To determine

The linear atomic density of atom.

Answer to Problem 61AAP

The linear atomic density of atom is $1.202\times {10}^{13}\text{atoms}/{\text{mm}}^2$.

Explanation of Solution

Draw the figure for the plane of $\left(100\right)$.

Figure (1)

Write the expression for the planer density of the atom.

    ${\rho }_1=\frac{n}{a^2}$                                                             (I)

Here, the number of the atoms is $n$ and the lattice constant is $a$.

Conclusion:

Substitute $2\text{atoms}$ for $n$ and $0.40788\text{nm}$ for $a$ in Equation (I).

    $\begin{array}{c}{\rho }_1=\frac{2\text{atoms}}{{\left(0.40788\text{nm}\right)}^2}\\ =\frac{2\text{atoms}}{{\left(0.40788\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\right)}^2}\\ =1.202\times {10}^{13}\text{atoms}/{\text{mm}}^2\end{array}$

Thus, the linear atomic density of atom is $1.202\times {10}^{13}\text{atoms}/{\text{mm}}^2$.

(b)

To determine

The linear atomic density of atom.

Answer to Problem 61AAP

The linear atomic density of atom is $8.50\times {10}^{12}\text{atoms}/{\text{mm}}^2$.

Explanation of Solution

Draw the figure for the plane of $\left(110\right)$.

Figure (2)

Write the expression for the planer density of the atom.

    ${\rho }_1=\frac{n}{\sqrt{2}{a}^2}$                                                             (II)

Here, the number of the atoms is $n$ and the lattice constant is $a$.

Conclusion:

Substitute $2\text{atom}$ for $n$ and $0.40788\text{nm}$ for $a$ in Equation (II).

    $\begin{array}{c}{\rho }_1=\frac{2\text{atom}}{\sqrt{2}{\left(0.40788\text{nm}\right)}^2}\\ =\frac{2\text{atom}}{\sqrt{2}{\left(0.40788\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\right)}^2}\\ =8.50\times {10}^{12}\text{atoms}/{\text{mm}}^2\end{array}$

Thus, the linear atomic density of atom is $8.50\times {10}^{12}\text{atoms}/{\text{mm}}^2$.

(c)

To determine

The linear atomic density of atom.

Answer to Problem 61AAP

The linear atomic density of atom is $1.963\times {10}^{13}\text{atoms}/{\text{mm}}^2$.

Explanation of Solution

Draw the figure for the plane of $\left(111\right)$.

Figure (3)

Write the expression for the planer density of the atom.

    ${\rho }_1=\frac{n}{\frac{\sqrt{6}}{4}{a}^2}$                                                             (III)

Here, the number of the atoms is $n$ and the lattice constant is $a$.

Conclusion:

Substitute $2\text{atom}$ for $n$ and $0.40788\text{nm}$ for $a$ in Equation (III).

    $\begin{array}{c}{\rho }_1=\frac{2\text{atom}}{\frac{\sqrt{6}}{4}{\left(0.40788\text{nm}\right)}^2}\\ =\frac{2\text{atom}}{\frac{\sqrt{6}}{4}{\left(0.40788\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\right)}^2}\\ =1.963\times {10}^{13}\text{atoms}/{\text{mm}}^2\end{array}$

Thus, the linear atomic density of atom is $1.963\times {10}^{13}\text{atoms}/{\text{mm}}^2$.