Chapter 3.15, Problem 60AAP

(a)

To determine

The linear atomic density of atom.

Answer to Problem 60AAP

The linear atomic density of atom is $12.017\times {10}^{12}\text{atoms}/{\text{mm}}^2$.

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:

Here, the number of atoms for $\text{BCC}$ for direction $\left(100\right)$ is $1\text{atom}$.

Substitute $1\text{atom}$ for $n$ and $0.28846\text{nm}$ for $a$ in Equation (I).

    $\begin{array}{c}{\rho }_1=\frac{1\text{atoms}}{{\left(0.28846\text{nm}\right)}^2}\\ =\frac{1\text{atoms}}{{\left(0.28846\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\right)}^2}\\ =12.017\times {10}^{12}\text{atoms}/{\text{mm}}^2\end{array}$

Thus, the linear atomic density of atom is $12.017\times {10}^{12}\text{atoms}/{\text{mm}}^2$.

(b)

To determine

The linear atomic density of atom.

Answer to Problem 60AAP

The linear atomic density of atom is $16.99\times {10}^{12}\text{atoms}/{\text{mm}}^2$.

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:

Here, the number of atoms for $\text{BCC}$ for direction $\left(110\right)$ is $2\text{atom}$.

Substitute $2\text{atom}$ for $n$ and $0.28846\text{nm}$ for $a$ in Equation (II).

    $\begin{array}{c}{\rho }_1=\frac{2\text{atom}}{\sqrt{2}{\left(0.28846\text{nm}\right)}^2}\\ =\frac{2\text{atom}}{\sqrt{2}{\left(0.28846\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\right)}^2}\\ =16.99\times {10}^{12}\text{atoms}/{\text{mm}}^2\end{array}$

Thus, the linear atomic density of atom is $16.99\times {10}^{12}\text{atoms}/{\text{mm}}^2$.

(c)

To determine

The linear atomic density of atom.

Answer to Problem 60AAP

The linear atomic density of atom is $20.81\times {10}^{12}\text{atoms}/{\text{mm}}^2$.

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.

    ${\rho }_1=\frac{n}{\frac{\sqrt{3}}{2}{a}^2}$                                                             (III)

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

Conclusion:

Here, the number of atoms for $\text{BCC}$ for direction $\left(111\right)$ is $\frac{3}{2}\text{atoms}$.

Substitute $\frac{3}{2}\text{atoms}$ for $n$ and $0.28846\text{nm}$ for $a$ in Equation (III).

    $\begin{array}{c}{\rho }_1=\frac{\frac{3}{2}\text{atoms}}{\frac{\sqrt{3}}{2}{\left(0.28846\text{nm}\right)}^2}\\ =\frac{\frac{3}{2}\text{atoms}}{\frac{\sqrt{3}}{2}{\left(0.28846\text{nm}\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\right)}^2}\\ =20.81\times {10}^{12}\text{atoms}/{\text{mm}}^2\end{array}$

Thus, the linear atomic density of atom is $20.81\times {10}^{12}\text{atoms}/{\text{mm}}^2$.