Chapter 2, Problem 2.14P

To determine

(a)

The comparison between the linear density of the $\left[100\right]$ and $\left[111\right]$ directions in the BCC metal iron with lattice parameter of $0.286\text{nm}$.

Answer to Problem 2.14P

The linear atom density in $\left[100\right]$ direction is $3.5\text{atoms}/\text{nm}$ whereas the linear density in $\left[111\right]$ direction is $4.04\text{atoms}/\text{nm}$.

Explanation of Solution

Given:

Lattice parameter is $0.286\text{nm}$.

Formula used:

The formula for Linear Atomic Density is given by,

$LAD=\frac{C_{\text{L}}}{L}$      ......... (I)

Here, LAD is the linear atomic density, $C_{\text{L}}$ is the contribution of atoms with centers on a length of line $L$.

In BCC, the length of line is given by,

$L=a\sqrt{3}$      ......... (II)

Here, $L$ is the length and $a$ is lattice parameter.

Calculation:

In $\left[100\right]$ direction, there are $0.5$ atom at $\left(0,0,0\right)$ and $0.5$ atom at $\left(1,0,0\right)$. Therefore, the total atom is $1$.

Substitute $1$ atom for $C_{\text{L}}$ and $0.286\text{nm}$ for lattice parameter.

$\begin{array}{c}LAD=\frac{1\text{atom}}{0.286\text{nm}}\\ =3.5\text{atoms}/\text{nm}\end{array}$

In $\left[111\right]$ direction, there is $0.5$ atom each at $\left(0,0,0\right)$ and $\left(1,1,1\right)$, and $1$ atom at $\left(1,0,0\right)$. Therefore, the total number of atoms is $2$.

Substitute $0.286\text{nm}$ for $a$ in equation (II).

$\begin{array}{c}L=0.286\text{nm}\times \sqrt{3}\\ =0.495\text{nm}\end{array}$

Substitute $2$ for $C_{\text{L}}$ and $0.495\text{nm}$ for $L$ in equation (II).

$\begin{array}{c}LAD=\frac{2\text{atom}}{0.495\text{nm}}\\ =4.04\text{atoms}/\text{nm}\end{array}$

Conclusion:

Therefore, the linear atom density in $\left[100\right]$ direction is $3.5\text{atoms}/\text{nm}$ whereas the linear density in $\left[111\right]$ direction is $4.04\text{atoms}/\text{nm}$.

To determine

(b)

The most closely packed direction in the BCC structure.

Answer to Problem 2.14P

The most closely packed direction in the BCC structure is $\left[111\right]$.

Explanation of Solution

Calculation:

The linear atom density in $\left[100\right]$ direction is $3.5\text{atoms}/\text{nm}$ whereas the linear density in $\left[111\right]$ direction is $4.04\text{atoms}/\text{nm}$. The linear atom density in $\left[111\right]$ direction is greater than in $\left[100\right]$ direction. So, the $\left[111\right]$ is closely packed.

Conclusion:

Therefore, the most closely packed direction in the BCC structure is $\left[111\right]$.

To determine

(c)

The radius of an iron atom, if it is assumed that the atoms touch along the most closely packed direction.

Answer to Problem 2.14P

The radius of an iron atom if it is assumed that the atoms touch along the most closely packed direction is $0.1238\text{nm}$.

Explanation of Solution

Formula used:

The length of body diagonal $\left[111\right]$ in BCC is given by,

$L=4R$      ......... (III)

Here, $R$ is the atomic radius.

Calculation:

Equate equation (II) and equation (III).

$\begin{array}{c}a\sqrt{3}=4R\\ R=\frac{a\sqrt{3}}{4}\end{array}$

Substitute $0.286\text{nm}$ for $a$ in the above expression.

$\begin{array}{c}R=\frac{0.286\text{nm}\times \sqrt{3}}{4}\\ =0.1238\text{nm}\end{array}$

Conclusion:

Therefore, the radius of an iron atom, if it is assumed that the atoms touch along the most closely packed direction is $0.1238\text{nm}$.