Chapter 5.7, Problem 33SEP

To determine

The flux of $\text{Mn}$ atoms between the surface and plane $2\text{mm}$ deep.

Answer to Problem 33SEP

The flux of $\text{Mn}$ atoms between the surface and plane $2\text{mm}$ deep is $6.8625\times {10}^5{\text{atoms/m}}^2\cdot \text{s}$.

Explanation of Solution

Write the formula for lattice constant $\left(a\right)$ of unit of FCC iron.

  $a=\frac{4R}{\sqrt{2}}$        (I)

Here, the radius of the iron atom is $R$.

Write the formula for volume $\left(V\right)$ of the unit cell.

  $V=a^3$        (II)

Write the formula for Fick’s law of diffusion.

  $J=-D\frac{dC}{dx}$        (III)

Here, flux or flow of atoms is $J$ in $\frac{\text{atoms}}{{\text{m}}^2\cdot \text{s}}$, constant of diffusivity is $D$ in $\frac{{\text{m}}^2}{\text{s}}$ , and concentration gradient is $\frac{dC}{dx}$ in $\frac{\text{atoms}}{{\text{m}}^4}$.

Conclusion:

Refer Table 3.2, “Selected metals that have the BCC crystal structure at room temperature (20°C) and their lattice constants and atomic radii”.

The atomic radius of iron is $0.124\text{nm}$.

Substitute $0.124\text{}\text{nm}$ for $R$ in Equation (I).

  $\begin{array}{c}a=\frac{4\left(0.124\text{}\text{nm}\right)}{\sqrt{2}}\\ =0.351\text{nm}\end{array}$

Substitute $0.351\text{nm}$ for $a$ in Equation (II).

  $\begin{array}{c}V={\left(0.351\text{nm}\right)}^3\\ ={\left(0.351\text{nm}\times \frac{{\text{10}}^{-9}\text{m}}{1\text{nm}}\right)}^3\\ =4.32\times {10}^{-24}{\text{m}}^3\end{array}$

The number of atoms per unit volume is expressed as follows.

  $\begin{array}{c}\text{atoms per unit volume}=\frac{\text{no}\text{. of atoms in FCC crystal}}{\text{volume}\text{of unit cell}\left(V\right)}\\ =\frac{4\text{atoms}}{4.32\times {10}^{-24}{\text{m}}^3}\\ =9.25\times {10}^{28}{\text{atoms/m}}^3\end{array}$

The concentration of atoms at the surface $C_s$ is $0.6a\%$.

  $\begin{array}{c}{C}_s=0.6a\%\\ =\frac{0.6}{100}\left(9.25\times {10}^{28}{\text{atoms/m}}^3\right)\\ =5.55\times {10}^{26}{\text{atoms/m}}^3\end{array}$

The concentration of atom at the distance of $2\text{mm}$ below the surface $C_{2\text{mm}}$ is $0.1a\%$.

  $\begin{array}{c}{C}_{\text{2}\text{mm}}=0.1a\%\\ =\frac{0.1}{100}\left(9.25\times {10}^{28}{\text{atoms/m}}^3\right)\\ =9.25\times {10}^{25}{\text{atoms/m}}^3\end{array}$

Refer Table 5.2, “Diffusivities at 500°C and 1000°C for selected solute-solvent diffusion systems”.

The diffusivity of manganese in FCC iron at 500°C is $3\times {10}^{-24}{\text{m}}^2\text{/s}$.

Here,

  $\begin{array}{c}dC=C_{\text{2}\text{mm}}-C_s\\ =\left(9.25\times {10}^{25}{\text{atoms/m}}^3\right)-\left(5.55\times {10}^{26}{\text{atoms/m}}^3\right)\\ =-4.575\times {10}^{26}{\text{atoms/m}}^3\end{array}$

Substitute $3\times {10}^{-24}{\text{m}}^2\text{/s}$ for $D$, $-4.575\times {10}^{26}{\text{atoms/m}}^3$ for $dC$, and $2\text{mm}$ for $dx$ in Equation (III).

  $\begin{array}{c}J=-\left(3\times {10}^{-24}{\text{m}}^2\text{/s}\right)\frac{-4.575\times {10}^{26}{\text{atoms/m}}^3}{2\text{mm}}\\ =-\left(3\times {10}^{-24}{\text{m}}^2\text{/s}\right)\frac{-4.575\times {10}^{26}{\text{atoms/m}}^3}{2\text{mm}\times \frac{1\text{m}}{1000\text{m}}}\\ =6.8625\times {10}^5{\text{atoms/m}}^2\cdot \text{s}\end{array}$

Thus, the flux of $\text{Mn}$ atoms between the surface and plane $2\text{mm}$ deep is $6.8625\times {10}^5{\text{atoms/m}}^2\cdot \text{s}$.