Chapter 5, Problem 5.58P

Interpretation Introduction

Interpretation:

The concentration of $\text{Ga}$ at a depth of $2.0$ micrometer if the surface concentration of $\text{Ga si 1023}{\text{atoms/cm}}^3$ needs to be calculated.

Concept introduction:

The derived mathematical expression from Fick's second law is as follows:

  $\frac{c_s-c_x}{c_s-c_0}=\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\mathrm{..........}\left(1\right)$

Here, $c_s$ is the concentration of diffusion at the surface, $c_0$ is the concentration of diffusion at x, $c_0$ is the uniform concentration of diffused atoms in the material, $c_x$ is the concentration of atoms diffusing at x and D is diffusion coefficient.

Answer to Problem 5.58P

The value of $c_x\text{ si 3}\text{.55}\times {\text{10}}^{23}{\text{ atoms/cm}}^3$.

Explanation of Solution

Gallium $\left(\text{Ga}\right)$ content at any depth $x$ beneath the surface can be calculated using the expression derived from Fick's Second Law.

Mathematical expression of Fick's Second Law is as given below:

  $\frac{\partial c}{\partial t}=\frac{\partial }{\partial x}\left(D\frac{\partial c}{\partial x}\right)$

The derived mathematical expression from Fick's Second Law is

  $\frac{c_s-c_x}{c_s-c_0}=\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\mathrm{..........}\left(1\right)$

The diffusion coefficient $D\text{ is 7}\times {\text{10}}^{-13}c{m}^2/s$.

When time $t=3600s\left(1hr\right)$

Substitute the above values in Fick's equation $\left(1\right)$ as given below.

  $\begin{array}{c}\frac{c_s-c_x}{c_s-c_0}=\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\\ \frac{{10}^{23}-c_x}{{10}^{23}}=\text{erf}\left(\frac{2\times {10}^{-4}}{2\sqrt{\left(7\times {10}^{-13}\right)\left(3600\right)}}\right)\\ =\text{erf}\left(1.99\right)\end{array}$

The value of the error $\left(\frac{x}{2\sqrt{Dt}}\right)\text{ is 1}\text{.99}$, the corresponding value of the $\left(\text{erf}\frac{x}{2\sqrt{Dt}}\right)$ from table $\left(1\right)\text{ is 0}\text{.9953}$.

Error function values for Fick's second law
Argument of the error   $\frac{x}{2\sqrt{Dt}}$	Values of the error function   $\text{erf}\frac{x}{2\sqrt{Dt}}$
$0$   $$	$0$
$0.10$	$0.1125$
$0.20$	$0.2227$
$0.30$	$0.3286$
$0.40$	$0.4284$
$0.50$	$0.5205$
$0.60$	$0.6039$
$0.70$	$0.6778$
$0.80$	$0.7421$
$0.90$	$0.7969$
$1.00$	$0.8427$
$1.50$	$0.9961$
$2.00$	$0.9953$

  $\begin{array}{c}\left(\text{erf}\frac{x}{2\sqrt{Dt}}\right)=0.9953\\ \frac{{10}^{23}-c_x}{{10}^{23}}=0.9953\\ c_x=4.7\times {10}^{20}{\text{ atoms/cm}}^3\end{array}$

Therefore, the value of $c_x$ is $4.7\times {10}^{20}{\text{ atoms/cm}}^3$.

When time $t=7200s\left(2\text{ hr}\right)$

Substitute the above values in Fick's equation $\left(1\right)$ as given below.

  $\begin{array}{c}\frac{c_s-c_x}{c_s-c_0}=\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\\ \frac{{10}^{23}-c_x}{{10}^{23}}=\text{erf}\left(\frac{2\times {10}^{-4}}{2\sqrt{\left(7\times {10}^{-13}\right)\left(7200\right)}}\right)\\ =\text{erf}\left(0.995\right)\end{array}$

The value of the error $\left(\frac{x}{2\sqrt{Dt}}\right)\text{ is 0}\text{.995}$, the corresponding value of the $\left(\text{erf}\frac{x}{2\sqrt{Dt}}\right)$ from table $\left(1\right)\text{ is 0}\text{.8427}$.

  $\begin{array}{c}\left(\text{erf}\frac{x}{2\sqrt{Dt}}\right)=0.8427\\ \frac{{10}^{23}-c_x}{{10}^{23}}=0.8427\\ c_x=1.57\times {10}^{23}{\text{ atoms/cm}}^3\end{array}$

Therefore, the value of $c_x$ is. $1.57\times {10}^{23}{\text{ atoms/cm}}^3$

When time $t=10800s\left(\text{3 hr}\right)$

Substitute the above values in Fick's equation $\left(1\right)$ as given below.

  $\begin{array}{c}\frac{c_s-c_x}{c_s-c_0}=\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)\\ \frac{{10}^{23}-c_x}{{10}^{23}}=\text{erf}\left(\frac{2\times {10}^{-4}}{2\sqrt{\left(7\times {10}^{-13}\right)\left(10800\right)}}\right)\\ =\text{erf}\left(0.663\right)\end{array}$

The value of the error $\left(\frac{x}{2\sqrt{Dt}}\right)\text{ is 0}\text{.663 }$, the corresponding value of the $\left(\text{erf}\frac{x}{2\sqrt{Dt}}\right)$ from table $\left(1\right)\text{ is 0}\text{.645}$ (extrapolation of the value since $0.663$ lies between $0.6\text{ and 0}\text{.7}$ ).

  $\begin{array}{c}\left(\text{erf}\frac{x}{2\sqrt{Dt}}\right)=0.645\\ \frac{{10}^{23}-c_x}{{10}^{23}}=0.645\\ c_x=3.55\times {10}^{23}{\text{ atoms/cm}}^3\end{array}$

Conclusion

Therefore, the value of $c_x$ is. $3.55\times {10}^{23}{\text{ atoms/cm}}^3$