Chapter 5, Problem 5.71P

Interpretation Introduction

(a)

Interpretation:

A graph of concentration (c) with x in centimeter at a time for $\text{t}=5\min ,10\min \text{and}15\min$ should be plotted.

Concept introduction:

Fick's Law of diffusion: This law states that molar flux is directly proportional to concentration gradient. The law is stated as:

  $\text{J}\text{=}-\text{D}\frac{{\text{dC}}_{\text{A}}}{\text{dx}}$

Where,

J is the molar flux defined as the number of atoms passing per unit area per unit time.

D is the diffusion coefficient in ${\text{cm}}^{\text{2}}\text{/sec}$.

  $\frac{{\text{dC}}_{\text{A}}}{\text{dx}}$ is the concentration gradient in $\frac{\text{atoms}}{{\text{cm}}^{\text{3}}\text{.cm}}$.

Factors affection diffusion are as follows:

1. Temperature
2. Diffusion coefficient

The following equation is stated as:

  $\text{D}={\text{D}}_{\text{0}}\exp \left(\frac{\text{-Q}}{\text{RT}}\right)$

Where,

Q is the activation energy in calorie/ mole.

R is universal gas constant in $\frac{\text{cal}}{\text{mole}\text{K}}$.

T is the absolute temperature in kelvin.

  ${\text{D}}_{\text{0}}$ is constant for given diffusion system and its value is defined as $\frac{\text{1}}{\text{T}}=0$ or $\text{T}\text{=}\infty$.

Answer to Problem 5.71P

The required graph for concentration versus x is shown below:

Explanation of Solution

Given information:

The equation used is given as,

  $c\left(x,t\right)=\frac{Q}{\sqrt{\pi Dt}}\exp \left(\frac{-x^2}{4Dt}\right)$

Where,

Q is the initial surface concentration having unit of $\text{atoms}\text{/}{\text{cm}}^{\text{2}}$.

  ${\text{10}}^{14}\text{atoms}\text{/}{\text{cm}}^{\text{2}}$ of phosphorous at the surface of silicon having the background concentration of ${\text{10}}^{16}\text{atoms}\text{/}{\text{cm}}^3$.

Temperature for water is ${1100}^{°}\text{C}$.

Diffusion coefficient of phosphorous in silicon at a temperature of ${1100}^{°}\text{C}$ is $\text{6}\text{.5}\times {\text{10}}^{-13}{\text{cm}}^2/\sec$

t is the time in seconds.

Given equation for finding out the plot between of concentration (c) with x,

  $c\left(x,t\right)=\frac{Q}{\sqrt{\pi Dt}}\exp \left(\frac{-x^2}{4Dt}\right)$

Based on given data calculation of $c\left(x,t\right)$ considering $\text{t}=5\min$ ,

  $Q={\text{10}}^{14}\text{atoms}\text{/}{\text{cm}}^{\text{2}}$

  $\text{t}=5\min$

  $\text{D}\text{=}\text{6}\text{.5}\times {\text{10}}^{-13}{\text{cm}}^2/\sec$

Substituting the values,

  $c\left(x,5\right)=\frac{{\text{10}}^{14}}{\sqrt{\pi \times \text{6}\text{.5}\times {\text{10}}^{-13}\times 5}}\exp \left(\frac{-x^2}{4\times \text{6}\text{.5}\times {\text{10}}^{-13}\times 5}\right)$

  $\begin{array}{l}c\left(x,5\right)=3.12\times {10}^{19}\exp \left(\frac{-x^2}{4\times \text{6}\text{.5}\times {\text{10}}^{-13}\times 5}\right)\\ c\left(x,5\right)=3.12\times {10}^{19}\exp \left(7.69\times {\text{10}}^{10}\right)\text{atoms}\text{/}{\text{cm}}^{\text{2}}\end{array}$

The relationship of concentration at a time of 5 min is,

  $c\left(x,5\right)=3.12\times {10}^{19}\exp \left(7.69\times {\text{10}}^{10}\right)\text{atoms}\text{/}{\text{cm}}^{\text{2}}$.

Given equation for finding out the plot between of concentration (c) with x,

  $c\left(x,t\right)=\frac{Q}{\sqrt{\pi Dt}}\exp \left(\frac{-x^2}{4Dt}\right)$

Based on given data calculation of $c\left(x,t\right)$ considering $\text{t}=10\min$ ,

  $Q={\text{10}}^{14}\text{atoms}\text{/}{\text{cm}}^{\text{2}}$

  $\text{t}=10\min$

  $\text{D}\text{=}\text{6}\text{.5}\times {\text{10}}^{-13}{\text{cm}}^2/\sec$

Substituting the values,

  $c\left(x,10\right)=\frac{{\text{10}}^{14}}{\sqrt{\pi \times \text{6}\text{.5}\times {\text{10}}^{-13}\times 10}}\exp \left(\frac{-x^2}{4\times \text{6}\text{.5}\times {\text{10}}^{-13}\times 10}\right)$

  $\begin{array}{l}c\left(x,10\right)=2.212\times {10}^{19}\exp \left(\frac{-x^2}{4\times \text{6}\text{.5}\times {\text{10}}^{-13}\times 10}\right)\\ c\left(x,10\right)=2.212\times {10}^{19}\exp \left(3.8461\times {\text{10}}^{10}\right)\text{atoms}\text{/}{\text{cm}}^{\text{2}}\end{array}$

The relationship of concentration at a time of 10 min is,

  $c\left(x,10\right)=2.212\times {10}^{19}\exp \left(3.8461\times {\text{10}}^{10}\right)\text{atoms}\text{/}{\text{cm}}^{\text{2}}$.

Given equation for finding out the plot between of concentration (c) with x,

  $c\left(x,t\right)=\frac{Q}{\sqrt{\pi Dt}}\exp \left(\frac{-x^2}{4Dt}\right)$

Based on given data calculation of $c\left(x,t\right)$ considering $\text{t}=5\min$ ,

  $Q={\text{10}}^{14}\text{atoms}\text{/}{\text{cm}}^{\text{2}}$

  $\text{t}=15\min$

  $\text{D}\text{=}\text{6}\text{.5}\times {\text{10}}^{-13}{\text{cm}}^2/\sec$

Substituting the values,

  $c\left(x,15\right)=\frac{{\text{10}}^{14}}{\sqrt{\text{6}\text{.5}\times {\text{10}}^{-13}\times 15}}\exp \left(\frac{-x^2}{4\times \text{6}\text{.5}\times {\text{10}}^{-13}\times 15}\right)$

  $\begin{array}{l}c\left(x,15\right)=1.806\times {10}^{19}\exp \left(\frac{-x^2}{4\times \text{6}\text{.5}\times {\text{10}}^{-13}\times 15}\right)\\ c\left(x,15\right)=1.806\times {10}^{19}\exp \left(2.5641\times {\text{10}}^{10}\right)\text{atoms}\text{/}{\text{cm}}^{\text{2}}\end{array}$

The relationship of concentration at a time of 15 min is,

  $c\left(x,15\right)=1.806\times {10}^{19}\exp \left(2.5641\times {\text{10}}^{10}\right)\text{atoms}\text{/}{\text{cm}}^{\text{2}}$

Thus, based on given relation the concentration is dependent on x and t. The graph for the same considering time into consideration,

The graph is representing the linear slope of concentration versus x with respect to given time.

Interpretation Introduction

(b)

Interpretation:

Time required by the phosphorous concentration to make it equal to the concentration of boron at a depth of $1\text{μm}$ should be determined.

Concept introduction:

Fick's Law of diffusion: This law states that molar flux is directly proportional to concentration gradient. The law is stated as:

  $\text{J}\text{=}-\text{D}\frac{{\text{dC}}_{\text{A}}}{\text{dx}}$

Where,

J is the molar flux defined as the number of atoms passing per unit area per unit time.

D is the diffusion coefficient in ${\text{cm}}^{\text{2}}\text{/sec}$.

  $\frac{{\text{dC}}_{\text{A}}}{\text{dx}}$ is the concentration gradient in $\frac{\text{atoms}}{{\text{cm}}^{\text{3}}\text{.cm}}$.

Factors affection diffusion are as follows:

1. Temperature
2. Diffusion coefficient

The following equation is stated as:

  $\text{D}={\text{D}}_{\text{0}}\exp \left(\frac{\text{-Q}}{\text{RT}}\right)$

Where,

Q is the activation energy in calorie/ mole.

R is universal gas constant in $\frac{\text{cal}}{\text{mole}\text{K}}$.

T is the absolute temperature in kelvin.

  ${\text{D}}_{\text{0}}$ is constant for given diffusion system and its value is defined as $\frac{\text{1}}{\text{T}}=0$ or $\text{T}\text{=}\infty$.

Answer to Problem 5.71P

Thus, the value of Time required by the phosphorous concentration to make it equal to the concentration of boron at a depth of $1\text{μm}$ is $\text{t }=13821.92\text{hr}$.

Explanation of Solution

Given information:

The equation used is given as,

  $c\left(x,t\right)=\frac{Q}{\sqrt{\pi Dt}}\exp \left(\frac{-x^2}{4Dt}\right)$

Where,

Q is the initial surface concentration having unit of $\text{atoms}\text{/}{\text{cm}}^{\text{2}}$.

  ${\text{10}}^{14}\text{atoms}\text{/}{\text{cm}}^{\text{2}}$ of phosphorous at the surface of silicon having the background concentration of ${\text{10}}^{16}\text{atoms}\text{/}{\text{cm}}^3$.

Temperature for water is ${1100}^{°}\text{C}$.

Diffusion coefficient of phosphorous in silicon at a temperature of ${1100}^{°}\text{C}$ is $\text{6}\text{.5}\times {\text{10}}^{-13}{\text{cm}}^2/\sec$

t is the time in seconds.

Given equation for finding out the value of time (t) in seconds

  $\begin{array}{l}c\left(x,t\right)=\frac{Q}{\sqrt{\pi Dt}}\exp \left(\frac{-x^2}{4Dt}\right)\\ t=49758906\sec \\ \text{t }=\frac{49758906\sec }{3600\sec /\text{hr}}\\ \text{t }=13821.92\text{hr}\\ \end{array}$

Based on given data calculation of $c\left(x,t\right)$ considering $\text{t}=5\min$ ,

  $Q={\text{10}}^{14}\text{atoms}\text{/}{\text{cm}}^{\text{2}}$

  $c={10}^{16}\text{atoms}\text{/}{\text{cm}}^{\text{3}}$

  $\text{D}\text{=}\text{6}\text{.5}\times {\text{10}}^{-13}{\text{cm}}^2/\sec$

  $x=1\text{μm}\text{=}\text{1}\times {\text{10}}^{-6}\text{m}$

Substituting the values,

  ${10}^{16}=\frac{{\text{10}}^{14}}{\sqrt{\pi \times \text{6}\text{.5}\times {\text{10}}^{-13}\times t}}\exp \left(\frac{{\left(-{10}^{-6}\right)}^2}{4\times \text{6}\text{.5}\times {\text{10}}^{-13}\times t}\right)$

On solving the equation,

t = 49758906

Conversion of seconds to hours,

  $\begin{array}{l}\text{t }=\frac{49758906\sec }{3600\sec /\text{hr}}\\ \text{t }=13821.92\text{hr}\end{array}$

Thus, the value of Time required by the phosphorous concentration to make it equal to the concentration of boron at a depth of $1\text{μm}$ is $\text{t }=13821.92\text{hr}$.