Chapter 5, Problem 5.33P

Interpretation Introduction

(a)

Interpretation:

Atomic percent of $\text{Zn per cm}$ needs to be calculated.

Concept Introduction:

Fick's first law of equation is as follows:

Answer to Problem 5.33P

The concentration gradient in atomic percent $Zn\text{ per cm}$ is $-\text{2000 at \% Zn/cm}$.

Explanation of Solution

The concentration gradient in atomic percent $Zn$ per $cm$ can be calculated the using the Fick's First law equation as given below.

  $\begin{array}{c}J=-D\frac{dc}{dx}\\ =-D\frac{\Delta c}{\Delta x}\\ =-D\frac{c_i-c_s}{\Delta x}\end{array}$

  $\begin{array}{c}\text{concentraiton gradient}=\frac{c_i-c_s}{\Delta x}\\ =\frac{c_i-c_s}{\Delta x}\mathrm{...........}\left(1\right)\end{array}$

Here,

The 1^st phase composition in atomic percent $c_i\text{ is 20 atomic }\%Zn$.

The 2^nd phase composition in atomic percent is $c_s\text{ is 25 atomic \% Zn}$.

The thickness of concentration gradient $\Delta x\text{ is 0}\text{.025 mm}$.

Substitute the above value in the equaion $\left(1\right)$, we get

  $\begin{array}{c}\frac{c_i-c_s}{\Delta x}=\frac{\left(20-25\right)\text{at}\text{\% Zn}}{\left(\text{0}\text{.025mm}\right)\left(\text{0}\text{.1cm/mm}\right)}\\ =-2000\text{ at \% Zn/cm}\end{array}$

Hence, the concentration gradient in atomic percent $Zn\text{ per cm}$ is $-\text{2000 at \% Zn/cm}$.

Interpretation Introduction

(b)

Interpretation:

The concentration gradient $wt\text{ \% of Zn}$ needs to be calculated.

Concept Introduction:

The concentration gradient can be calculated as follows:

  $\text{Concentraiton gradient}=\frac{c_i-c_s}{\Delta x}$

Answer to Problem 5.33P

The value of concentration gradient in atomic $\%Zn\text{per cm}\text{is }-2032wt\%Zn/cm$.

Explanation of Solution

Calculate the weight $\%\text{ of }\text{Zn}$ on each side of material having concentration gradient of width $0.025mm$.

  $\text{Weight \% of Zn on phase 1}=\frac{\%\text{of Zn }\times \text{Atomic mass of Zn}}{\begin{array}{l}\%\text{of Zn }\times \text{Atomic mass of Zn}+\\ \%\text{of Cu }\times \text{Atomic mass of Cu}\end{array}}\mathrm{.......................}\left(2\right)$

The percentage of $\text{Zn in Zn}-\text{Cu alloy is 20\%}$.

The percentage of $\text{C}\text{}\text{u in Zn}-\text{Cu alloy is 80\%}$

Atomic weight of $\text{Cu}\text{and Zn}\text{is 63}\text{.55}\text{and 65}\text{.39}$ respectively obtained from atomic periodic table.

Substitute the above value in equation $\left(2\right)$ as given below.

  $\begin{array}{c}\text{Weight \% of Zn on phase 1 }\left(c_i\right)=\frac{\text{20×65}\text{.39 g/mol}}{\text{20×65}\text{.39 g/mol+80×63}\text{.54 g/mol}}\times 100\\ =20.46\\ \text{Weight \% of Zn on phase 2 }\left(c_s\right)=\frac{\text{25×65}\text{.39 g/mol}}{\text{25×65}\text{.39 g/mol+75×63}\text{.54 g/mol}}\times 100\\ =25.54\end{array}$

Substitute the above value of $c_i\text{ and }{c}_s$ in the equation $\left(1\right)$.

  $\begin{array}{c}\text{concentraiton gradient}=\frac{c_i-c_s}{\Delta x}\\ =\frac{\left(20.46\%-25.54\%\right)}{\text{0}\text{.0025 cm}}\\ =-2032\text{wt}\text{\%}\text{Zn/cm}\end{array}$

Hence, the value of concentration gradient in atomic $\%Zn\text{per cm}\text{is }-2032wt\%Zn/cm$.

Here minus sign indicates that flux is moving from higher concentration to lower concentration.

Interpretation Introduction

(c)

Interpretation:

The value of concentration gradient in $Zn{\text{atoms/cm}}^3.cm$ is to be calculated.

Concept Introduction:

The number of atoms of Zn per ${\text{cm}}^{\text{3}}$ can be calculated as follows:

  ${\text{Number of atoms of Zn per cm}}^3=\frac{\text{No}\text{. of atoms of Zn per unit cell}\times \text{fraction of Zn in lattice}}{{\left(\text{Lattice parameter}\right)}^3}\mathrm{.............}\left(3\right)$

Here, no. of atoms of $Zn$ per unit cell is $4$ atoms/cell (FCC system)

Answer to Problem 5.33P

The value of concentration gradient in $Zn{\text{atoms/cm}}^{\text{3}}\text{.cm}$ is $-1.68Zn{\text{atoms/cm}}^3\text{.cm}$.

Explanation of Solution

Calculate the numbers of atoms per $c{m}^3$

  ${\text{Number of atoms of Zn per cm}}^3=\frac{\begin{array}{l}\text{No}\text{. of atoms of Zn per unit cell}\times \\ \text{fraction of Zn in lattice}\end{array}}{{(\text{Lattice parameter})}^3}$

Here no. of atoms of $\text{Zn}$ per unit cell is $4$ atoms/cell (FCC system)

The fraction of $\text{Zn}$ in lattice is $0.2$ and $0.25$ for phase $1\left(c_1\right)$ and phase $2\left(c_2\right)$ respectively from part $\left(b\right)$ of same question.

Lattice parameter is $3.63\times {10}^{-8}cm$.

Substitute the above value in equation $\left(3\right)$

  $\begin{array}{c}{\text{Number of atoms of Zn per cm}}^3\text{ on phase 1}\left(c_i\right)=\frac{\begin{array}{l}\text{No}\text{. of atoms of Zn per unit cell}\times \\ \text{fraction of Zn in lattice}\end{array}}{{(\text{Lattice parameter})}^3}\\ =\frac{\left(4\text{ atoms/cell}\right)\left(0.2\right)}{{\left(3.63\times {10}^{-8}cm\right)}^3}\\ =0.0167\times {10}^{24}{\text{ Zn atoms/cm}}^3\end{array}$

Similarly,

  $\begin{array}{c}{\text{Number of atoms of Zn per cm}}^3\text{ on phase 1}\left(c_i\right)=\frac{\begin{array}{l}\text{No}\text{. of atoms of Zn per unit cell}\times \\ \text{fraction of Zn in lattice}\end{array}}{{(\text{Lattice parameter})}^3}\\ =\frac{\left(4\text{ atoms/cell}\right)\left(0.25\right)}{{\left(3.63\times {10}^{-8}cm\right)}^3}\\ =0.0209\times {10}^{24}{\text{ Zn atoms/cm}}^3\end{array}$

Substitute the above value of $c_i$ and $c_s$ in the equation $\left(1\right)$

  $\begin{array}{c}\text{concentration gradient}=\frac{c_i-c_s}{\Delta x}\\ =\frac{\left(0.0167\times {10}^{24}-0.0209\times {10}^{24}\right)}{0.0025cm}\\ =-1.68{\text{Zn atoms/cm}}^3.\text{cm}\end{array}$

Hence, the value of concentration gradient in $Zn{\text{atoms/cm}}^{\text{3}}\text{.cm}$ is $-1.68Zn{\text{atoms/cm}}^3\text{.cm}$.