Chapter 12, Problem 12.128P

a.

Interpretation Introduction

Interpretation:

The number of wafers that can be made from $\text{Si}$ ingot is to be calculated.

Concept introduction:

Density is defined as mass per unit volume. Mass and volume are physical quantities and the units of mass and volume are fundamental units. Density is the ratio of mass to the volume. The unit of volume is derived from the units of mass and volume. The SI unit of density is ${\text{kg/m}}^{\text{3}}$. The formula to calculate density is,

  $\text{Density}=\frac{\text{Mass}}{\text{Volume}}$        (1)

The formula to calculate the volume of the cylinder is as follows:

  $V=\pi r^{2}h$        (2)

Here,

$r$ is the radius of the cylinder.

$h$ is the height of the cylinder.

Answer to Problem 12.128P

The number of wafers that can be made from $\text{Si}$ ingot is $1.11\times {10}^3$.

Explanation of Solution

Rearrange the equation (1) to calculate the volume of $\text{Si}$ ingot.

  $\text{Volume}\text{of}\text{Si ingot}=\left(\frac{\text{Mass}}{\text{Density}}\right)$        (3)

Substitute $2.34{\text{g/cm}}^3$ for density and $4.00\text{kg}$ for mass in the equation (3) to calculate the volume of $\text{Si}$ ingot.

  $\begin{array}{c}\text{Volume}\text{of Si}\text{ingot}=\left(\frac{4.00\text{kg}}{2.34{\text{g/cm}}^3}\right)\left(\frac{1000\text{g}}{1\text{kg}}\right)\\ =1709.402{\text{cm}}^3\end{array}$

The diameter of $\text{Si}$ ingot is $5.20\text{in}$. Therefore the radius of $\text{Si}$ ingot is calculated as follows:

$\begin{array}{c}\text{Radius of}\text{Si ingot}=\left(\frac{5.20\text{in}}{2}\right)\\ =2.60\text{in}\left(\frac{2.54\text{cm}}{\text{1}\text{in}}\right)\\ =6.604\text{ cm}\end{array}$

Rearrange the equation (2) to calculate the height of $\text{Si}$ ingot.

  $h=\frac{V}{\pi r^2}$        (4)

Substitute $1709.402{\text{cm}}^3$ for $V$, $3.14159$ for $\pi$ and $6.604\text{ cm}$ for $r$ in the equation (4) to calculate the height of $\text{Si}$ ingot.

  $\begin{array}{c}h=\frac{1709.402{\text{cm}}^3}{\left(3.14159\right){\left(6.604\text{ cm}\right)}^2}\\ =12.476\text{ cm}\end{array}$

The thickness of the wafer is $1.12\times {10}^{-4}\text{m}$. Therefore the number of the wafer is calculated as follows:

  $\begin{array}{c}\text{Number of wafer}=\left(\frac{12.476\text{ cm}}{1.12\times {10}^{-4}\text{m}}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =1113.94\\ =1.11\times {10}^3.\end{array}$

Conclusion

$1.11\times {10}^3$ $\text{Si}$ ingot can be made from $4.00\text{kg}$ cylindrical ingot that is $5.20\text{in}$ in diameter.

(b)

Interpretation Introduction

Interpretation:

The mass of a wafer is to be calculated.

Concept introduction:

The formula to calculate the mass of the cylinder is as follows:

  $M=\pi r^{2}hd$        (5)

Here,

$M$ is the mass of the cylinder.

$r$ is the radius of the cylinder.

$h$ is the height of the cylinder.

$d$ is the density of the cylinder.

Answer to Problem 12.128P

The mass of a wafer is $3.59\text{g}$.

Explanation of Solution

Substitute $1.12\times {10}^{-4}\text{m}$ for $h$, $3.14159$ for $\pi$ and $6.604\text{ cm}$ for $r$, $2.34{\text{ g/cm}}^3$ for $d$ in the equation (5) to calculate the mass of wafer.

  $\begin{array}{c}M=\left(3.14159\right){\left(6.604\text{ cm}\right)}^2\left(1.12\times {10}^{-4}\text{m}\right)\left(\frac{100\text{cm}}{\text{1}\text{m}}\right)\left(2.34{\text{ g/cm}}^3\right)\\ =3.5909\text{ g}\\ \approx 3.59\text{g}\text{.}\end{array}$

Conclusion

The shape of the wafer is cylindrical. The mass of a wafer is calculated as $3.59\text{g}$.

(c)

Interpretation Introduction

Interpretation:

The balanced equation for the removal of the oxide layer on the wafer is to be written.

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass. Also, the amounts of substances in a balanced chemical reaction are stoichiometrically equivalent to each other.

Displacement reactions are those in which substances on both sides of the equation remain the same but the atoms exchange places in order to form the product. Displacement reactions can be further classified as a double displacement reaction and a single displacement reaction. In double displacement reaction, both the reactants are the compounds and in single displacement reaction, one of the reactants is an element. The general representation of a double displacement reaction is:

  $\text{AB+CD}\to \text{AD}+\text{CB}$

Answer to Problem 12.128P

The balanced equation for the removal of the oxide layer on the wafer is as follows:

  ${\text{SiO}}_{\text{2}}\left(s\right)+4\text{HF}\left(g\right)\to {\text{SiF}}_4\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)$

Explanation of Solution

Silicon combines with oxygen to form silicon dioxide. The coating of silicon dioxide interferes with the function of the wafer and is removed by gaseous hydrogen fluoride.

Sulfur displaces the hydrogen from hydrogen fluoride and leads to the formation of silicon tetrafluoride and water. The chemical equation for the reaction is as follows:

  ${\text{SiO}}_{\text{2}}\left(s\right)+4\text{HF}\left(g\right)\to {\text{SiF}}_4\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)$

Conclusion

Displacement reactions are those in which substances on both sides of the equation remain the same but the atoms exchange places in order to form the product.

(d)

Interpretation Introduction

Interpretation:

The moles of $\text{HF}$ required per wafer is to be calculated.

Concept introduction:

Stoichiometry of a reaction is utilized to determine the amount of any species in the reaction by the relationship between the reactants and products.

Consider the general reaction,

  $\text{A}+2\text{B}\to 3\text{C}$

One mole of $\text{A}$ reacts with two moles of $\text{B}$ to produce three moles of $\text{C}$. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

Answer to Problem 12.128P

The moles of $\text{HF}$ required per wafer is $3.84\times {10}^{-3}\text{mol}$.

Explanation of Solution

The formula to calculate the moles of $\text{HF}$ is as follows:

  $\text{Moles of}\text{HF}=\frac{\text{mass of}\text{Si}}{\text{molar mass of Si}}\left(\frac{0.750\%}{100\%}\right)\left(\frac{4\text{mol}\text{HF}}{1\text{mol}\text{Si}}\right)$        (6)

Substitute $3.5909\text{g}$ for the mass of $\text{Si}$, $28.09\text{g/mol}$ for molar mass of $\text{Si}$ in the equation (6).

  $\begin{array}{c}\text{Moles of}\text{HF}=\frac{3.5909\text{g}}{28.09\text{g/mol}}\left(\frac{0.750\%}{100\%}\right)\left(\frac{4\text{mol}\text{HF}}{1\text{mol}\text{Si}}\right)\\ =3.83506\times {10}^{-3}\text{mol}\\ \approx 3.84\times {10}^{-3}\text{mol}\text{.}\end{array}$

Conclusion

The number of moles of water required by one mole of wafer is $3.84\times {10}^{-3}\text{mol}$.