Q12.123CP - Chemistry: Molecular Nature Of Matter And Change

Question

In making computer chips, a 4.00-kg cylindrical ingot of ultrapure n-type doped silicon that is 5.20 inches in diameter is sliced into wafers $\mathbf{1}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{-4}$ m thick. 

(a) Assuming no waste, how many wafers can be made? 

(b) What is the mass of a wafer (d of Si = $\mathbf{2}\mathbf{.}\mathbf{34}\mathbf{g}\mathbf{/}{\mathbf{cm}}^3$ ; V of a cylinder = ${\mathbf{\pi r}}^2\mathbf{h}$ )?

(c) A key step in making p-n junctions for the chip is the chemical removal of the oxide layer on the wafer through treatment with gaseous HF. Write a balanced equation for this reaction. 

(d) If 0.750% of the Si atoms are removed during the treatment in part (c), how many moles of HF is required per wafer, assuming 100% reaction yield?

Step-by-Step Solution

Answer

1. Total number of Wafers $\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{18}\mathbf{\times }{\mathbf{10}}^6$ from the 4kg cylindrical ingot of ultrapure n-type doped silicon.
2. Mass of wafers $\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{-3}\mathbf{g}$ from the 4kg cylindrical ingot of ultrapure n-type doped silicon.
3. The oxide layer on the semi-conductor by HF forms a water molecule and fluoride ion. The reaction is as follows:

                                 $6\mathbf{HF}\mathbf{+}\mathbf{SiO}\mathbf{2}\to {\mathbf{SiF}}^{-}\mathbf{6}\mathbf{+}2\mathbf{H}\mathbf{3}{\mathbf{O}}^{+}$

 d.  From the above reaction, it shows that 6 moles of HF react with the one mole of ${\mathbf{SiO}}_{\mathbf{2}}$. Removing  the 0.75% of Si it requires the 4.5moles of the HF.

Step 1: Definition

Density can be defined as the ratio of mass to volume. Density is inversely proportional to the volume of the unit cell as the mass of the atom is constant.

 

Density equation for the unit cell:

 $\mathbf{\rho }\mathbf{=}\frac{M}{V}$

 

ρ= Density,  

M = mass 

V=Volume

Step 2: Numerical Explanation

Mass of Cylindrical ingot of ultrapure n-type doped silicon  $\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{kg}$

Diameter of the cylindrical cone  $\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{20}\mathbf{inch}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{21}\mathbf{cm}$

The radius of the cylindrical cone  $\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{61}\mathbf{cm}$

Diameter of the wafers sliced from the silicon  $\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{-4}\mathbf{m}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{12}\mathbf{\times }{\mathbf{10}}^{-5}\mathbf{cm}$

The radius of the wafers sliced from the silicon $\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{56}\mathbf{\times }{\mathbf{10}}^{-5}\mathbf{cm}$

a.

 

$\begin{array}{c}\mathrm{Number} \mathrm{of} \mathrm{wafers}=\frac{\mathrm{Radius} \mathrm{of} \mathrm{Cylindrical} \mathrm{Cone}}{\mathrm{Radius} \mathrm{of} \mathrm{Wafers}}\\ \mathrm{Number} \mathrm{of} \mathrm{wafers}=\frac{6.61}{0.56\times {10}^{-5}\mathrm{cm}}\\ \mathrm{Number} \mathrm{of} \mathrm{wafers}=1.18\times {10}^6\end{array}$

Number of Wafers =  $\mathbf{1}\mathbf{.}\mathbf{18}\mathbf{\times }{\mathbf{10}}^6$

b.

Mass of cylindrical cone of silicon  $\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{kg}$

Number of wafers $\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{18}\mathbf{\times }{\mathbf{10}}^6$  

 

  $\begin{array}{l}\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{wafers}\mathbf{=}\frac{\mathrm{Mass} \mathrm{of} \mathrm{Cylindrical} \mathrm{Cone} \mathrm{Silicon}}{\mathrm{Number} \mathrm{of} \mathrm{Wafers}}\\ \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{wafers}\mathbf{=}\frac{\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^3}{\mathbf{1}\mathbf{.}\mathbf{18}\mathbf{\times }{\mathbf{10}}^6}\\ \mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{wafers}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{-3}\mathbf{g}\end{array}$

 

Mass of Wafers  $\mathbf{3}\mathbf{.}\mathbf{38}\mathbf{\times }{\mathbf{10}}^{-3}\mathbf{g}$

c.

The p-n junction involves the formation of positive and negative semi-conductors to form a conductor to carry current.

The oxide layer on the semi-conductor by HF forming a water molecule and fluoride ion. The reaction as follows:

 $\mathbf{6}\mathbf{HF}\mathbf{+}\mathbf{SiO}\mathbf{2}\mathbf{\to }{\mathbf{SiF}}^{-}\mathbf{6}\mathbf{+}\mathbf{2}\mathbf{H}\mathbf{3}{\mathbf{O}}^{+}$

d.

From the above reaction, it shows that 6 moles of HF react with the one mole of $\mathbf{SiO}\mathbf{2}$. For removing the 0.75% of Si it requires the 4.5moles of the HF.

 $\begin{array}{c}\mathbf{Number}\mathbf{of}\mathbf{Moles}\mathbf{of}\mathbf{HF}\\ \Rightarrow \mathbf{0}\mathbf{.}\mathbf{75}\mathbf{\%}\mathbf{of}\mathbf{ }\mathbf{HF}\mathbf{=}\mathbf{6}\mathbf{\times }\frac{\mathbf{0}\mathbf{.}\mathbf{75}}{100}\\ \Rightarrow \mathbf{0}\mathbf{.}\mathbf{75}\mathbf{\%}\mathbf{of}\mathbf{ }\mathbf{HF}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{moles}\end{array}$