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Chapter 12: Problem 130

Silicon has the diamond structure with a unit cell edge length of $543 \mathrm{pm}$ and eight atoms per unit cell. (a) How many silicon atoms are there in \(1 \mathrm{~cm}^{3}\) of material? (b) Suppose you dope that $1 \mathrm{~cm}^{3}$ sample of silicon with 1 ppm of phosphorus that will increase the conductivity by a factor of a million. How many milligrams of phosphorus are required?

Short Answer

Expert verified
In 1 cm³ of silicon, there are approximately \(4.995\times10^{22}\) silicon atoms, with a mass of 2.33 g. To dope this sample with 1 ppm phosphorus and increase conductivity by a factor of a million, about \(2.33\times10^{-3}\,mg\) of phosphorus is required.

Step by step solution

01

Calculate the volume of one unit cell of silicon

First, we need to find the volume of one unit cell of silicon. Since the unit cell is a cube, we can calculate the volume using the formula: Volume = Edge_length³ The edge length is given in picometers (pm), so we'll first convert it to centimeters (cm): \(Edge\,length\,(in\,cm) = \frac{543\,pm}{1\times10^{12}\,pm}= 5.43\times10^{-8}\,cm\) Now, we can find the volume of a single unit cell of silicon: \(Volume\,of\,unit\,cell = (5.43\times10^{-8}\,cm)^3 = 1.601\times10^{-22}\,cm^3\)
02

Calculate the number of unit cells in 1 cm³ of silicon

Next, we'll find out how many unit cells are present in 1 cm³ of silicon. To do this, we'll divide 1 cm³ by the volume of a single unit cell: \(Number\,of\,unit\,cells = \frac{1\,cm^3}{1.601\times10^{-22}\,cm^3} = 6.244\times10^{21}\)
03

Find the number of silicon atoms in 1 cm³ of silicon

Now that we have the number of unit cells present in 1 cm³ of silicon, we can find the total number of silicon atoms by multiplying the number of unit cells by the number of atoms per unit cell (which is 8): \(Number\,of\,silicon\,atoms = (6.244\times10^{21})\times8 = 4.995\times10^{22}\) So, there are approximately \(4.995\times10^{22}\) silicon atoms in 1 cm³ of the material.
04

Calculate the mass of 1 cm³ of silicon

To find the mass of 1 cm³ of silicon, we will use the mass of one silicon atom (which is the atomic weight of silicon divided by Avogadro's number) and multiply it by the number of silicon atoms in 1 cm³: \(Mass\,of\,1\,silicon\,atom = \frac{28.0855\,g/mol}{6.022\times10^{23}\,atoms/mol} = 4.6637\times10^{-23}\,g\) Mass of 1 cm³ of silicon: \(Mass\,of\,1\,cm^3\,silicon = (4.995\times10^{22})\times(4.6637\times10^{-23}\,g) = 2.33\,g\)
05

Determine the mass of phosphorus needed for doping

We are given that 1 ppm of phosphorus is required to increase the conductivity by a factor of a million. To find how many milligrams of phosphorus are required for doping: \(Mass\,of\,phosphorus = 1\,ppm\,of\,2.33\,g = \frac{2.33\,g}{10^6} = 2.33\times10^{-6}\,g\) Convert the mass of phosphorus from grams to milligrams: \(Mass\,of\,phosphorus\,(mg) = 2.33\times10^{-6}\,g\times\frac{1000\,mg}{1\,g} = 2.33\times10^{-3}\,mg\) So, approximately \(2.33\times10^{-3}\,mg\) of phosphorus is required for doping 1 cm³ of silicon.

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