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Chapter 12: Problem 123

Silicon has the diamond structure with a unit cell edge length of \(5.43 \AA\) 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 cm鲁 of silicon material, there are \(5.0 \times 10^{23}\) silicon atoms. To dope the silicon sample with 1 ppm concentration of phosphorus, 0.02329 mg of phosphorus is required.

Step by step solution

01

Find the volume of a silicon unit cell.

The unit cell of silicon has a cubic shape with an edge length of 5.43 脜. First, we need to convert this to cm: Edge length (in cm) = \(5.43 \times 10^{-8}\) cm Now, we can find the volume of the unit cell: Volume of unit cell = edge length鲁 = \((5.43 \times 10^{-8})^3\) = \(1.6021 \times 10^{-22} \mathrm{~cm}^{3}\)
02

Determine the number of unit cells in 1 cm鲁

To find the number of unit cells in 1 cm鲁 of silicon, divide the volume of 1 cm鲁 by the volume of a unit cell: Number of unit cells = \(\frac{1 \mathrm{~cm}^{3}}{1.6021 \times 10^{-22} \mathrm{~cm}^{3}}\) = \(6.25 \times 10^{22}\)
03

Calculate the total number of silicon atoms in 1 cm鲁

Since there are 8 silicon atoms per unit cell, we can find the total number of silicon atoms in 1 cm鲁 by multiplying the number of unit cells by 8: Total number of silicon atoms = \(8 \times 6.25 \times 10^{22}\) = \(5.0 \times 10^{23}\)
04

Calculate the mass of 1 cm鲁 of silicon

Now we can find the mass of 1 cm鲁 of silicon sample. The molar mass of silicon is approximately 28 g/mol. We know that 1 mole of silicon consists of \(6.022 \times 10^{23}\) atoms. So, we can find the mass of the total number of silicon atoms that we calculated in step 3: Mass of 1 cm鲁 of silicon = \(\frac{5.0 \times 10^{23} \ \text{atoms} \times 28 \ \text{g/mol}}{6.022 \times 10^{23} \ \text{atoms/mol}}\) = 23.29 g
05

Find the mass of phosphorus required for 1 ppm concentration

We are asked to dope the silicon sample with 1 ppm (parts per million) of phosphorus, which means that we need to add 1 mg of phosphorus per 1g of silicon. Mass of phosphorus required = \(\frac{23.29\text{g}}{10^{6} \text{g/mg}}\) = 0.00002329 g So, 0.00002329 g or 0.02329 mg of phosphorus is required to dope the 1 cm鲁 sample of silicon with a 1 ppm concentration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Crystal Lattice
The concept of a crystal lattice is fundamental in understanding the structure of crystalline solids like silicon. A crystal lattice is an arrangement of atoms, molecules, or ions in a highly ordered microscopic structure, forming a lattice that extends in all directions. For silicon, which has a diamond cubic crystal structure, each unit cell鈥攁 repeating unit that forms the crystal鈥攃ontains eight atoms arranged in a way that resembles two interpenetrating face-centered cubic lattices.

The edges of a silicon unit cell measure 5.43 angstroms, and these precise measurements are crucial when determining how many atoms are in a given volume. For any calculations involving unit cells, remember that understanding their three-dimensional arrangement assists in visualizing how many unit cells fill a certain space, such as one cubic centimeter, and subsequently how many atoms are present.
Doping Semiconductors
Doping semiconductors is a process where impurities are intentionally introduced to a semiconductor to enhance its conductivity. The introduction of a dopant, such as phosphorus, into a silicon crystal lattice creates additional free charge carriers that greatly increase the material's electrical conductivity.

The proprietary level of doping is measured in parts per million (ppm), indicating how many dopant atoms are added relative to the number of semiconductor atoms. Even with a minuscule amount of phosphorus, the conductivity can increase by factors of millions. This principle is at the heart of creating various electronic components, from microchips to solar cells, where precise control over the semiconductor properties is paramount.
Molar Mass
The molar mass, often expressed in grams per mole (g/mol), represents the mass of one mole of any substance. For elements, the molar mass in grams per mole is numerically equal to the atomic mass of the element as listed on the periodic table.

Silicon's molar mass is approximately 28 g/mol, essential for converting between the number of atoms and the mass of the substance. During calculations involving molar mass, it serves as a conversion factor that links the macroscopic mass of a sample to its microscopic atomic count. For instance, knowing the molar mass allows the calculation of how many atoms are in a 1 cm鲁 silicon sample and how much mass of a dopant, such as phosphorus, is needed for a specific doping concentration.

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Most popular questions from this chapter

Which of the following statements does not follow from the fact that the alkali metals have relatively weak metal-metal bonding? (a) The alkali metals are less dense than other metals. (b) The alkali metals are soft enough to be cut with a knife. (c) The alkali metals are more reactive than other metals. (d) The alkali metals have higher melting points than other metals. (e) The alkali metals have low ionization energies.

Red light-emitting diodes are made from GaAs and GaP solid solutions, \(\mathrm{GaP}_{x} \mathrm{As}_{1-x}\) (see Exercise 12.75). The original red LEDs emitted light with a wavelength of \(660 \mathrm{~nm}\). If we assume that the band gap varies linearly with composition between \(x=0\) and \(x=1\), estimate the composition (the value of \(x\) ) that is used in these LEDs.

At room temperature and pressure RbI crystallizes with the NaCl-type structure. (a) Use ionic radii to predict the length of the cubic unit cell edge. (b) Use this value to estimate the density. (c) At high pressure the structure transforms to one with a CsCl-type structure. (c) Use ionic radii to predict the length of the cubic unit cell edge for the highpressure form of RbI. (d) Use this value to estimate the density. How does this density compare with the density you calculated in part (b)?

A particular form of cinnabar (HgS) adopts the zinc blende structure. The length of the unit cell edge is \(5.852 \AA\). (a) Calculate the density of \(\mathrm{HgS}\) in this form. (b) The mineral tiemmanite ( \(\mathrm{HgSe}\) ) also forms a solid phase with the zinc blende structure. The length of the unit cell edge in this mineral is \(6.085 \AA\). What accounts for the larger unit cell length in tiemmanite? (c) Which of the two substances has the higher density? How do you account for the difference in densities?

Proteins are naturally occurring polymers formed by condensation reactions of amino acids, which have the general structure In this structure, \(-\mathrm{R}\) represents \(-\mathrm{H},-\mathrm{CH}_{3}\), or another group of atoms; there are 20 different natural amino acids, and each has one of 20 different \(R\) groups. (a) Draw the general structure of a protein formed by condensation polymerization of the generic amino acid shown here. (b) When only a few amino acids react to make a chain, the product is called a "peptide" rather than a protein; only when there are 50 amino acids or more in the chain would the molecule be called a protein. For three amino acids (distinguished by having three different \(R\) groups, R1, R2, and \(R 3\) ), draw the peptide that results from their condensation reactions. (c) The order in which the \(\mathrm{R}\) groups exist in a peptide or protein has a huge influence on its biological activity. To distinguish different peptides and proteins, chemists call the first amino acid the one at the " \(\mathrm{N}\) terminus" and the last one the one at the "C terminus." From your drawing in part (b) you should be able to figure out what " \(\mathrm{N}\) terminus" and " \(\mathrm{C}\) terminus" mean. How many different peptides can be made from your three different amino acids?
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