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Which of the following substances, when added in trace amounts to silicon, would produce a \(p\) -type semiconductor: (a) sulfur, (b) arsenic, (c) lead, (d) boron, (e) gallium arsenide, (f) gallium? Explain.

Short Answer

Expert verified
Boron and Gallium can create a p-type semiconductor when added to silicon in trace amounts.

Step by step solution

01

Valence Electrons of Silicon

Silicon, the base material in this scenario, has four valence electrons in its outermost shell. This information will be used to compare with the valence electrons of the given substances.
02

Valence Electrons of the Tracers

Determine the number of valence electrons for each of the given substances. Sulfur has 6, Arsenic has 5, Lead has 4, Boron has 3, Gallium Arsenide is a compound and Gallium itself has 3.
03

Identify p-type Semiconductor Materials

In order to create a p-type semiconductor, we need to introduce a substance that has one less valence electron than Silicon. In our case, Boron and Gallium, both having 3 valence electrons, meet the criteria. Therefore, adding either Boron or Gallium to silicon can create a p-type semiconductor.

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Key Concepts

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

Silicon Doping
Silicon doping involves intentionally introducing impurities into pure silicon, a process that transforms its electrical properties. This is crucial in creating semiconductors, which are fundamental components of modern electronics.
When we talk about doping in semiconductors, we essentially mean adding a substance in small amounts to silicon to alter its conductance. For a substance to effectively dope silicon and create a specific type of semiconductor, it should have a different number of valence electrons than silicon.
To create a p-type semiconductor, we utilize substances with fewer valence electrons than silicon, such as Boron or Gallium. This creates "holes" in the silicon structure—a lack of electrons where one should be, allowing for the flow of positive charge. This controlled introduction of elements forms the basis of tailoring silicon for specific electronic applications.
Valence Electrons
Valence electrons are the outermost electrons of an atom and determine how an element interacts with others. In the context of semiconductors, these electrons play a key role.
Silicon, for instance, has four valence electrons. These electrons form covalent bonds with another silicon atom, creating a stable and structured lattice. To create a p-type semiconductor, we need to add atoms with fewer valence electrons than silicon, usually three.
Boron and Gallium, with three valence electrons, are perfect candidates. When inserted into the silicon lattice, they create a vacancy or "hole" where a silicon atom would normally have an electron. This hole acts as a positive charge carrier, facilitating the creation of a p-type semiconductor.
Semiconductor Materials
Semiconductor materials are critical in the world of electronics. They possess electrical conductivity between a conductor and an insulator, making them versatile. The conductivity of semiconductors can be modified through a process called doping.
Silicon, a widely used semiconductor material, forms the backbone of modern electronic devices due to its abundant availability and favorable properties.
  • Intrinsic Semiconductors: Pure silicon with balanced electron and hole concentrations.
  • Extrinsic Semiconductors: Silicon modified by doping to form p-type or n-type semiconductors.
Through doping with elements like Boron or Gallium, silicon becomes a p-type semiconductor. This modification is vital for the development of electronic components like diodes and transistors, allowing for advancements in technology.

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Most popular questions from this chapter

Indicate several ways in which the valence-bond method is superior to Lewis structures in describing covalent bonds.

The 60 -cycle alternating electric current (AC) commonly used in households changes direction 120 times per second. That is, in a one-second time period a terminal at an electric outlet is positive 60 times and negative 60 times. In direct electric current (DC), the flow between terminals is in one direction only. A rectifer is a device that converts alternating to direct current. One type of rectifier is the \(p-n\) junction rectifier. It is commonly incorporated in adapters required to operate electronic devices from ordinary house current. In the operation of this rectifier, a \(p\) -type semiconductor and an \(n\) -type semiconductor are in contact along a boundary, or junction. Each semiconductor is connected to one of the terminals in an AC electrical outlet. Describe how this rectifier works. That is, show that when the semiconductors are connected to the terminals in an AC outlet, half the time a large flow of charge occurs and half the time essentially no charge flows across the \(p-n\) junction.

The bond angle in \(\mathrm{H}_{2} \mathrm{Se}\) is best described as (a) between \(109^{\circ}\) and \(120^{\circ} ;\) (b) less than in \(\mathrm{H}_{2} \mathrm{S} ;\) (c) less than in \(\mathrm{H}_{2} \mathrm{S},\) but not less than \(90^{\circ} ;(\mathrm{d})\) less than \(90^{\circ}\)

Draw a Lewis structure(s) for the nitrite ion, \(\mathrm{NO}_{2}^{-}\) Then propose a bonding scheme to describe the \(\sigma\) and the bonding in this ion. What conclusion can you reach about the number and types of \(\pi\) molecular orbitals in this ion? Explain.

Methyl nitrate, \(\mathrm{CH}_{3} \mathrm{NO}_{3}\), is used as a rocket propellant. The skeletal structure of the molecule is \(\mathrm{CH}_{3} \mathrm{ONO}_{2}\). The N and three O atoms all lie in the same plane, but the \(\mathrm{CH}_{3}\) group is not in the same plane as the \(\mathrm{NO}_{3}\) group. The bond angle \(\mathrm{C}-\mathrm{O}-\mathrm{N}\) is \(105^{\circ},\) and the bond angle \(\mathrm{O}-\mathrm{N}-\mathrm{O}\) is \(125^{\circ} .\) One nitrogen-to-oxygen bond length is \(136 \mathrm{pm},\) and the other two are \(126 \mathrm{pm}\) (a) Draw a sketch of the molecule showing its geometric shape. (b) Label all the bonds in the molecule as \(\sigma\) or \(\pi\), and indicate the probable orbital overlaps involved. (c) Explain why all three nitrogen-to-oxygen bond lengths are not the same.

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