Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

State whether silicon would form \(n\) -type or \(p\) -type semiconductors with the following elements: Ga, Sb, Al, As.

Short Answer

Expert verified
Ga, Al form \(p\)-type; Sb, As form \(n\)-type semiconductors.

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Understand Group Electron Count

Silicon belongs to group 14 of the periodic table and has 4 valence electrons. The process of forming either \(n\)-type or \(p\)-type semiconductors depends on whether the added element contributes more or fewer electrons than silicon.
02

Analyze Group of Elements

Identify the group number of each element in the periodic table: - Ga (Gallium) belongs to group 13, with 3 valence electrons. - Sb (Antimony) belongs to group 15, with 5 valence electrons. - Al (Aluminum) belongs to group 13, with 3 valence electrons. - As (Arsenic) belongs to group 15, with 5 valence electrons.
03

Determine Semiconductor Types for Each Element

- Ga and Al: Since these elements have fewer valence electrons (3) than silicon (4), they create a \(p\)-type semiconductor, incorporating holes into silicon's structure.- Sb and As: These elements have more valence electrons (5), which allows them to donate electrons to silicon, forming an \(n\)-type semiconductor, increasing electron concentration.

Key Concepts

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

n-type semiconductors
In the world of semiconductors, when we talk about \( n \)-type semiconductors, we're referring to materials that have an abundance of electrons. This happens when an impurity is added, or "doped," into a pure semiconductor like silicon, which itself is in group 14 of the periodic table. If you add elements from group 15, like antimony (Sb) or arsenic (As), these atoms bring an extra electron, as they have five valence electrons compared to silicon's four.
This extra electron can easily start moving around, which is great for conducting electricity. Basically, in \( n \)-type semiconductors, the extra electrons serve as the charge carriers, making it easier for electricity to flow. Because of this increased electron concentration, \( n \)-type materials are very efficient at conducting electric currents.
  • These materials are used in many electronic devices such as diodes and transistors.
  • They are crucial for the functioning of modern computing technologies.
p-type semiconductors
On the flip side, \( p \)-type semiconductors are just as crucial, but they operate a bit differently. When creating a \( p \)-type semiconductor, you add elements from group 13 of the periodic table, such as gallium (Ga) or aluminum (Al). These elements have only three valence electrons, which is one less than silicon's four.
This lack of an electron creates what are known as "holes" in the silicon lattice. These holes act like positive charge carriers because they allow other electrons to move in, effectively "filling" the holes and creating a flow of charge. The movement of these holes is a bit like musical chairs; as electrons jump to fill one hole, they leave another behind.
  • \( p \)-type materials are widely used along with \( n \)-type materials to create p-n junctions, crucial for the operation of diodes and transistors.
  • The interaction between holes and electrons is fundamental to semiconductor technology.
Valence electrons
Understanding valence electrons is key when discussing semiconductors. Valence electrons are the outermost electrons of an atom and are critical in determining how atoms interact or bond with each other.
Silicon, which is in group 14 of the periodic table, has four valence electrons. These electrons play a major role in its ability to form semiconductors. Their ability to share or transfer electrons defines the electrical properties of materials.
For instance, when silicon is doped with elements like phosphorus or boron, it changes the behavior of the silicon network:
  • Elements with more valence electrons (like phosphorus) make silicon more conductive, leading to \( n \)-type behavior.
  • Elements with fewer valence electrons (like boron) create \( p \)-type characteristics by forming holes.
Silicon
Silicon is a fundamental material in the technology and electronics industries, primarily because of its excellent semiconductor properties. It belongs to the group 14 in the periodic table and naturally has four valence electrons.
In its pure form, silicon is an insulator. However, by adding impurities, we can manipulate its electronic characteristics. This process of doping with either group 13 or group 15 elements transforms silicon from merely a passive material into an active one capable of conducting electricity.
  • Silicon's crystalline structure is suitable for creating p-n junctions, which are essential for forming diodes and transistors.
  • It's favoured in the industry due to its abundance and cost-effectiveness.
Periodic table groups
The periodic table is a blueprint for understanding atomic behavior, including how elements interact to form compounds like semiconductors. Each "group" in the periodic table is a vertical column that signifies elements with similar properties and the same number of valence electrons.
For example, group 13 consists of elements like aluminum and gallium, which have three valence electrons. Group 14, where silicon resides, has four, and group 15 includes elements like phosphorus and arsenic with five.
  • These groups help predict how elements will interact, especially in semiconductor doping.
  • The additional electron in group 15 elements makes them ideal for \( n \)-type semiconductors, while the deficit in group 13 makes them ideal for \( p \)-type semiconductors.
Understanding these group trends is pivotal for grasping how semiconductors can be engineered to meet specific technological needs.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In basic solution, aluminum metal is a strong reducing agent and is oxidized to \(\mathrm{AlO}_{2}^{-}\). Give balanced equations for the reaction of \(\mathrm{Al}\) in basic solution with the following: (a) \(\mathrm{NaNO}_{3}\), to give ammonia; (b) water, to give hydrogen; (c) \(\mathrm{Na}_{2} \mathrm{SnO}_{3}\), to give metallic tin.

A \(0.450-\mathrm{g}\) sample of steel contains manganese as an impurity. The sample is dissolved in acidic solution and the manganese is oxidized to the permanganate ion \(\mathrm{MnO}_{4}^{-}\). The \(\mathrm{MnO}_{4}^{-}\) ion is reduced to \(\mathrm{Mn}^{2+}\) by reacting with \(50.0 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{FeSO}_{4}\) solution. The excess \(\mathrm{Fe}^{2+}\) ions are then oxidized to \(\mathrm{Fe}^{3+}\) by \(22.4 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} .\) Calculate the percent by mass of manganese in the sample.

Describe some of the properties of aluminum that make it one of the most versatile metals known.

The following are two reaction schemes involving magnesium. Scheme I: When magnesium burns in oxygen, a white solid (A) is formed. A dissolves in \(1 M\) \(\mathrm{HCl}\) to give a colorless solution (B). Upon addition of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to \(\mathrm{B}\), a white precipitate is formed (C). On heating, \(\mathrm{C}\) decomposes to \(\mathrm{D}\) and a colorless gas is generated (E). When \(\mathrm{E}\) is passed through limewater [an aqueous suspension of \(\left.\mathrm{Ca}(\mathrm{OH})_{2}\right]\), a white precipitate appears (F). Scheme II: Magnesium reacts with \(1 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to produce a colorless solution (G). Treating \(\mathrm{G}\) with an excess of \(\mathrm{NaOH}\) produces a white precipitate (H). H dissolves in \(1 \mathrm{M} \mathrm{HNO}_{3}\) to form a colorless solution. When the solution is slowly evaporated, a white solid (I) appears. On heating I, a brown gas is given off. Identify A-I, and write equations representing the reactions involved.

Why is potassium usually not prepared electrolytically from one of its salts?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free