/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Gallium arsenide, GaAs, has gain... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 40

Gallium arsenide, GaAs, has gained widespread use in semiconductor devices that convert light and electrical signals in fiberoptic communications systems. Gallium consists of \(60 . \%^{69} \mathrm{Ga}\) and \(40 . \%^{71} \mathrm{Ga}\). Arsenic has only one naturally occurring isotope, \({ }^{75}\) As. Gallium arsenide is a polymeric material, but its mass spectrum shows fragments with the formulas GaAs and \(\mathrm{Ga}_{2} \mathrm{As}_{2} .\) What would the distribution of peaks look like for these two fragments?

Short Answer

Expert verified
For GaAs fragment, the distribution of peaks will have a 60% abundance for \(^{69}\)Ga - \(^{75}\)As and a 40% abundance for \(^{71}\)Ga - \(^{75}\)As. For Ga鈧侫s鈧� fragment, the distribution of peaks will have a 36% abundance for \(^{69}\)Ga - \(^{69}\)Ga - \(^{75}\)As - \(^{75}\)As, a 24% abundance for \(^{69}\)Ga - \(^{71}\)Ga - \(^{75}\)As - \(^{75}\)As, a 24% abundance for \(^{71}\)Ga - \(^{69}\)Ga - \(^{75}\)As - \(^{75}\)As, and a 16% abundance for \(^{71}\)Ga - \(^{71}\)Ga - \(^{75}\)As - \(^{75}\)As.

Step by step solution

01

Determine the possible combinations of isotopes in GaAs fragment

Let's list down all possible combinations of Gallium and Arsenic isotopes in GaAs fragment: 1. \(^{69}\)Ga - \(^{75}\)As 2. \(^{71}\)Ga - \(^{75}\)As
02

Calculate the relative abundances for GaAs fragment

We multiply the abundances of the individual isotopes in each combination: 1. \(^{69}\)Ga - \(^{75}\)As: 0.60 * 1 = 0.60 or 60% 2. \(^{71}\)Ga - \(^{75}\)As: 0.40 * 1 = 0.40 or 40%
03

Determine the possible combinations of isotopes in Ga鈧侫s鈧� fragment

Let's list down all possible combinations of Gallium and Arsenic isotopes in Ga鈧侫s鈧� fragment: 1. \(^{69}\)Ga - \(^{69}\)Ga - \(^{75}\)As - \(^{75}\)As 2. \(^{69}\)Ga - \(^{71}\)Ga - \(^{75}\)As - \(^{75}\)As 3. \(^{71}\)Ga - \(^{69}\)Ga - \(^{75}\)As - \(^{75}\)As 4. \(^{71}\)Ga - \(^{71}\)Ga - \(^{75}\)As - \(^{75}\)As
04

Calculate the relative abundances for Ga鈧侫s鈧� fragment

We multiply the abundances of the individual isotopes in each combination: 1. \(^{69}\)Ga - \(^{69}\)Ga - \(^{75}\)As - \(^{75}\)As: (0.60 * 0.60) * (1 * 1) = 0.36 or 36% 2. \(^{69}\)Ga - \(^{71}\)Ga - \(^{75}\)As - \(^{75}\)As: (0.60 * 0.40) * (1 * 1) = 0.24 or 24% 3. \(^{71}\)Ga - \(^{69}\)Ga - \(^{75}\)As - \(^{75}\)As: (0.40 * 0.60) * (1 * 1) = 0.24 or 24% 4. \(^{71}\)Ga - \(^{71}\)Ga - \(^{75}\)As - \(^{75}\)As: (0.40 * 0.40) * (1 * 1) = 0.16 or 16% Now we have the distribution of peaks for both fragments: For GaAs fragment: 1. \(^{69}\)Ga - \(^{75}\)As: 60% 2. \(^{71}\)Ga - \(^{75}\)As: 40% For Ga鈧侫s鈧� fragment: 1. \(^{69}\)Ga - \(^{69}\)Ga - \(^{75}\)As - \(^{75}\)As: 36% 2. \(^{69}\)Ga - \(^{71}\)Ga - \(^{75}\)As - \(^{75}\)As: 24% 3. \(^{71}\)Ga - \(^{69}\)Ga - \(^{75}\)As - \(^{75}\)As: 24% 4. \(^{71}\)Ga - \(^{71}\)Ga - \(^{75}\)As - \(^{75}\)As: 16%

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Isotope Distribution
When we use mass spectrometry to analyze a substance, we often look at isotopes to understand the distribution of different elements within the material. Isotopes are variations of elements that have different numbers of neutrons. This concept is essential in interpreting mass spectra because each isotope contributes to the peaks observed.
In the case of gallium arsenide,
  • Gallium (Ga) has two main isotopes -
    • \(^{69}\text{Ga}\), with a natural abundance of 60%, and
    • \(^{71}\text{Ga}\), with a natural abundance of 40%.
  • Arsenic (As), on the other hand, is typically found as \(^{75}\text{As}\), which is the only naturally occurring isotope.
This means that when we look at a mass spectrum of gallium arsenide (GaAs), we will see two main peaks corresponding to the different combinations of these isotopes:
  • \(^{69}\text{Ga} - ^{75}\text{As}\) at 60% abundance due to \(^{69}\text{Ga}'s\) higher abundance.
  • \(^{71}\text{Ga} - ^{75}\text{As}\) at 40% abundance.
Understanding isotope distribution helps us predict and interpret the peaks in mass spectrometry, crucial for identifying compounds like GaAs.
Gallium Arsenide
Gallium arsenide (\(\text{GaAs}\)) is a compound of gallium and arsenic with noteworthy properties that make it invaluable in electronics. It鈥檚 a favorite for semiconductor devices because it surpasses silicon in several ways.
  • GaAs boasts superior electron mobility, resulting in faster electronic devices that are essential for high-frequency communication systems.
  • It also has a direct band gap, meaning it can efficiently convert electric energy into light, which is pivotal for optoelectronic applications.
  • Moreover, GaAs is more resistant to radiation damage, which is beneficial for space and military applications.
Mass spectrometry of GaAs involves analyzing fragments like GaAs and \(\text{Ga}_2\text{As}_2\). This helps in understanding the composition and quality of the material, essential for producing reliable semiconductor devices.
Semiconductor Materials
Semiconductor materials form the foundation of modern electronics. A semiconductor, like gallium arsenide, has electrical properties between conductors and insulators, making them crucial for controlling electronic circuits.
  • Silicon is the most common semiconductor but GaAs is preferred in high-speed and optoelectronic applications due to its unique properties such as higher electron mobility.
  • Semiconductors have adjustable conductivity, which is possible due to doping - the process of adding impurities to form p-type or n-type materials. This adjustability allows semiconductors to perform various functions inside electronic devices.
  • Mass spectrometry aids in quality control of semiconductors by analyzing the isotopic composition and purity of materials like GaAs, ensuring they meet the stringent requirements of electronic manufacturers.
Understanding these properties and processes ensures the successful application of semiconductor materials in devices from smartphones to solar panels.

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

Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz: \(3 \mathrm{NaHCO}_{3}(a q)+\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q) \longrightarrow\) $$ 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) $$ a. What mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) should be used for every \(1.0 \times 10^{2} \mathrm{mg}\) \(\mathrm{NaHCO}_{3} ?\) b. What mass of \(\mathrm{CO}_{2}(g)\) could be produced from such a mixture?

Consider the following reaction: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If a container were to have 10 molecules of \(\mathrm{O}_{2}\) and 10 molecules of \(\mathrm{NH}_{3}\) initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

Ascorbic acid, or vitamin \(\mathrm{C}\left(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{O}_{0}\right)\), is an essential vitamin. It cannot be stored by the body and must be present in the diet. What is the molar mass of ascorbic acid? Vitamin C tablets are taken as a dietary supplement. If a typical tablet contains \(500.0 \mathrm{mg}\) vitamin \(\mathrm{C}\), what amount (moles) and what number of molecules of vitamin \(\mathrm{C}\) does it contain?

Bauxite, the principal ore used in the production of aluminum, has a molecular formula of \(\mathrm{Al}_{2} \mathrm{O}_{3}=2 \mathrm{H}_{2} \mathrm{O}\). a. What is the molar mass of bauxite? b. What is the mass of aluminum in \(0.58\) mol bauxite? c. How many atoms of aluminum are in \(0.58 \mathrm{~mol}\) bauxite? d. What is the mass of \(2.1 \times 10^{24}\) formula units of bauxite?

Calculate the molar mass of the following substances. a. \(\mathrm{H}\) b. Q N \(\mathrm{N}\) c. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation
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.