Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 29

Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at its melt- ing temperature (645C). Assume an energy for defect formation of 1.86 eV.

Short Answer

Expert verified
Answer: The fraction of lattice sites with Schottky defects in cesium chloride at its melting temperature (645°C) is approximately 0.095 or 9.5%.

Step by step solution

01

Convert the given temperature in Celsius to Kelvin

To do calculations with temperature involving Boltzmann's statistics, we need to convert the given temperature from Celsius to Kelvin. To convert 645°C to Kelvin, add 273.15: Temperature (T) = 645 + 273.15 K T = 918.15 K
02

Convert the energy for defect formation to Joules

We have the energy for defect formation given in eV, but in order to use it in Boltzmann's constant calculations, we need to convert it to Joules. To convert 1.86 eV to Joules, multiply it by the conversion factor (1.60218×10^{-19} J/eV): Energy (E) = 1.86 eV × 1.60218×10^{-19} J/eV E = 2.979×10^{-19} J
03

Use Boltzmann's statistics to calculate the fraction of lattice sites with Schottky defects

According to Boltzmann's statistics, the fraction of lattice sites with Schottky defects (n) can be calculated using the following formula: n = exp(-E/(kT)) Where E is the energy for defect formation (in Joules), k is Boltzmann's constant (8.617×10^{-5} eV/K or 1.38×10^{-23} J/K), and T is the absolute temperature (in Kelvin). We can now plug in the known values (E – in Joules, k – in J/K and T – in K) into the formula: n = exp(-2.979×10^{-19} J/(1.38×10^{-23} J/K × 918.15 K)) Calculating the expression gives: n = exp(-2.979×10^{-19} J/(1.266×10^{-20} J)) n = exp(-2.356)
04

Calculate the numerical value for the fraction of lattice sites with Schottky defects

Now we can find the numerical value for n: n = exp(-2.356) n ≈ 0.095 Therefore, the fraction of lattice sites with Schottky defects in cesium chloride at its melting temperature (645°C) is approximately 0.095 or 9.5%.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

One crystalline form of silica (SiO2) has a cu- bic unit cell, and from x-ray diffraction data it is known that the cell edge length is 0.700 nm. If the measured density is 2.32 g/cm3 , how many Si4 and O2 ions are there per unit cell?

Express cach of the following expressions using a single positive index: (a) \(x^{4} x^{7}\) (b) \(x^{2}(-x)\) (c) \(\frac{x^{2}}{x}\). (d) \(\frac{x^{-2}}{x^{-1}}\) (c) \(\left(x^{-2}\right)^{4}\) (f) \(\left(x^{-25} x^{-3.5}\right)^{2}\)

(a) Suppose that CaO is added as an impurity to Li2O. If the Ca2- substitutes for Li, what kind of vacancies would you expect to form? How many of these vacancies are created for every Ca2- added? (b) Suppose that CaO is added as an impurity to CaCl2. If the O2substitutes for Cl, what kind of vacancies would you expect to form? How many of these vacancies are created for every O2 added?

For a ceramic compound, what are the two char- acteristics of the component ions that determine the crystal structure?

A circular specimen of \(\mathrm{MgO}\) is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is \(5560 \mathrm{~N}\left(1250 \mathrm{lb}_{\mathrm{f}}\right)\), the flexural strength is \(105 \mathrm{MPa}(15,000 \mathrm{psi})\), and the separation between load points is \(45 \mathrm{~mm}\) (1.75 in.).
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Kinematics Physics

Read Explanation

Famous Physicists

Read Explanation

Fundamentals of Physics

Read Explanation

Turning Points in Physics

Read Explanation

Energy Physics

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.

Sign-up for free