Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 4 - Imperfections in Solids - Questions and Problems - Page 135: 4.5

Answer

$Q_{v} = 1.40 eV/atom $

Work Step by Step

Given: Nickel (Ni) where the equilibrium number of vacancies at 1123 K is $4.7 \times 10^{22} m^{-3}$ Atomic weight - 58.69 g/mol density - 8.80 g/$cm^{3}$ Required: energy for vacancy formation Solution: Using Equation 4.2: $N = \frac{N_{A}p_{Ni}}{A_{Ni}} = {\frac{(6.022 \times 10^{23} atoms/mol)(8.80 g/cm^{3})}{58.69 g/mol}} = 9.03 \times 10^{22} atoms/cm^{3} = 9.03 \times 10^{28} atoms/m^{3} $ Using Equation 4.1 and taking its natural logarithm: $ ln N_{v} = ln N (-\frac{Q_{v}}{kT}) $ $Q_{v} = -kT ln(\frac{N_{v}}{N}) = -(8.62 \times 10^{-5} eV/atom-K)(1123 K)ln [\frac{4.7 \times 10^{22} m^{-3}}{9.03 \times 10^{28}}] = 1.40 eV/atom $
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