## Materials Science and Engineering: An Introduction

$N_{v}=1.16\times 10^{23}vacancies.m^{-3}$
given that the equilibrium fraction of lattice sites that are vacant in silver (Ag) at $700^\circ$ is $\frac{N_{v}}{N}=2 \times 10^{6}$ So $N_{v}=2 \times 10^{6}\times N$ Value of N for Ag is $N=\frac{\rho_{Ag}\times N_{A}}{A_{Ag}}$ $N=\frac{10.35\times 6.023\times 10^{23}}{107.87}$ $N=5.78\times 10^{28}atom.m^{-3}$ $N_{v}=2 \times 10^{6}\times5.78\times 10^{28}atom.m^{-3}$ $N_{v}=1.16\times 10^{23}vacancies.m^{-3}$