Answer
$N_{v} = 3.52 \times 10^{24} m^{3}$
Work Step by Step
Given:
Gold (Au) at 900°C (1173 K)
energy for vacancy formation is 0.98 eV/atom
density is $18.63 g/cm^{3}$
(196.9 g/mol)
Required:
number of vacancies per cubic meter
Solution:
Using Equations 4.1 and 4.2,
$ N_{v} = N exp (-\frac{Q_{v}}{kT}) = \frac{N_{A}p_{Au}}{A_{Au}} exp(-\frac{Q_{v}}{kT}) $
Substituting the given values,
$N_{v} = {\frac{(6.022 \times 10^{23} atoms/mol)(18.63 g/cm^{3})}{196.9 g/mol}} exp [-\frac{0.98 eV/atom}{(8.62 \times 10^{-5} eV/atom-K)(1173 K)}] = 3.52 \times 10^{18} cm^{3}= 3.52 \times 10^{24} m^{3}$
