Answer
$D= 4.80 \times 10^{-13} m^{2}/s$
Work Step by Step
Given:
activation energy for copper in silver - 193,000 J/mol
D at 1000 K - $1.0 \times 10^{-14} m^{2}/s$
Required:
diffusion coefficient at 1200 K
Solution:
Using Equation 5.8, solve for $D_{0}$:
$D_{0}= D exp (\frac{Q_{d}}{RT}) = (1.0 \times 10^{-14} m^{2}/s)exp[\frac{(193,000 J/mol)}{(8.31 J/mol.K)(1000 K)}] = 1.22 \times 10^{-4} m^{2}/s$
Using Equation 5.8 to solve for $D$:
$D= D exp (\frac{Q_{d}}{RT}) = (1.22 \times 10^{-4} m^{2}/s)exp[-\frac{(193,000 J/mol)}{(8.31 J/mol.K)(1200 K)}] \\=4.80 \times 10^{-13} m^{2}/s$
