Question 5.6: Tungsten Thorium Diffusion Couple Consider a diffusion coupl......

The lattice parameter of BCC tungsten is 3.165 Å. Thus, the number of tungsten atoms/cm³ is
$\frac{W \ atoms}{cm³} = \frac{2 \ atoms/cell}{(3.165 × 10^{-8} \ cm)³/cell} = 6.3 × 10^{22}$
In the tungsten-1 at% thorium alloy, the number of thorium atoms is
$c_{Th} = (0.01)(6.3  × 10^{22}) = 6.3 × 10^{20} \ \frac{Th \ atoms}{cm^3}$
In the pure tungsten, the number of thorium atoms is zero. Thus, the concentration gradient is
$\frac{Δc}{Δx} = \frac{0 \ – \ 6.3 × 10^{20} \ \frac{Th \ atoms}{cm^3}}{0.01 \ cm}=-6.3 × 10^{22} \ \frac{Th \ atoms}{cm^3  ⋅  cm}$
a. Volume diffusion
$D = 1.0 \ \frac{cm^2}{s} \exp [ \frac{-120,000 \ \frac{cal}{mol}}{(1.987 \ \frac{cal}{mol  ⋅  K})(2273 \ K)}] = 2.89 × 10^{-12}  cm^2 /s$
$J = -D \frac{Δc}{Δx}=-(2.89 × 10^{-12} \ \frac{cm^2}{s})(-6.3 × 10^{22} \ \frac{Th \ atoms}{cm^3  ⋅  cm})$
$= 1.82 × 10^{11} \ \frac{Th \ atoms}{cm^2  ⋅  s}$
b. Grain boundary diffusion
$D = 0.74 \ \frac{cm^2}{s} \exp [ \frac{-90,000 \ \frac{cal}{mol}}{(1.987 \ \frac{cal}{mol  ⋅  K})(2273 \ K)}] = 1.64 × 10^{-9}  cm^2 /s$
$J =-(1.64 × 10^{-9} \ \frac{cm^2}{s})(-6.3 × 10^{22} \ \frac{Th \ atoms}{cm^3  ⋅  cm})=1.03 × 10^{14} \ \frac{Th \ atoms}{cm^2  ⋅  s}$
c. Surface diffusion
$D = 0.47 \ \frac{cm^2}{s} \exp [ \frac{-66,400 \ \frac{cal}{mol}}{(1.987 \ \frac{cal}{mol  ⋅  K})(2273 \ K)}] = 1.94 × 10^{-7}  cm^2 /s$
$J =-(1.94 × 10^{-7} \ \frac{cm^2}{s})(-6.3 × 10^{22} \ \frac{Th \ atoms}{cm^3  ⋅  cm})=1.22 × 10^{16} \ \frac{Th \ atoms}{cm^2  ⋅  s}$