Tungsten Thorium Diffusion Couple
Consider a diffusion couple between pure tungsten and a tungsten alloy containing 1 at% thorium. After several minutes of exposure at 2000°C, a transition zone of 0.01 cm thickness is established. What is the flux of thorium atoms at this time if diffusion is due to (a) volume diffusion, (b) grain boundary diffusion, and (c) surface diffusion? (See Table 5-2.)
Table 5-2 The effect of the type of diffusion for thorium in tungsten and for selfdiffusion in silve | ||||
Diffusion Type | Diffusion Coefficient (D) | |||
Thorium in Tungsten | Silver in Silver | |||
D_{0} (cm^2/s) | Q (cal/mol) | D_{0} (cm^2/s) | Q (cal/mol) | |
Surface | 0.47 | 66,400 | 0.068 | 8,900 |
Grain boundary | 0.74 | 90,000 | 0.24 | 22,750 |
Volume | 1.00 | 120,000 | 0.99 | 45,700 |
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}