diffusion example problems

# Suppose there is 02 atomic of cu at the surface of an

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Suppose there is 0.2 atomic % of Cu at the surface of an Al sheet and 0.1 atomic % Cflux of Cu atoms from the surface inward at 500ºC? Al is FCC with a lattice parameter of 0.405 nm; Remember FCC means 4 atoms/cell and D 500 = 4 x 10 -10 cm 2 /sec V cell = (0.405 nm) 3 /cell = 6.64 x 10 -23 cm 3 /cell a/o distance 0.2 0.1 1 mm Sketch of physical situation

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Let’s calculate the calculate the concentration change by subtracting the final minus the initial. C = C f - C i C f = 6.64 x 10 -23 cm 3 /cell (0.001)(4 atoms/cell) = 6.02 x 10 19 Cu atoms/cm 3 C i = 6.64 x 10 -23 cm 3 /cell (0.002)(2 atoms/cell) = 1.2 x 10 20 Cu atoms/cm 3
C = C f - C i = 6.02 x 10 19 Cu atoms/cm 3 - 1.2 x 10 20 Cu atoms/cm 3 = - 5.98 x 10 19 cu atoms/cm 3 C/ x = - 5.98 x 10 19 Cu atoms/cm 3 0.1 cm = - 5.98 x 10 20 Cu atoms/cm 4

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J 500 = - (4 x 10 -10 cm 2 /sec) (– 5.98 x 10 20 Cu atoms/cm 4 ) = 2.39 x 10 11 Cu atoms/cm 2 sec
Another diffusion problem. A 4 cm diameter, 0.5 mm thick spherical container made of BCC Fe holds N at 700ºC. The concentration at the inner surface is 0.05 atomic % N and at the outer surface it is 0.002 atomic % N. Calculate the number of grams of N that are lost from the container per hour. 2 cm 0.5 mm 700ºC concentration at the inner surface is 0.05 atomic % N concentration at the outer surface is 0.002 atomic % N N 2 BCC Fe

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a/o N distance 0.05 0.002 0.5 mm Sketch of physical situation Key equations: J = - D c/ x D = D o exp[-Q/RT] T (ºK) = 700 + 273 = 973 D o = 0.0047 cm 2 /sec Q = 18,300 cal/mol See table 5.1
D = 0.0047 cm 2 /sec x -18,300 cal/mol (1.987 cal/molºK) x 973 ºK exp = 3.64 x 10 -7 cm 2 /sec

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