diffusion example problems

# C 005 0002 atomic 0048 atomic n but we need it in

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c = 0.05 – 0.002 atomic % = 0.048 atomic % N But, we need it in atoms/cm3: Lattice paramater of BCC Fe = 0.2866 nm For BCC there are 2 atoms per cell Therefore we get 0.85 x 10 23 Fe atoms/cm 3 Now we have already calculated the fractional change in composition of N: 0.00048 x 0.85 x 10 23 = 4.08 x 10 19 N atoms/cm 3 = c

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C/ x = 4.08 x 10 19 N atoms/cm 3 0.05 cm = 8.16 x 10 20 N atoms/cm 4 J 973 = - (3.64 x 10 -7 cm 2 /sec) (8.16 x 10 20 N atoms/cm 4 ) = -2.97 x 10 14 N atoms/cm 2 sec
Now we have to account for the flux through the total surface area. Area of surface = 4 π r 2 = 50.26 cm 2 Time = 3600 sec So, in one hour (multiply flux by area by time): = 5.37 x 10 19 N atoms 1 mole of N = 6.02 x 10 23 N = 14 g N 5.37 x 1019 N atoms 6.02 x 1023 N x 14 g = 1.25 x 10 -3 g
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• Spring '10
• Barnard

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