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3 years ago

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A diffusion couple, made by welding a thin onecentimeter square slab of pure metal A to a similar slab of pure metal B, was give

n a diffusion anneal at an elevated temperature and then cooled to room temperature. On chemically analyzing successive layers of the specimen, cut parallel to the weld interface, it was observed that, at one position, over a distance of 5000 nm, the atom fraction of metal A, NA, changed from 0.30 to 0.35. Assume that the number of atoms per m3 of both pure metals is 9 x 10^28. First determine the concentration gradient dnA/dx. Then if the diffusion coefficient, at the point in question and annealing temperature, was 2 10^-14 m^2/s.
Required:
Determine the number of A atoms per second that would pass through this cross-section at the annealing temperature.

1 answer:

bezimeni [28]3 years ago

7 0

Answer:

The value is $H = 18*10^{2} \ Atom / sec$

Explanation:

From the question we are told that

The atom fraction of metal A at point G is $A = 0.30 \ m$

The atom fraction of metal A at a distance 5000nm from G is $A_2 = 0.35$

The number of atoms per $m^3$ is $N_h = 9 * 10^{28}$

The diffusion coefficient is $D = 2* 10^{-14 } m^2/s$

Generally of the concentration of atoms of metal A at G is

$N_A = A * N_h$

=> $N_A = 0.3 * 9 * 10^{28}$

=> $N_A = 2.7 * 10^{28} 2.7 atoms/m^3$

Generally of the concentration of atoms of metal A at a distance 5000nm from G is

$D = 0.35 *9 * 10^{28}$

=> $D = 3.15 * 10^{28} \ atoms / m^3$

The concentration gradient is mathematically represented as

$\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{5000nm - 0 }$

=> $\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{[5000 *10^{-9}] - 0 }$

=> $\frac{dN_A}{dx} = 9 * 10^{20} / m^4$

Generally the flux of the atoms per unit area according to Fick's Law is mathematically represented as

$J = -D* \frac{d N_A}{dx}$

=> $J = -2* 10^{-14 * 9 * 10^{20}$

=> $J = 18*10^{6}\ atoms\ crossing\ /m^2 s$

Generally if the cross-section area is $a = 1 cm^2 = 10^{-4} \ m^2$

Generally the number of atom crossing the above area per second is mathematically is

$H = 18*10^{6} * 10^{-4}$

=> $H = 18*10^{2} \ Atom / sec$

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