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If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature of 1100˚C for 8 hours, what is the depth below the surface at which the concentration is 1016 atoms/cm3 if the surface concentration is maintained at 1018 atoms/cm3? Use D = 2 x 10-12 cm2/s for aluminum diffusing in silicon at 1100˚C.

User Sage
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Answer:

8.354 nanometers

Step-by-step explanation:

To treat a diffusive process in function of time and distance we need to solve 2nd Ficks Law. This a partial differential equation, with certain condition the solution looks like this:


\frac{C_(s)-C{x}}{C_(s)-C_(o)}=erf(x/2√(D*t))

Where Cs is the concentration in the surface of the solid

Cx is the concentration at certain deep X

Co is the initial concentration of solute in the solid

and erf is the error function

Then we solve right side,


\frac{C_(s)-C{x}}{C_(s)-C_(o)}=(1018atoms/cm3-1016atoms/cm3)/(1018atoms/cm3)=0.001964

And we need to look up the inverse error function of 0.001964 resulting in: 0.00174055

Then we solve for x:


x=0.00174055*2*√(D*t) =0.00174055*2*\sqrt{2*10^(-12)cm^(2)/s*8h*3600s/h}=8.35464*10^(-7)cm

User Janusz Skonieczny
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