Chemistry, 14.02.2020 00:04 mikego5

If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature of 1100°C for 206 minutes, what is the depth below the surface at which the concentration is 1016 atoms/cm3? if the surface concentration is 1018 atoms/cm^3?

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the depth below the silicon surface at which the aluminum concentration would be $10^{16}$ atoms/ $cm^{3}$ is $5.723*10^{-6}$

Explanation:

The concentrations of a diffusing element in a silicon wafer at any particular position below the surface and the time it takes to reach that concentration has a relationship given by Fick's second law

$\frac{C_{s}-C_{s}}{C_{x}-C_{0}}=erf(\frac{x}{2\sqrt{Dt} } ) ...(1)$

Where

The concentration of the carbon on the gear surface is $C_{s}$

The initial uniform (normal) concentration of carbon in the gear is $C_{0}$

The concentration of carbon at a distance x below the gear surface after time t is $C_{x}$

Distance below the gear surface is x

Diffusivity of carbon at a given temperature is D

Time is t

erf is a mathematical function called error function

The impurity diffusion is a method in which an impurity is diffused into a silicon wafer at a very high temperature.The given values are

$C_{s}$ = $10^{18}$ atoms/ $cm^{3}$

$C_{0}$ = 0 atoms/ $cm^{3}$

$C_{x}$ = $10^{16}$ atoms/ $cm^{3}$

t = 206 minutes = 12360 s [1 minutes= 60 seconds ]

For an aluminum diffusing in silicon at 11100°C Diffusivity is $=2*10^{-12} cm^{2}/s$

Substituting these values we have

$\frac{10^{18}-10^{16}}{10^{18}-0}[\frac{x}{2\sqrt{2*10^{-12}cm^{2}/s[\frac{10^{-4}m^{2}}{1cm^{2}} ]*12360s} } ]$

$0.99 =erf[\frac{x}{3.145*10^{-6}} ]$

Let assume $R = [\frac{x}{3.145*10^{-6}} ]$

Hence

erf(R) = 0.99

$R = [\frac{x}{3.145*10^{-6}} ] ...(2)$

To determine for what number the error function is 0.99

The table on the first uploaded image is a table that error function for some number

Where z is the same thing as R

calculating the R (i.e z) value whose error function is 0.99 , using the interpolation method we have

$\frac{R-1.8}{1.9-1.8}=\frac{0.9900-0.9891}{0.9928-0.9891}$

$\frac{R-1.8}{0.1}=0.2432$

R = 1.82

Substituting the value of R in the equation 2 we have

$1.82 = [\frac{x}{3.145*10^{-6}} ]$

$x = 1.82*3.145*10^{-6}$

x = 5.723*10^-6m

So therefore the value of x is $5.723*10^{-6}$

Hence the depth below the silicon surface at which the aluminum concentration would be $10^{16}$ atoms/ $cm^{3}$ is $5.723*10^{-6}$

If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperatu

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1.solid

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Answer from: Quest

You know how the first one if filled up like the objects do the same but with the other categories

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If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperatu...