The Gaussian error function can be approximated for small values of x as erf(x)≈x. Suppose a silicon wafer is exposed to aluminum vapor at 1300∘ C. At what distance (in centimeters) will the concentration of Al be 35% of the surface concentration in 164 hours? Assume the diffusivity of Al in Si is 10−10 cm2/s at 1300∘ C.
Give your answer in scientific notation, up to two decimal points in unit centimeters.
To solve this problem, we can use Fick's second law of diffusion, which relates the concentration profile with time and distance:
∂C/∂t = D * ∂²C/∂x²
Where:
- C is the concentration of aluminum in the silicon wafer
- t is time
- D is the diffusivity of aluminum in silicon
- x is the distance from the surface of the wafer
In this case, we are given that the concentration at the surface of the wafer is 100% (or 1), and we need to find the distance at which the concentration is 35% (or 0.35) after 164 hours.
The Gaussian error function comes into play when we consider that the concentration profile is not linear but described by the error function. However, for small values of x, erf(x) can be approximated as x.
To find the distance x, we need to rearrange Fick's second law in terms of x:
∂²C/∂x² = (1 / (D * ∂t)) * ∂C/∂t
Since we are given the diffusivity D and the time ∂t, we can integrate this equation using the initial condition C(x=0) = 1 and the boundary condition C(x=infinity) = 0.35.
∫[1 to 0.35] dx / (1 / (D * ∂t)) = -∫[0 to x] ∂C/∂t dx
∫[1 to 0.35] dx * (D * ∂t) = -∫[0 to x] ∂C/∂t dx
(D * ∂t) * (0.35 - 1) = ∫[0 to x] ∂C/∂t dx
(D * ∂t) * (-0.65) = ∫[0 to x] ∂C/∂t dx
0.65 * D * ∂t = C(x=0) - C(x)
Since C(x) = 0.35, we can substitute these values into the equation:
0.65 * D * ∂t = 1 - 0.35
Simplifying:
0.65 * D * ∂t = 0.65
Now we can solve for x by substituting the values for D and ∂t given in the problem and rearranging the equation:
x = 0.65 * D * ∂t / 0.65
x = (0.65 * 10^(-10) cm²/s) * (164 hours) / 0.65
x ≈ 10^(-10) cm²/s * 164 hours
x ≈ 1.64 * 10^(-10) cm²·h/s
To convert the units from cm²·h/s to cm, we need to multiply by the square root of time since the diffusion is a function of the square root of time:
x ≈ (1.64 * 10^(-10) cm²·h/s) * sqrt(164 hours)
Now we can calculate the numerical value:
x ≈ 1.64 * 10^(-10) cm²·h/s * 12.806
x ≈ 2.09984 * 10^(-9) cm²·h/s
Therefore, the distance x at which the concentration of Al will be 35% of the surface concentration in 164 hours is approximately 2.10 * 10^(-9) cm.