Question

A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contains 1 Sb atom per 108 Si atoms and the other surface contains 500 Sb atoms per 108 Si atoms. The lattice parameter for Si is 5.4307 Å (Appendix A). Calculate the concentration gradient in (a) at% Sb per cm; and (b) Sb atoms cm ? cm3 .

Answer 1

Answer:

- 0.0249% Sb/cm

$-1.2465 * 10^9 \frac{atoms}{cm^3.cm}$

Explanation:

Given that:

One surface contains 1 Sb atom per  10⁸  Si atoms and the other surface contains 500 Sb atoms per  10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb  per cm can be calculated as follows:

The difference in concentration = $\delta_c$

The distance $\delta_x$ = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing  1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per   10⁸ silicon atoms can be calculated as follows:

$\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%$

= - 0.0249% Sb/cm

b) The concentration $(c_1)$ of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

$c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}$

Lattice parameter = 5.4307 Å;  To cm ; we have

= $5.4307A^0* \frac{10^{-8}cm}{ A^0}$

$c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}$

= $0.00499*10^{17}atoms/cm^3$

The concentration $(c_2)$ of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

$c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}$

   =  $\frac{4*10^{-3}}{1.601*10^{-22}}$

   = $2.4938*10^{17}atoms/cm^3$

Finally, to calculate the concentration gradient

$(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}$

$(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}$

$= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}$