Question

Estimate the proportion of boron impurity which will increase the conductivity of a pure silicon sample by a factor of 100. Assume that each boron atom creates a hole and the concentration of holes in pure silicon at the same temperature is 7 × 10¹⁵ holes per cubic metre. Density of silicon 5 × 10²⁸ atoms per cubic metre.

Answer 1

Initially, the total number of charge carriers per cubic metre is given by
n_i = 2 × 7 × 10¹⁵
$\Rightarrow$n_i = 14 × 10¹⁵

Finally, the total number of charge carriers per cubic metre is given by
n_f = 14 × 10¹⁷/m³

We know that the product of the concentrations of holes and conduction electrons remains almost the same.
Let x be the number of holes.
Thus,
$$\begin{gathered} (7\times {10}^{15})\times (7\times {10}^{15})=x\times (14\times {10}^{17}-x) \\ \Rightarrow 14x\times {10}^{17}-x^2=49\times {10}^{30} \\ \Rightarrow x^2-14x\times {10}^{17}-49\times {10}^{30}=0 \\ \Rightarrow x=\frac{14\times {10}^{17}\pm \sqrt{(14{)}^2\times {10}^{34}+4\times 49\times {10}^{30}}}{2} \\ \Rightarrow x=\frac{14\times {10}^{17}\pm \sqrt{(14{)}^2\times {10}^{34}+4\times 49\times {10}^{30}}}{2} \\ \Rightarrow x=\frac{28.0007}{2}\times {10}^{17}=14.00035\times {10}^{17} \end{gathered}$$
This is equal to the increased number of holes or the number of atoms of boron added.
Number of atoms of boron added = $(14.00035\times {10}^{17}-7\times {10}^{15})=1386.035\times {10}^{15}$
Now, 1386.035 × 10¹⁵ atoms are added per 5 × 10²⁸ atoms of Si in 1 m³.
Therefore, 1 atom of boron is added per $\frac{5\times {10}^{28}}{1386.035\times {10}^{15}}$ atoms of Si in 1 m³.
Proportion of boron impurity is $3.607\times {10}^{-3}\times {10}^{13}=3.607\times {10}^{10}$.