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 × 10^{15} holes per cubic metre. Density of silicon 5 × 10^{28} atoms per cubic metre.

#### Solution

Initially, the total number of charge carriers per cubic metre is given by

n_{i} = 2 × 7 × 10^{15}

\[\Rightarrow\] n_{i} = 14 × 10^{15}

Finally, the total number of charge carriers per cubic metre is given by

n_{f}_{ }= 14 × 10^{17}/m^{3}

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,

\[(7 \times {10}^{15} ) \times (7 \times {10}^{15} ) = x \times (14 \times {10}^{17} - x)\]

\[ \Rightarrow 14x \times {10}^{17} - x^2 = 49 \times {10}^{30} \]

\[ \Rightarrow x^2 - 14x \times {10}^{17} - 49 \times {10}^{30} = 0\]

\[ \Rightarrow x = \frac{14 \times {10}^{17} \pm \sqrt{(14 )^2 \times {10}^{34} + 4 \times 49 \times {10}^{30}}}{2}\]

\[ \Rightarrow x = \frac{14 \times {10}^{17} \pm \sqrt{(14 )^2 \times {10}^{34} + 4 \times 49 \times {10}^{30}}}{2}\]

\[ \Rightarrow x = \frac{28 . 0007}{2} \times {10}^{17} = 14 . 00035 \times {10}^{17}\]

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 × 10^{15} atoms are added per 5 × 10^{28} atoms of Si in 1 m^{3}.

Therefore, 1 atom of boron is added per \[\frac{5 \times {10}^{28}}{1386 . 035 \times {10}^{15}}\] atoms of Si in 1 m^{3}.

Proportion of boron impurity is \[3 . 607 \times {10}^{- 3} \times {10}^{13} = 3 . 607 \times {10}^{10}\]