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

Explanation:

At first, the total amount of charge carriers per cubic metre is shown as ni = $n_{1}=2 \times 7 \times 10^{15}$

$\Rightarrow n_{1}=14 \times 10^{15}$

Lastly, the total amount of charge carriers per cubic metre is shown as $n_{\mathrm{f}}=14 \times 10^{17} / \mathrm{m}^{2}$

It is known that the product of holes concentrations and the electrons conduction stays the same.

Let x represent the number of holes.

So,$(7 \times 1015) \times(7 \times 1015)=(14 \times 1017-x) \times x$

$\Rightarrow \mathrm{x}=1017 \times 28.00072=14.00035 \times 1017$

The number of holes is equal to the increased holes number or the boron atoms number added.

Added number of boron atoms = $(14.00035 \times 1015 \times 1017-7)=10^{15} \times 1386.035$

Now, per $5 \times 10^{2 a}$ atoms are added in 1 m3.

Hence, 1 boron atom is added per $5 \times 10281386.035 \times 1015$ Si atoms in 1 m3.

Boron impurity’s proportion is $3.607 \times 10-3 \times 1013=3.607 \times 1010$.