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 $7\times {10}^{15}$ holes per cubic metre. Density of silicon is $5\times {10}^{28}$ atoms per cubic metre.

Answer 1

Total number of charge carriers initially

$=2\times 7\times {10}^{15}$

$=14\times {10}^{15}$/Cubic meter

Finally the total number of charge carriers

$=14\times {10}^{17}/m^3$

We know the product of the concentrations of holes and conduction electrons remains almost the same.

Let 'x' be the number of holes.

So, $(7\times {10}^{15})\times (7\times {10}^{15})$

$=x\times (14\times {10}^{17}-x)$

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

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

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

$=\frac{14\times {10}^{17}\pm \sqrt{{10}^{34}+4\times 49\times {10}^{30}}}{2}$

$=\frac{28.0007}{2}\times {10}^{17}=14.00035\times {10}^{17}$

= Increased in number of holes or the number of atoms of Boron added.

Now, $1386.035\times {10}^{15}$ atoms are added per $5\times {10}^{28}$ atoms of Si in 1$m^3.$

$\therefore$ 1 atom of Boron is added per

$\frac{5\times {10}^{28}}{1386.035\times {10}^{15}}$

$=3.607\times {10}^{-3}\times {10}^{13}$

$=3.607\times {10}^{10}$