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 meter. Density of silicon is $5\times {10}^{28}$ atoms per cubic meter.

• A. $1.8\times {10}^5$
• B. $3.571\times {10}^{10}$
• C. $2.580\times {10}^{10}$
• D. $4.576\times {10}^{10}$

Answer 1

The correct option is B $3.571\times {10}^{10}$

In a pure silicon, $n_e=n_h$,

$\therefore$ Total number of charge carries initially
$=2\times 7\times {10}^{15}=14\times {10}^{15}{\text{m}}^{-3}$

As the conductivity should be increased by a factor of $100$,

Final total number of charge carries
$=14\times {10}^{15}\times 100=14\times {10}^{17}{\text{m}}^{-3}$

Let $x$ be the hole concentration after doping, which is also equal to boron concentration.

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

So, $(7\times {10}^{15})\times (7\times {10}^{15})=x\times (14\times {10}^{17}-x)$

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

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

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

$\Rightarrow x\approx 14\times {10}^{17}{\text{m}}^{-3}$

Thus, the proportion of boron impurity added is given by,

$\frac{5\times {10}^{28}}{14\times {10}^{17}}=3.571\times {10}^{10}$

Therefore, $1$ boron atom is added for every $3.571\times {10}^{10}$ atoms of silicon.

Hence, option $\left(B\right)$ is correct.