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,

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

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

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

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

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

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

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 × 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 \[3 . 607 \times  {10}^{- 3}  \times  {10}^{13}    =   3 . 607 \times  {10}^{10}\]