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प्रश्न
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.
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उत्तर
Initially, the total number of charge carriers per cubic metre is given by
ni = 2 × 7 × 1015
\[\Rightarrow\] ni = 14 × 1015
Finally, the total number of charge carriers per cubic metre is given by
nf = 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 \[3 . 607 \times {10}^{- 3} \times {10}^{13} = 3 . 607 \times {10}^{10}\]
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