A) \[1.8\times {{10}^{5}}\]
B) \[5.8\times {{10}^{5}}\]
C) \[6.8\times {{10}^{5}}\]
D) \[8.8\times {{10}^{5}}\]
Correct Answer: A
Solution :
[a] : In pure semiconductor electron-hole pair \[=7\times {{10}^{15}}{{m}^{-3}}\] Initially total charge carrier\[{{n}_{initial}}={{n}_{h}}+{{n}_{e}}\] \[=14\times {{10}^{15}}\] After doping donor impurity \[{{N}_{D}}=\frac{5\times {{10}^{28}}}{{{10}^{7}}}=5\times {{10}^{21}}\]and\[{{n}_{e}}=\frac{{{N}_{D}}}{2}=2.5\times {{10}^{21}}\] So,\[{{n}_{final}}={{n}_{h}}+{{n}_{e}}\Rightarrow {{n}_{final}}\approx {{n}_{e}}\approx 2.5\times {{10}^{21}}\] \[(\because {{n}_{e}}>>{{n}_{h}})\] \[Factor=\frac{{{n}_{final}}-{{n}_{initial}}}{{{n}_{initial}}}=\frac{2.5\times {{10}^{21}}-14\times {{10}^{15}}}{14\times {{10}^{15}}}\] \[\approx \frac{2.5\times {{10}^{21}}}{14\times {{10}^{15}}}=1.8\times {{10}^{5}}\]
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