A) \[2.8 \times \,1{{0}^{5}}\]
B) \[3.1\,\,\times \,\,{{10}^{2}}\]
C) \[4.2\,\,\times \,\,{{10}^{5}}\]
D) \[1.8\,\,\times \,\,{{10}^{5}}\]
Correct Answer: D
Solution :
In pure semiconductor electron-hole pair \[=\,\,\,7\times {{10}^{15}}/{{m}^{3}}\] \[{{n}_{initial}}=\,\,{{n}_{h}}+\,\,{{n}_{e}}=\,\,14\times 1{{0}^{15}}\] after doping donor Impurity \[{{\operatorname{N}}_{D}}=\,\,\frac{5\times {{10}^{28}}}{{{10}^{7}}}\,\,=\,\,5\times {{10}^{21}}\,\,and\,\,{{n}_{e}}=\,\,\frac{{{N}_{D}}}{2}\,\,=\,\,2.5\times 1{{0}^{21}}\]So, \[{{n}_{final}}={{n}_{h}}+{{n}_{e}}\] \[\Rightarrow \,\,\,\,\operatorname{n}final\approx {{n}_{e}}\approx \,\,2.5\times 1{{0}^{21}}\,\,(\because \,\,{{n}_{e}}>>{{n}_{h}})\] \[Factor=\frac{{{n}_{final}}-{{n}_{initial}}}{{{n}_{initial}}}\] \[=\,\,\frac{2.5\times {{10}^{21}}-14\times {{10}^{15}}}{14\times {{10}^{15}}}\approx \frac{2.5\times {{10}^{21}}}{14\times {{10}^{15}}}=1.8\times {{10}^{5}}\]
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