A) \[2.8\times {{10}^{5}}\]
B) \[3.1\times {{10}^{2}}\]
C) \[4.2\times {{10}^{5}}\]
D) \[1.8\times {{10}^{5}}\]
Correct Answer: D
Solution :
[d] In pure semiconductor electron-hole pair \[=7\times {{10}^{15}}/{{m}^{3}}\] \[{{n}_{initial}}={{n}_{h}}+{{n}_{e}}=14\times {{10}^{15}}\] 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 \[=\frac{{{n}_{final}}-{{n}_{initial}}}{{{n}_{initial}}}\] \[=\frac{2.5\times {{10}^{21}}-14\times {{10}^{15}}}{14\times {{10}^{15}}}\approx \frac{2.5\times {{10}^{21}}}{14\times {{10}^{15}}}=1.8\times {{10}^{5}}\]You need to login to perform this action.
You will be redirected in
3 sec