A) 21/6
B) 5
C) 3/8
D) 2
Correct Answer: D
Solution :
When some potential difference is applied across a piece of intrinsic semiconductor current flows in it due to both electron and holes ie, \[I={{I}_{e}}+{{I}_{h}}\] Moreover, \[{{I}_{e}}=eA{{n}_{e}}{{v}_{e}}\] ... (i) \[{{I}_{h}}=eA{{n}_{h}}{{v}_{h}}\] ...(ii) Now, dividing Eqs. (i) and (ii), we get \[\frac{{{I}_{e}}}{{{I}_{h}}}=\frac{{{n}_{e}}}{{{n}_{h}}}\frac{{{v}_{e}}}{{{v}_{h}}}\] Or \[\frac{(4/5)I}{(1/5)I}=\frac{{{n}_{e}}}{{{n}_{h}}}\times \frac{2{{v}_{h}}}{{{v}_{h}}}\] \[\left[ \begin{matrix} Given & {{v}_{e}}=2{{v}_{h}} \\ {} & {{I}_{e}}=\frac{4}{5}I \\ and & {{I}_{h}}=\frac{1}{5}I \\ \end{matrix} \right]\] \[\Rightarrow \] \[\frac{{{n}_{e}}}{{{n}_{h}}}=2\]You need to login to perform this action.
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