A) n-type and its resistivity is 0.034 \[\Omega \text{-}m\]
B) n-type and its resistivity is 0.34 \[\Omega \text{-}m\]
C) p-type and its resistivity is 0.034\[\Omega \text{-}m\]
D) p-type and its resistivity is 0.34\[\Omega \text{-}m\]
Correct Answer: B
Solution :
Given, \[{{n}_{e}}=8\times {{10}^{18}}{{m}^{-3}}\] and \[{{n}_{h}}=5\times {{10}^{18}}{{m}^{-3}}\] and \[{{\mu }_{e}}=2.3\,{{m}^{2}}\text{/}V\text{-}s\] and \[{{\mu }_{h}}=0.01\,\,{{m}^{2}}{{m}^{2}}\text{/}V\text{-}s\] \[\because \,{{n}_{e}}>{{n}_{h}},\] so semiconductor is n-type. Now, conductivity\[\sigma =\frac{\text{1}}{\text{Resistivity}\,\text{( }\!\!\rho\!\!\text{ )}}\] \[=e\,({{n}_{e}}{{\mu }_{e}}+{{n}_{h}}{{\mu }_{h}})\] \[\Rightarrow \frac{1}{\rho }=1.6\times {{10}^{-19}}[8\times {{10}^{18}}\times 2.3+5\times {{10}^{18}}\times 0.01]\] \[\Rightarrow \] \[\rho =0.34\,\,\Omega \text{-}m\]You need to login to perform this action.
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