A) n-type and its resistivity is 0.34 \[\Omega \]-m
B) p-type and its resistivity is 0.034 \[\Omega \]-m
C) n-type and its resistivity is 0.034 \[\Omega \]-m
D) p-type and its resistivity is 3.4 \[\Omega \]-m
Correct Answer: A
Solution :
Key Idea: In n-type semiconductors electrons are majority carriers and holes are minority carriers. \[{{\text{n}}_{e}}=8\times {{10}^{8}}/{{m}^{3}},\] \[{{n}_{h}}=5\times {{10}^{18}}/{{m}^{3}},\] \[{{\mu }_{e}}=2.3{{m}^{2}}/V-s.\] \[{{\mu }_{h}}=0.01{{m}^{2}}/V-s.\] As \[{{n}_{e}}>{{n}_{h}},\]so semiconductor is n-type. Also conductivity \[(\sigma )=\frac{1}{\operatorname{Re}sistivity(\rho )}\] \[=e({{n}_{e}}{{\mu }_{e}}+{{n}_{h}}{{\mu }_{h}})\] \[\Rightarrow \] \[\frac{1}{\rho }=1.6\times {{10}^{-19}}\] \[[8\times {{10}^{18}}\times 2.3\times 5\times {{10}^{8}}\times 0.01]\] \[\Rightarrow \] \[=0.34\,\Omega -\text{m}\]
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