A) \[1.6\times {{10}^{-6}}\Omega m\]
B) \[1.6\times {{10}^{-5}}\Omega m\]
C) \[1.6\times {{10}^{-8}}\Omega m\]
D) \[1.6\times {{10}^{-7}}\Omega m\]
Correct Answer: B
Solution :
1. \[\frac{\rho L}{A}=R=\frac{V}{i}=\frac{V}{neA{{V}_{d}}}\] \[\rho =\frac{E}{L}\frac{{}}{ne{{V}_{d}}}=\frac{\left( \frac{5}{0.1} \right)}{\left( 8\times {{10}^{28}} \right)\left( 1.6\times {{10}^{-19}} \right)\left( 2.5\times {{10}^{-4}} \right)}=\left( \frac{50}{32} \right){{10}^{-5}}\]
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