A) \[{{10}^{-5}}\Omega m\]
B) \[{{10}^{-6}}\Omega m\]
C) \[{{10}^{-7}}\Omega m\]
D) \[{{10}^{-8}}\Omega m\]
Correct Answer: D
Solution :
\[\rho =\frac{m}{n{{e}^{2}}\tau }\] \[=1.67\times {{10}^{-8}}\Omega m\]You need to login to perform this action.
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