A) \[0.3\,mm/\sec \]
B) \[0.1\,mm/\sec \]
C) \[0.2\,mm/\sec \]
D) \[0.2\,cm/\sec \]
Correct Answer: B
Solution :
Density of \[Cu=9\times {{10}^{3}}kg/{{m}^{3}}\] (mass of 1 m3 of Cu) \[6.0\times {{10}^{23}}\] atoms has a mass \[=\text{ }63\times {{10}^{3}}kg\] \ Number of electrons per \[{{m}^{3}}\]are \[=\frac{6.0\times {{10}^{23}}}{63\times {{10}^{-3}}}\times 9\times {{10}^{3}}=8.5\times {{10}^{28}}\] Now drift velocity \[={{v}_{d}}=\frac{i}{neA}\] \[=\frac{1.1}{8.5\times {{10}^{28}}\times 1.6\times {{10}^{-19}}\times \pi \times {{(0.5\times {{10}^{-3}})}^{2}}}\] \[=0.1\times {{10}^{-3}}m\,/sec\]
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