A) 0.3 mm/s
B) 0.5 mm/s
C) 0.1 mm/s
D) 0.2 mm/s
Correct Answer: C
Solution :
Number of atoms in 63 g of copper = Avogadro number \[=6\times {{10}^{23}}\] Volume of 63 g of copper \[=\frac{63}{density}=\frac{63}{9}\] \[=7\text{ }c{{m}^{3}}\] \[n=\frac{6.02\times {{10}^{23}}}{7}per\,c{{m}^{3}}\] \[=\frac{6.02\times {{10}^{29}}}{7}per\,c{{m}^{3}}\] Area \[A=\pi {{r}^{2}}\] \[=\pi {{(0.5\times {{10}^{-3}})}^{2}}{{m}^{2}}\] Drift velocity is\[v=\frac{1}{neA}\] \[=0.1\times {{10}^{-3}}m/s\] \[=0.1\text{ }mm/s\]
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