Answer:
The drift velocity of the electrons in this case is c) 5.52 x 10-4 m/s
Explanation:
Hi
First of all, we need to find the volume occupied by 63.3g of copper, so [tex]V=\frac{m}{\rho}= \frac{0.0633Kg}{8,900kg/m^{3}} =7.11 \times 10^{-6} m^{3}[/tex], then if we assume each atom of copper contributes with one free electron to the material body [tex]n=\frac{N_{A}}{V}=\frac{6.02 \times 10^{23} electrons}{7.11 \times 10^{-6} m^{3}} =8.46 \times 10^{28} electrons/m^{3}[/tex].
Finally, we apply [tex]v_{d}=\frac{I}{nqA}[/tex], where A is area so, [tex]A=\pi (\frac{d}{2})^{2}= \pi (\frac{0.0016m}{2})^{2}=2 \times 10^{-6} m^{2}[/tex], thus [tex]v_{d}=\frac{I}{nqA}=\frac{15C/s}{(8.46 \times 10^{28} m^{-3})(1.60 \times 10^{-19} C)(2 \times 10^{-6} m^{2})}=5.50 \times 10^{-4} m/s[/tex].
As we can see c. 5.52 x 10-4 m/s is the nearest one.
