For copper,$\rho$ = 8.93 g/cm³ and M = 63.5 g/mol.Assuming one free electron per copper atom,what is the drift velocity of electrons in a copper wire of radius 0.625 mm carrying a current of 3 A? A)3.54 $\times$ 10^-4 m/s B)1.80 $\times$ 10^-4 m/s C)4.26 $\times$ 10^-4 m/s D)7.52 $\times$ 10^-4 m/s E)2.46 $\times$ 10^-4 m/s

Question

Quizplus · Accepted Answer

First, we need to find the number density of copper atoms in the wire:

$n = \frac{ho}{\mu m} = \frac{8.93\,g/cm^3}{63.5\,g/mol \cdot \frac{\pi}{6} (0.625\,mm)^3} = 8.46 \cdot 10^{28}/m^3$

Next, we can use the equation for drift velocity:

$v_d = \frac{I}{neA}$

where $I$ is the current, $n$ is the number density of charge carriers (in this case, the number density of copper atoms), $e$ is the elementary charge, $A$ is the cross-sectional area of the wire, and $v_d$ is the drift velocity.

Substituting in the values, we get:

$v_d = \frac{3A}{neA} \cdot \frac{1}{ho} \cdot \frac{M}{1 	ext{ mol}} = \frac{3}{neho} M$

Plugging in the values for $n$, $ho$, and $M$, we get:

$v_d = \frac{3}{8.46 \cdot 10^{28}/m^3 \cdot 8.93\,g/cm^3 \cdot 63.5\,g/mol} \cdot 6.022 \cdot 10^{23} \cdot 63.5\,g/mol \cdot \frac{1\,m}{100\,cm} = 1.80 \cdot 10^{-4}\,m/s$

So the answer is B.

For Copper ρ\rhoρ = 893 G/cm3 and M = 63 ×\times× 10-4 M/s B)1

For Copper $\rho$ = 893 G/cm³ and M = 63 $\times$ 10^-4 M/s
B)1