Copper has one conduction electron per atom. Its density is `8.89 g//cm^3` and its atomic mass. `63.54 g//mol`. If a copper wire of diameter `1.0 mm` carries a current of `2.0 A`, what is the drift speed of the electrons in the wire?

Explanation

To find the drift speed of electrons in a copper wire, we can use the formula that relates current (I), drift speed (v_d), the number of charge carriers per unit volume (n), and the cross-sectional area (A) of the wire:

\[ I = n \cdot A \cdot e \cdot v_d \]

Where:
- \( I \) = current (in amperes)
- \( n \) = number of free electrons per unit volume (in m³)
- \( A \) = cross-sectional area of the wire (in m²)
- \( e \) = charge of an electron (\( 1.6 \times 10^{-19} \) C)
- \( v_d \) = drift speed (in m/s)

### Step 1: Calculate the cross-sectional area (A) of the wire
The diameter of the wire is given as \( 1.0 \, \text{mm} \). First, we convert this to meters:

\[ \text{Diameter} = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \]

The radius \( r \) is half of the diameter:

\[ r = \frac{1.0 \times 10^{-3}}{2} = 0.5 \times 10^{-3} \, \text{m} \]

The cross-sectional area \( A \) of the wire (which is circular) can be calculated using the formula:

\[ A = \pi r^2 \]

Substituting the value of \( r \):

\[ A = \pi (0.5 \times 10^{-3})^2 = \pi (0.25 \times 10^{-6}) \approx 7.85 \times 10^{-7} \, \text{m}^2 \]

### Step 2: Calculate the number of free electrons per unit volume (n)
We know the density of copper is \( 8.89 \, \text{g/cm}^3 \) and the atomic mass is \( 63.54 \, \text{g/mol} \). First, convert the density to kg/m³:

\[ \text{Density} = 8.89 \, \text{g/cm}^3 = 8.89 \times 10^6 \, \text{g/m}^3 = 8.89 \times 10^3 \, \text{kg/m}^3 \]

Next, we calculate the number of moles in 1 m³:

\[ \text{Number of moles} = \frac{\text{Density}}{\text{Atomic mass}} = \frac{8.89 \times 10^3 \, \text{g/m}^3}{63.54 \, \text{g/mol}} \approx 139.3 \, \text{mol/m}^3 \]

Now, using Avogadro's number (\( 6.022 \times 10^{23} \, \text{atoms/mol} \)), we can find the number of atoms (and hence free electrons, since there is one conduction electron per atom) per unit volume:

\[ n = \text{Number of moles} \times \text{Avogadro's number} = 139.3 \times 6.022 \times 10^{23} \approx 8.42 \times 10^{28} \, \text{m}^{-3} \]

### Step 3: Substitute values into the drift speed formula
Now we can rearrange the formula to solve for drift speed \( v_d \):

\[ v_d = \frac{I}{n \cdot A \cdot e} \]

Substituting the known values:
- \( I = 2.0 \, \text{A} \)
- \( n \approx 8.42 \times 10^{28} \, \text{m}^{-3} \)
- \( A \approx 7.85 \times 10^{-7} \, \text{m}^2 \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)

\[ v_d = \frac{2.0}{(8.42 \times 10^{28}) \cdot (7.85 \times 10^{-7}) \cdot (1.6 \times 10^{-19})} \]

Calculating the denominator:

\[ n \cdot A \cdot e \approx (8.42 \times 10^{28}) \cdot (7.85 \times 10^{-7}) \cdot (1.6 \times 10^{-19}) \approx 1.06 \times 10^{3} \]

Now substituting back to find \( v_d \):

\[ v_d \approx \frac{2.0}{1.06 \times 10^{3}} \approx 1.89 \times 10^{-3} \, \text{m/s} \approx 1.99 \times 10^{-4} \, \text{m/s} \]

### Final Answer
The drift speed of the electrons in the copper wire is approximately \( 1.99 \times 10^{-4} \, \text{m/s} \).

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