Assume that each atom of copper contributes one free electron. Density of Cu is 9 g/cm`""^(3)` and atomic weight is 63 g. If current flowing through a Cu wire of 1 mm diameter is 11 A, drift velocity of clectrons will be

Options

A. 0.1 mm/s
B. 0.2 mm/s
C. 0.3 mm/s
D. 0.5 mm/s

Answer

To find the drift velocity of electrons in a copper wire, we can follow these steps:

### Step 1: Calculate the mass of 1 m³ of copper
Given:
- Density of copper (ρ) = 9 g/cm³ = 9 x 10³ kg/m³ (since 1 g/cm³ = 1000 kg/m³)

So, the mass of 1 m³ of copper is:
\[ \text{Mass} = \text{Density} = 9 \times 10^3 \text{ kg/m}^3 \]

### Step 2: Calculate the number of atoms per m³ of copper
Using Avogadro's number (N_A = 6.022 x 10²³ atoms/mol) and the atomic weight of copper (A = 63 g/mol), we can find the number of atoms per m³.

First, we convert the atomic weight into kg:
\[ A = 63 \text{ g/mol} = 0.063 \text{ kg/mol} \]

Now, the number of moles in 1 m³ of copper is:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Atomic weight}} = \frac{9 \times 10^3 \text{ kg/m}^3}{0.063 \text{ kg/mol}} \approx 1.42 \times 10^5 \text{ mol/m}^3 \]

Now, we can find the number of atoms per m³:
\[ \text{Number of atoms per m}^3 = \text{Number of moles} \times N_A = 1.42 \times 10^5 \text{ mol/m}^3 \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 8.57 \times 10^{28} \text{ atoms/m}^3 \]

### Step 3: Calculate the cross-sectional area of the wire
Given the diameter of the wire (d = 1 mm = 1 x 10⁻³ m), the radius (r) is:
\[ r = \frac{d}{2} = \frac{1 \times 10^{-3}}{2} = 0.5 \times 10^{-3} \text{ m} \]

The cross-sectional area (A) of the wire is:
\[ A = \pi r^2 = \pi (0.5 \times 10^{-3})^2 \approx 7.85 \times 10^{-7} \text{ m}^2 \]

### Step 4: Calculate the drift velocity (V_d)
The formula for drift velocity is given by:
\[ V_d = \frac{I}{n \cdot e \cdot A} \]
Where:
- I = current = 11 A
- n = number of charge carriers per m³ = 8.57 \times 10^{28} \text{ electrons/m}^3
- e = charge of an electron = 1.6 \times 10^{-19} C
- A = cross-sectional area = 7.85 \times 10^{-7} m²

Substituting the values:
\[ V_d = \frac{11}{(8.57 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (7.85 \times 10^{-7})} \]

Calculating the denominator:
\[ n \cdot e \cdot A \approx (8.57 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (7.85 \times 10^{-7}) \approx 1.07 \times 10^{3} \]

Now calculating drift velocity:
\[ V_d \approx \frac{11}{1.07 \times 10^{3}} \approx 0.0103 \text{ m/s} \approx 10.3 \text{ mm/s} \]

### Final Answer:
The drift velocity of electrons in the copper wire is approximately **10.3 mm/s**.

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