A current of 30 ampere is flowing through a wire of cross-sectional area `2 mm^(2)`. Calculate the drit velocity of electrons. Assuming the temperature of the wire to be `27^(@)C`, also calculate the rms velocity at this temperature. Which velocity is larger ? Given that Boltzman's constant `=1.38xx10^(-23) J K^(-1)`, density of copper `8.9 g cm^(-3)`, atomic mass of copper =63.

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert given values to SI units
- Cross-sectional area (A) = 2 mm² = 2 × 10^(-6) m²
- Density of copper = 8.9 g/cm³ = 8.9 × 10^3 kg/m³
- Atomic mass of copper = 63 g/mol = 63 × 10^(-3) kg/mol
- Current (I) = 30 A
- Boltzmann's constant (k) = 1.38 × 10^(-23) J/K

### Step 2: Calculate the number of free electrons per unit volume (N)
To find the number of free electrons per unit volume (N), we first need to calculate the number of moles in 1 m³ of copper:

\[ \text{Number of moles in 1 m³} = \frac{\text{Density}}{\text{Atomic mass}} = \frac{8.9 \times 10^3 \text{ kg/m}^3}{63 \times 10^{-3} \text{ kg/mol}} = \frac{8.9 \times 10^3}{63 \times 10^{-3}} \approx 141.27 \text{ mol} \]

Next, we convert moles to atoms using Avogadro's number (N_A = 6.022 × 10²³ atoms/mol):

\[ \text{Number of atoms in 1 m³} = 141.27 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 8.5 \times 10^{28} \text{ atoms/m}^3 \]

Since each copper atom contributes one free electron, we have:

\[ N \approx 8.5 \times 10^{28} \text{ electrons/m}^3 \]

### Step 3: Calculate the drift velocity (V_D)
Using the formula for drift velocity:

\[ I = N \cdot e \cdot A \cdot V_D \]

Where:
- \( e \) (charge of an electron) = 1.6 × 10^(-19) C

Rearranging to find \( V_D \):

\[ V_D = \frac{I}{N \cdot e \cdot A} \]

Substituting the values:

\[ V_D = \frac{30 \text{ A}}{(8.5 \times 10^{28} \text{ electrons/m}^3) \cdot (1.6 \times 10^{-19} \text{ C}) \cdot (2 \times 10^{-6} \text{ m}^2)} \]

Calculating \( V_D \):

\[ V_D \approx \frac{30}{(8.5 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (2 \times 10^{-6})} \approx 1.1 \times 10^{-3} \text{ m/s} \]

### Step 4: Calculate the RMS velocity (V_rms)
The RMS velocity for electrons at a given temperature is given by:

\[ V_{rms} = \sqrt{\frac{3kT}{m_0}} \]

Where:
- \( k \) = Boltzmann's constant
- \( T \) = Temperature in Kelvin (27°C = 300 K)
- \( m_0 \) = mass of an electron = 9.11 × 10^(-31) kg

Substituting the values:

\[ V_{rms} = \sqrt{\frac{3 \cdot (1.38 \times 10^{-23} \text{ J/K}) \cdot 300 \text{ K}}{9.11 \times 10^{-31} \text{ kg}}} \]

Calculating \( V_{rms} \):

\[ V_{rms} \approx \sqrt{\frac{1.242 \times 10^{-20}}{9.11 \times 10^{-31}}} \approx 3.44 \times 10^{2} \text{ m/s} \]

### Step 5: Compare the velocities
Now we compare \( V_D \) and \( V_{rms} \):

- \( V_D \approx 1.1 \times 10^{-3} \text{ m/s} \)
- \( V_{rms} \approx 3.44 \times 10^{2} \text{ m/s} \)

Clearly, \( V_{rms} \) is much larger than \( V_D \).

### Final Answers
- Drift velocity \( V_D \approx 1.1 \times 10^{-3} \text{ m/s} \)
- RMS velocity \( V_{rms} \approx 3.44 \times 10^{2} \text{ m/s} \)
- The RMS velocity is larger than the drift velocity.