In a copper sample, that drift velocity of electrons in 100 V/m electric field is 2.63m/s. The conductivity of copper is 5.5 x 10-7 Ω-1 m-1. Calculate mobility, electron density and relaxation time for electrons (mass of electron = 9.1 x 10-31 kg).

ANSWER : The following is the answer one by one :

1. Mobility of Electrons:

The formula for electron mobility (μ) in a conductor is given by:

\[ \mu = \frac{v_d}{E} \]

Where:

- \( \mu \) = electron mobility

- \( v_d \) = drift velocity of electrons

- \( E \) = electric field strength

Given:

- Drift velocity (\( v_d \)) = 2.63 m/s

- Electric field (\( E \)) = 100 V/m

Let's calculate the mobility of electrons using the formula:

\[ \mu = \frac{2.63 \, \text{m/s}}{100 \, \text{V/m}} \]

\[ \mu = 0.0263 \, \text{m}^2/\text{V s} \]

2. Electron Density:

The relationship between conductivity (\( \sigma \)), electron density (\( n \)), and electron charge (\( e \)) is given by:

\[ \sigma = n \cdot e \cdot \mu \]

Where:

- \( \sigma \) = conductivity

- \( n \) = electron density

- \( e \) = charge of an electron (\( 1.6 \times 10^{-19} \) C)

Given:

- Conductivity (\( \sigma \)) = \( 5.5 \times 10^{-7} \, \Omega^{-1} \text{m}^{-1} \)

- Charge of an electron (\( e \)) = \( 1.6 \times 10^{-19} \, \text{C} \)

- Electron mobility (\( \mu \)) = 0.0263 \( \text{m}^2/\text{V s} \)

We can rearrange the formula to solve for electron density (\( n \)):

\[ n = \frac{\sigma}{e \cdot \mu} \]

Let's plug in the values:

\[ n = \frac{5.5 \times 10^{-7} \, \Omega^{-1} \text{m}^{-1}}{1.6 \times 10^{-19} \, \text{C} \cdot 0.0263 \, \text{m}^2/\text{V s}} \]

\[ n \approx 1.98 \times 10^{28} \, \text{m}^{-3} \]

3. Relaxation Time for Electrons:

The relaxation time (\( \tau \)) of electrons is given by:

\[ \tau = \frac{m}{n \cdot e^2 \cdot \sigma} \]

Where:

- \( \tau \) = relaxation time

- \( m \) = mass of an electron

- \( n \) = electron density

- \( e \) = charge of an electron

- \( \sigma \) = conductivity

Given:

- Mass of an electron (\( m \)) = \( 9.1 \times 10^{-31} \, \text{kg} \)

- Electron density (\( n \)) ≈ \( 1.98 \times 10^{28} \, \text{m}^{-3} \)

- Charge of an electron (\( e \)) = \( 1.6 \times 10^{-19} \, \text{C} \)

- Conductivity (\( \sigma \)) = \( 5.5 \times 10^{-7} \, \Omega^{-1} \text{m}^{-1} \)

Let's calculate the relaxation time (\( \tau \)):

\[ \tau = \frac{9.1 \times 10^{-31} \, \text{kg}}{1.98 \times 10^{28} \, \text{m}^{-3} \cdot (1.6 \times 10^{-19} \, \text{C})^2 \cdot 5.5 \times 10^{-7} \, \Omega^{-1} \text{m}^{-1}} \]

\[ \tau \approx 2.03 \times 10^{-14} \, \text{s} \]

So, after calculations:

- Electron mobility (\( \mu \)) ≈ 0.0263 \( \text{m}^2/\text{V s} \)

- Electron density (\( n \)) ≈ \( 1.98 \times 10^{28} \, \text{m}^{-3} \)

- Relaxation time (\( \tau \)) ≈ \( 2.03 \times 10^{-14} \, \text{s} \)

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