The resistivity of material of a wire is `2xx10^(-8) Omegam` and the number of free electrons per unit volume is `9xx10^(28)` per `m^3` . Calculate the relaxation time which is defined as the average time interval between two successive collisions.
Also calculate the mean free path of electron (distance covered by electron between two successive collisions) if the drift velocity of electrons is `1.7 xx10^(-4)m//s`.

Explanation

To solve the given problem, we will follow these steps:

### Step 1: Identify the given values
- Resistivity, \( \rho = 2 \times 10^{-8} \, \Omega \cdot m \)
- Number of free electrons per unit volume, \( n = 9 \times 10^{28} \, m^{-3} \)
- Charge of an electron, \( e = 1.6 \times 10^{-19} \, C \)
- Mass of an electron, \( m_e = 9.1 \times 10^{-31} \, kg \)
- Drift velocity, \( v_d = 1.7 \times 10^{-4} \, m/s \)
...