Find the time of relation between collision and free path of electrons in copper at room temperature .Given resistance of copper `= 1.5xx 10^(-8) Omega m` number density of electron in copper `= 8.5 xx 10^(28)m^(-3)`
charge on electron `= 1.6 xx 10^(19)C`,
mass of electrons ` = 9.1 xx 10^(-19) kg`
Drift velocity of free electron `= 1.6 xx 10^(-4)ms^(-1)`

Explanation

To find the time relation between collision (τ) and mean free path (λ) of electrons in copper at room temperature, we will use the given data and relevant formulas.

### Step 1: Write down the given data
- Resistivity of copper (ρ) = \(1.5 \times 10^{-8} \, \Omega \, m\)
- Number density of electrons (n) = \(8.5 \times 10^{28} \, m^{-3}\)
- Charge on electron (e) = \(1.6 \times 10^{-19} \, C\)
- Mass of electron (m) = \(9.1 \times 10^{-31} \, kg\)
- Drift velocity of free electrons (v_d) = \(1.6 \times 10^{-4} \, m/s\)

### Step 2: Calculate the relaxation time (τ)
The formula for resistivity in terms of relaxation time is given by:
\[ \rho = \frac{m}{n e^2 \tau} \]
Rearranging this gives:
\[ \tau = \frac{m}{n e^2 \rho} \]

### Step 3: Substitute the values into the formula
Substituting the known values:
- \(m = 9.1 \times 10^{-31} \, kg\)
- \(n = 8.5 \times 10^{28} \, m^{-3}\)
- \(e = 1.6 \times 10^{-19} \, C\)
- \(\rho = 1.5 \times 10^{-8} \, \Omega \, m\)

Calculating \(e^2\):
\[ e^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, C^2 \]

Now substituting into the formula for τ:
\[ \tau = \frac{9.1 \times 10^{-31}}{(8.5 \times 10^{28})(2.56 \times 10^{-38})(1.5 \times 10^{-8})} \]

### Step 4: Calculate τ
Calculating the denominator:
\[ (8.5 \times 10^{28})(2.56 \times 10^{-38})(1.5 \times 10^{-8}) = 3.264 \times 10^{-17} \]

Now calculating τ:
\[ \tau = \frac{9.1 \times 10^{-31}}{3.264 \times 10^{-17}} \approx 2.78 \times 10^{-14} \, s \]

### Step 5: Calculate the mean free path (λ)
The mean free path (λ) is given by:
\[ \lambda = v_d \cdot \tau \]
Substituting the values:
\[ \lambda = (1.6 \times 10^{-4} \, m/s) \cdot (2.78 \times 10^{-14} \, s) \]

### Step 6: Calculate λ
Calculating λ:
\[ \lambda = 4.448 \times 10^{-18} \, m \]

### Final Results
- Relaxation time (τ) ≈ \(2.78 \times 10^{-14} \, s\)
- Mean free path (λ) ≈ \(4.448 \times 10^{-18} \, m\)