In conductor, if the number of conduction electrons per unit volume is `8.5xx10^(28) m^(-3)` and mean free time is 25 fs (femto seconds), it's approximate resistivity is : `(m_(e)=9.1xx10^(-31) kg)`

Options

A. `10^(-6) Omegam`
B. `10^(-7) Omegam`
C. `10^(-8) Omegam`
D. `10^(-5) Omegam`

Answer

To find the resistivity of the conductor, we can use the formula:

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

Where:
- \( \rho \) is the resistivity,
- \( m \) is the mass of the electron,
- \( n \) is the number of conduction electrons per unit volume,
- \( e \) is the charge of the electron,
- \( \tau \) is the mean free time (relaxation time).

### Step 1: Identify the given values
- Number of conduction electrons per unit volume, \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \)
- Mean free time, \( \tau = 25 \, \text{fs} = 25 \times 10^{-15} \, \text{s} \)
- Mass of the electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- Charge of the electron, \( e = 1.6 \times 10^{-19} \, \text{C} \)

### Step 2: Substitute the values into the resistivity formula
Now, substituting the values into the resistivity formula:

\[ \rho = \frac{9.1 \times 10^{-31} \, \text{kg}}{(8.5 \times 10^{28} \, \text{m}^{-3}) \times (1.6 \times 10^{-19} \, \text{C})^2 \times (25 \times 10^{-15} \, \text{s})} \]

### Step 3: Calculate \( e^2 \)
First, calculate \( e^2 \):

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

### Step 4: Calculate the denominator
Now calculate the denominator:

\[ n e^2 \tau = (8.5 \times 10^{28}) \times (2.56 \times 10^{-38}) \times (25 \times 10^{-15}) \]

Calculating step by step:

1. Calculate \( n e^2 \):
\[ n e^2 = 8.5 \times 10^{28} \times 2.56 \times 10^{-38} = 2.176 \times 10^{-9} \]

2. Now multiply by \( \tau \):
\[ n e^2 \tau = 2.176 \times 10^{-9} \times 25 \times 10^{-15} = 5.44 \times 10^{-13} \]

### Step 5: Calculate resistivity
Now substitute back into the resistivity formula:

\[ \rho = \frac{9.1 \times 10^{-31}}{5.44 \times 10^{-13}} \approx 1.67 \times 10^{-18} \, \Omega \cdot \text{m} \]

### Final Result
Thus, the approximate resistivity of the conductor is:

\[ \rho \approx 1.67 \times 10^{-8} \, \Omega \cdot \text{m} \]