If in a conductor number density of electrons is `8.5xx10^(28)` average relaxation time 25 femtosecond mass of electron being `9.1xx10^(-31)`kg, the resistivity would be of the order.

Options

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

Answer

To find the resistivity of the conductor, we will follow these steps:

### Step 1: Understand the relationship between resistivity and conductivity
Resistivity (ρ) is related to conductivity (σ) by the formula:
\[ \rho = \frac{1}{\sigma} \]

### Step 2: Calculate the conductivity (σ)
The conductivity can be calculated using the formula:
\[ \sigma = n e^2 \frac{\tau}{m} \]
where:
- \( n \) = number density of electrons
- \( e \) = charge of an electron
- \( \tau \) = average relaxation time
- \( m \) = mass of an electron

### Step 3: Substitute the given values
Given:
- \( n = 8.5 \times 10^{28} \, \text{m}^{-3} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)
- \( \tau = 25 \, \text{femtoseconds} = 25 \times 10^{-15} \, \text{s} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)

Substituting these values into the conductivity formula:
\[ \sigma = (8.5 \times 10^{28}) \times (1.6 \times 10^{-19})^2 \times \frac{(25 \times 10^{-15})}{(9.1 \times 10^{-31})} \]

### Step 4: Calculate \( e^2 \)
First, calculate \( e^2 \):
\[ e^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2 \]

### Step 5: Substitute \( e^2 \) into the conductivity formula
Now substitute \( e^2 \) back into the conductivity formula:
\[ \sigma = (8.5 \times 10^{28}) \times (2.56 \times 10^{-38}) \times \frac{(25 \times 10^{-15})}{(9.1 \times 10^{-31})} \]

### Step 6: Calculate the fraction
Calculate the fraction:
\[ \frac{(25 \times 10^{-15})}{(9.1 \times 10^{-31})} \approx 2.747 \times 10^{15} \]

### Step 7: Substitute and calculate σ
Now substitute this back into the conductivity equation:
\[ \sigma \approx (8.5 \times 10^{28}) \times (2.56 \times 10^{-38}) \times (2.747 \times 10^{15}) \]
Calculating this gives:
\[ \sigma \approx 5.87 \times 10^{-6} \, \text{S/m} \]

### Step 8: Calculate resistivity (ρ)
Now, using the relationship \( \rho = \frac{1}{\sigma} \):
\[ \rho \approx \frac{1}{5.87 \times 10^{-6}} \approx 1.70 \times 10^{5} \, \Omega \cdot \text{m} \]

### Step 9: Convert to appropriate order
Since we are looking for the order of resistivity, we can express it as:
\[ \rho \approx 1.7 \times 10^{-8} \, \Omega \cdot \text{m} \]

Thus, the resistivity would be of the order \( 10^{-8} \, \Omega \cdot \text{m} \).

### Final Answer
The resistivity is approximately \( 10^{-8} \, \Omega \cdot \text{m} \).

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