When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is `2.5 xx 10^(-4) ms ^(-1)`. If the electron density in the wire is ` 8 xx 10^(28) m^(-3)`, the resistivity of the material is close to :

Options

A. `1.6 xx 10^(-7) Omegam`
B. `1.6 xx 10^(-6) Omegam`
C. `1.6 xx 10^(-5) omegam`
D. `1.6 xx 10^(-8) Omegam`

Answer

To find the resistivity of the material, we can use the relationship between current, drift velocity, electron density, and resistivity. Here’s the step-by-step solution:

### Step 1: Write down the formula for current (I)
The current (I) flowing through a conductor can be expressed as:
\[ I = n \cdot e \cdot v_d \cdot A \]
where:
- \( n \) = electron density (in \( m^{-3} \))
- \( e \) = charge of an electron (\( 1.6 \times 10^{-19} \, C \))
- \( v_d \) = drift velocity (in \( m/s \))
- \( A \) = cross-sectional area of the wire (in \( m^2 \))

### Step 2: Relate current to potential difference (V) and resistance (R)
According to Ohm's law:
\[ V = I \cdot R \]
And resistance (R) can be expressed as:
\[ R = \frac{\rho L}{A} \]
where:
- \( \rho \) = resistivity (in \( \Omega \cdot m \))
- \( L \) = length of the wire (in \( m \))

### Step 3: Substitute R in Ohm's law
Substituting the expression for resistance into Ohm's law gives:
\[ V = I \cdot \frac{\rho L}{A} \]
Rearranging this, we get:
\[ \rho = \frac{V \cdot A}{I \cdot L} \]

### Step 4: Substitute the expression for current
Substituting the expression for current \( I \) into the resistivity equation:
\[ \rho = \frac{V \cdot A}{n \cdot e \cdot v_d \cdot A \cdot L} \]
The cross-sectional area \( A \) cancels out:
\[ \rho = \frac{V}{n \cdot e \cdot v_d \cdot L} \]

### Step 5: Plug in the values
Now we can substitute the known values into the equation:
- \( V = 5 \, V \)
- \( n = 8 \times 10^{28} \, m^{-3} \)
- \( e = 1.6 \times 10^{-19} \, C \)
- \( v_d = 2.5 \times 10^{-4} \, m/s \)
- \( L = 0.1 \, m \)

Substituting these values:
\[ \rho = \frac{5}{(8 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (2.5 \times 10^{-4}) \cdot (0.1)} \]

### Step 6: Calculate the resistivity
Calculating the denominator:
\[ n \cdot e \cdot v_d \cdot L = (8 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (2.5 \times 10^{-4}) \cdot (0.1) \]
\[ = 8 \times 1.6 \times 2.5 \times 0.1 \times 10^{28 - 19 - 4} \]
\[ = 8 \times 1.6 \times 2.5 \times 0.1 \times 10^{5} \]
\[ = 3.2 \times 2.5 \times 10^{5} \]
\[ = 8.0 \times 10^{5} \]

Now substituting back into the resistivity equation:
\[ \rho = \frac{5}{8.0 \times 10^{5}} \]
\[ = 6.25 \times 10^{-6} \, \Omega \cdot m \]

### Conclusion
The resistivity of the material is approximately:
\[ \rho \approx 6.25 \times 10^{-6} \, \Omega \cdot m \]