A block of pure silicon at `300 K` has a length of `10 cm` and an area of `1.0 cm^(2)`. A battery of emf `2V` is connected across it. The mobility of electron is `0.14 m^(2) v^(-1) S^(-1)` and their number density is `1.5 xx 10^(16) m^(-3)`. The mobility of holes is `0.05 m^(2) v^(-1) S^(-1)`.
The electron current is

Options

A. `6.72xx10^(-4)` A
B. `6.72xx10^(-5)` A
C. `6.72xx10^(-6)` A
D. `6.72xx10^(-7)` A

Explanation

To find the electron current in the given silicon block, we can follow these steps:

### Step 1: Calculate the Electric Field (E)
The electric field (E) in the silicon block can be calculated using the formula:

\[ E = \frac{V}{L} \]

Where:
- \( V \) is the voltage (emf of the battery) = 2 V
- \( L \) is the length of the silicon block = 10 cm = 0.1 m

Substituting the values:

\[ E = \frac{2 \, \text{V}}{0.1 \, \text{m}} = 20 \, \text{V/m} \]

### Step 2: Calculate the Drift Velocity (v_d)
The drift velocity of the electrons can be calculated using the formula:

\[ v_d = \mu_n \cdot E \]

Where:
- \( \mu_n \) is the mobility of electrons = 0.14 m²/V·s
- \( E \) is the electric field calculated in Step 1 = 20 V/m

Substituting the values:

\[ v_d = 0.14 \, \text{m}^2/\text{V·s} \cdot 20 \, \text{V/m} = 2.8 \, \text{m/s} \]

### Step 3: Calculate the Electron Current (I)
The electron current can be calculated using the formula:

\[ I = n \cdot A \cdot v_d \cdot e \]

Where:
- \( n \) is the number density of electrons = \( 1.5 \times 10^{16} \, \text{m}^{-3} \)
- \( A \) is the cross-sectional area = 1.0 cm² = \( 1.0 \times 10^{-4} \, \text{m}^2 \)
- \( v_d \) is the drift velocity calculated in Step 2 = 2.8 m/s
- \( e \) is the charge of an electron = \( 1.6 \times 10^{-19} \, \text{C} \)

Substituting the values:

\[ I = (1.5 \times 10^{16} \, \text{m}^{-3}) \cdot (1.0 \times 10^{-4} \, \text{m}^2) \cdot (2.8 \, \text{m/s}) \cdot (1.6 \times 10^{-19} \, \text{C}) \]

Calculating this step-by-step:

1. Calculate \( n \cdot A \):
\[ n \cdot A = 1.5 \times 10^{16} \cdot 1.0 \times 10^{-4} = 1.5 \times 10^{12} \]

2. Calculate \( n \cdot A \cdot v_d \):
\[ n \cdot A \cdot v_d = 1.5 \times 10^{12} \cdot 2.8 = 4.2 \times 10^{12} \]

3. Finally, calculate the current \( I \):
\[ I = 4.2 \times 10^{12} \cdot 1.6 \times 10^{-19} = 6.72 \times 10^{-7} \, \text{A} \]

### Final Answer
The electron current is:

\[ I = 6.72 \times 10^{-7} \, \text{A} \]

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