A potential difference of `2.5 V` is applied across the faces of a germanium crystal plate. The face area of the crystal is `1 cm^(2)` and its thickness is `1.0 mm`. The free electron concentration in germanium is `2 xx 10^(19)m^(-3)` and the electron and holes mobilities are `0.33m^(2)/V s `and `0.17 m^(2)/V` s respectively. The current across the plate will be

Options

A. `0.2 A `
B. `0.4 A`
C. `0.6 A`
D. `0.8 A`

Answer

To find the current across the germanium crystal plate, we can follow these steps:

### Step 1: Convert given values to standard units
- The potential difference \( V = 2.5 \, \text{V} \).
- The face area \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \).
- The thickness \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \).
- The free electron concentration \( n = 2 \times 10^{19} \, \text{m}^{-3} \).
- The electron mobility \( \mu_e = 0.33 \, \text{m}^2/\text{V s} \).
- The hole mobility \( \mu_h = 0.17 \, \text{m}^2/\text{V s} \).

### Step 2: Calculate the conductivity \( \sigma \)
The conductivity \( \sigma \) can be calculated using the formula:
\[ \sigma = n \cdot e \cdot (\mu_e + \mu_h) \]
Where \( e \) is the charge of an electron, approximately \( 1.6 \times 10^{-19} \, \text{C} \).

Substituting the values:
\[ \sigma = (2 \times 10^{19}) \cdot (1.6 \times 10^{-19}) \cdot (0.33 + 0.17) \]
Calculating \( \mu_e + \mu_h \):
\[ \mu_e + \mu_h = 0.33 + 0.17 = 0.5 \, \text{m}^2/\text{V s} \]
Now substituting back:
\[ \sigma = (2 \times 10^{19}) \cdot (1.6 \times 10^{-19}) \cdot 0.5 \]
\[ \sigma = 2 \cdot 1.6 \cdot 0.5 = 1.6 \, \text{S/m} \]

### Step 3: Calculate the resistance \( R \)
The resistance \( R \) can be calculated using the formula:
\[ R = \frac{d}{\sigma \cdot A} \]
Substituting the values:
\[ R = \frac{1 \times 10^{-3}}{1.6 \cdot 1 \times 10^{-4}} \]
Calculating:
\[ R = \frac{1 \times 10^{-3}}{1.6 \times 10^{-4}} = \frac{10}{1.6} \approx 6.25 \, \Omega \]

### Step 4: Calculate the current \( I \)
Using Ohm's law, \( V = I \cdot R \), we can find the current \( I \):
\[ I = \frac{V}{R} \]
Substituting the values:
\[ I = \frac{2.5}{6.25} \]
Calculating:
\[ I = 0.4 \, \text{A} \]

### Final Answer
The current across the plate is \( 0.4 \, \text{A} \).

---