The length of a germanium rod is 0.58 cm and its area of cross-section is `1mm^(2)`. If for germanium `n_(i)=2.5xx10^(19)m^(-3),mu_(h)=0.19 m^(2)//V-s,mu_(e)=0.39 m^(2)//V-s`, then the resistance of the rod will be-

Options

A. `2.5kOmega`
B. `5.0kOmega`
C. `7.5kOmega`
D. `10.0kOmega`

Answer

To find the resistance of the germanium rod, we can follow these steps:

### Step 1: Convert the given dimensions to standard units
- Length of the germanium rod, \( L = 0.58 \, \text{cm} = 0.0058 \, \text{m} \)
- Area of cross-section, \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)

### Step 2: Calculate the intrinsic carrier concentration and mobility
Given:
- Intrinsic carrier concentration, \( n_i = 2.5 \times 10^{19} \, \text{m}^{-3} \)
- Hole mobility, \( \mu_h = 0.19 \, \text{m}^2/\text{V-s} \)
- Electron mobility, \( \mu_e = 0.39 \, \text{m}^2/\text{V-s} \)

### Step 3: Calculate the conductivity \( \sigma \)
The conductivity \( \sigma \) can be calculated using the formula:
\[ \sigma = n_i \cdot e \cdot (\mu_e + \mu_h) \]
Where \( e \) (the charge of an electron) is approximately \( 1.6 \times 10^{-19} \, \text{C} \).

Substituting the values:
\[ \sigma = 2.5 \times 10^{19} \cdot 1.6 \times 10^{-19} \cdot (0.39 + 0.19) \]
\[ \sigma = 2.5 \times 10^{19} \cdot 1.6 \times 10^{-19} \cdot 0.58 \]
Calculating this gives:
\[ \sigma = 2.5 \times 1.6 \times 0.58 \, \text{S/m} \]
\[ \sigma = 2.32 \, \text{S/m} \]

### Step 4: Calculate the resistivity \( \rho \)
The resistivity \( \rho \) is the reciprocal of conductivity:
\[ \rho = \frac{1}{\sigma} = \frac{1}{2.32} \approx 0.431 \, \Omega \cdot \text{m} \]

### Step 5: Calculate the resistance \( R \)
Using the formula for resistance:
\[ R = \rho \cdot \frac{L}{A} \]
Substituting the values:
\[ R = 0.431 \cdot \frac{0.0058}{1 \times 10^{-6}} \]
Calculating this gives:
\[ R = 0.431 \cdot 5800 \approx 2500.8 \, \Omega \]

### Step 6: Convert to kilohms
\[ R \approx 2.5 \, k\Omega \]

### Final Answer
The resistance of the germanium rod is approximately \( 2.5 \, k\Omega \).

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