The intrinsic carrier density in germanium crystal at 300 K is `2.5xx10^(13)` per `cm^(3)` if the electron density in an N-type germanium crystal at 300 K be `0.5xx10^(17)` per `cm^(3)` the hle density (per `cm^(3)`) in this N-type crystal at 300 K would be expected around-

Options

A. `2.5xx10^(13)`
B. `5xx10^(6)`
C. `1.25xx10^(10)`
D. `0.2xx10^(4)`

Answer

To find the hole density in an N-type germanium crystal at 300 K, we can use the relationship between the intrinsic carrier density, electron density, and hole density. Here’s a step-by-step solution:

### Step 1: Understand the given values
- Intrinsic carrier density (n_i) in germanium at 300 K:
\[ n_i = 2.5 \times 10^{13} \text{ cm}^{-3} \]
- Electron density (n_e) in N-type germanium at 300 K:
\[ n_e = 0.5 \times 10^{17} \text{ cm}^{-3} \]

### Step 2: Use the relationship between electron density, hole density, and intrinsic carrier density
In an N-type semiconductor, the relationship between the electron density (n_e), hole density (p), and intrinsic carrier density (n_i) is given by:
\[ n_e \cdot p = n_i^2 \]
From this equation, we can express the hole density (p) as:
\[ p = \frac{n_i^2}{n_e} \]

### Step 3: Calculate \( n_i^2 \)
First, we need to calculate \( n_i^2 \):
\[ n_i^2 = (2.5 \times 10^{13})^2 = 6.25 \times 10^{26} \text{ cm}^{-6} \]

### Step 4: Substitute the values into the equation for hole density
Now we can substitute \( n_i^2 \) and \( n_e \) into the equation for hole density:
\[ p = \frac{6.25 \times 10^{26}}{0.5 \times 10^{17}} \]

### Step 5: Simplify the expression
Calculating the above expression:
\[ p = \frac{6.25 \times 10^{26}}{0.5 \times 10^{17}} = \frac{6.25}{0.5} \times 10^{26 - 17} = 12.5 \times 10^{9} \text{ cm}^{-3} \]

### Step 6: Final answer
Thus, the hole density (p) in the N-type germanium crystal at 300 K is:
\[ p = 1.25 \times 10^{10} \text{ cm}^{-3} \]

### Summary
The hole density in the N-type germanium crystal at 300 K is approximately:
\[ \boxed{1.25 \times 10^{10} \text{ cm}^{-3}} \]