A semiconductor is known to have an electron concentration of `5 xx 10^(13)//cm^(3)` and hole concentration of `8 xx 10^(12)//cm^(3)`. The semiconductor is

Options

A. n-type
B. p-type
C. intrinsic
D. insulator

Answer

To determine the type of semiconductor based on the given electron and hole concentrations, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Electron concentration (n) = \(5 \times 10^{13} \, \text{cm}^{-3}\)
- Hole concentration (p) = \(8 \times 10^{12} \, \text{cm}^{-3}\)

2. **Compare Electron and Hole Concentrations**:
- We need to compare the values of n and p to determine the type of semiconductor.
- Here, \(n = 5 \times 10^{13} \, \text{cm}^{-3}\) and \(p = 8 \times 10^{12} \, \text{cm}^{-3}\).

3. **Determine the Dominant Carrier**:
- Since \(n > p\) (i.e., \(5 \times 10^{13} > 8 \times 10^{12}\)), the concentration of electrons is greater than that of holes.

4. **Classify the Semiconductor**:
- In semiconductors, if the electron concentration is greater than the hole concentration, the semiconductor is classified as an N-type semiconductor.
- Conversely, if the hole concentration were greater than the electron concentration, it would be classified as a P-type semiconductor.

5. **Conclusion**:
- Based on the comparison, we conclude that the semiconductor is N-type.

### Final Answer:
The semiconductor is N-type.