The concentration of electrons in a semiconductor is `3 xx 10^(13)//cm^3` and hole concentration is` 5 xx 10^(14) //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 concentrations of electrons and holes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Electron concentration, \( n_e = 3 \times 10^{13} \, \text{cm}^{-3} \)
- Hole concentration, \( n_h = 5 \times 10^{14} \, \text{cm}^{-3} \)

2. **Compare Electron and Hole Concentrations**:
- We need to compare the values of \( n_e \) and \( n_h \).
- Here, \( n_h = 5 \times 10^{14} \) is greater than \( n_e = 3 \times 10^{13} \).

3. **Determine Semiconductor Type**:
- In semiconductors, if the concentration of holes (\( n_h \)) is greater than the concentration of electrons (\( n_e \)), the semiconductor is classified as **p-type**.
- Conversely, if the concentration of electrons is greater than that of holes, it would be classified as **n-type**.

4. **Conclusion**:
- Since \( n_h > n_e \), we conclude that the semiconductor is **p-type**.

### Final Answer:
The semiconductor is **p-type**.

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