A crystal of intrinsic silicon at room temperature has a carrier concentration of `1.6xx10^(16)m^(-3)` . If the donor concentration level is `4.8xx10^(20) m^(-3)` , then the concentration of holes in the semiconductor is

Options

A. `53xx10^(12)m^(-3)`
B. `4xx10^(11)m^(-3)`
C. `4xx10^(12)m^(-3)`
D. `5.3xx10^(11)m^(-3)`

Answer

To find the concentration of holes (p) in the semiconductor, we can use the mass action law, which states:

\[ n_i^2 = n \cdot p \]

Where:
- \( n_i \) is the intrinsic carrier concentration,
- \( n \) is the concentration of donor atoms,
- \( p \) is the concentration of holes.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Intrinsic carrier concentration \( n_i = 1.6 \times 10^{16} \, \text{m}^{-3} \)
- Donor concentration \( n = 4.8 \times 10^{20} \, \text{m}^{-3} \)

2. **Use the Mass Action Law:**
- From the mass action law, we can rearrange the equation to find \( p \):
\[ p = \frac{n_i^2}{n} \]

3. **Calculate \( n_i^2 \):**
- Calculate \( n_i^2 \):
\[ n_i^2 = (1.6 \times 10^{16})^2 = 1.6 \times 1.6 \times 10^{32} = 2.56 \times 10^{32} \, \text{m}^{-6} \]

4. **Substitute Values into the Equation:**
- Now substitute \( n_i^2 \) and \( n \) into the equation for \( p \):
\[ p = \frac{2.56 \times 10^{32}}{4.8 \times 10^{20}} \]

5. **Simplify the Expression:**
- Divide the coefficients and subtract the exponents:
\[ p = \frac{2.56}{4.8} \times 10^{32 - 20} = \frac{2.56}{4.8} \times 10^{12} \]

6. **Calculate the Coefficient:**
- Calculate \( \frac{2.56}{4.8} \):
\[ \frac{2.56}{4.8} = 0.5333 \approx 0.53 \]

7. **Final Calculation:**
- Therefore,
\[ p \approx 0.53 \times 10^{12} = 5.3 \times 10^{11} \, \text{m}^{-3} \]

### Conclusion:
The concentration of holes in the semiconductor is approximately \( 5.3 \times 10^{11} \, \text{m}^{-3} \).