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 in the semiconductor, we can use the relationship between the intrinsic carrier concentration (Ni), the concentration of electrons (n), and the concentration of holes (p) in a semiconductor. The relationship is given by the equation:

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

Where:
- \( n \) is the concentration of electrons,
- \( p \) is the concentration of holes,
- \( N_i \) is the intrinsic carrier concentration.

### Step-by-Step Solution:

1. **Identify the given values:**
- Intrinsic carrier concentration, \( N_i = 1.6 \times 10^{16} \, m^{-3} \)
- Donor concentration (which gives the concentration of electrons), \( n = 4.8 \times 10^{20} \, m^{-3} \)

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

3. **Use the relationship to find the concentration of holes \( p \):**
\[ p = \frac{N_i^2}{n} \]
Substituting the values we have:
\[ p = \frac{2.56 \times 10^{32}}{4.8 \times 10^{20}} \]

4. **Perform the division:**
\[ p = \frac{2.56}{4.8} \times 10^{32 - 20} = \frac{2.56}{4.8} \times 10^{12} \]
\[ p = 0.5333 \times 10^{12} = 5.33 \times 10^{11} \, m^{-3} \]

5. **Final Result:**
The concentration of holes in the semiconductor is approximately:
\[ p \approx 5.3 \times 10^{11} \, m^{-3} \]