The electron density of intrinsic semiconductor at room temperature is `10^(16)m^(-3)` . When doped with a trivalent impurity , the electron density is decreased to `10^(14)m^(-3)` at the same temperature . The majority carrier density is

Options

A. `10^(16)m^(-3)`
B. `10^(18)m^(-3)`
C. `10^(21)m^(-3)`
D. `10^(20)m^(-3)`

Answer

To solve the problem step by step, we will use the information provided about the intrinsic semiconductor and the effects of doping with a trivalent impurity.

### Step-by-Step Solution:

1. **Identify the given values:**
- The intrinsic electron density (\(N_i\)) of the semiconductor at room temperature is \(10^{16} \, m^{-3}\).
- After doping with a trivalent impurity, the electron density (\(N_e\)) decreases to \(10^{14} \, m^{-3}\).

2. **Understand the impact of doping:**
- Doping with a trivalent impurity (like Aluminum) creates holes in the semiconductor. This means that the number of holes (\(N_h\)) will increase while the number of electrons (\(N_e\)) decreases due to recombination.

3. **Apply the Mass Action Law:**
- The Mass Action Law states that for semiconductors, the product of the electron density and hole density is constant at a given temperature:
\[ N_i^2 = N_e \cdot N_h \]
- Here, \(N_i\) is the intrinsic carrier concentration, \(N_e\) is the electron density, and \(N_h\) is the hole density.

4. **Plug in the values:**
- Substitute the known values into the equation:
\[ (10^{16})^2 = (10^{14}) \cdot N_h \]

5. **Calculate \(N_h\):**
- First, calculate \(10^{16}^2\):
\[ 10^{32} = 10^{14} \cdot N_h \]
- Now, solve for \(N_h\):
\[ N_h = \frac{10^{32}}{10^{14}} = 10^{32 - 14} = 10^{18} \, m^{-3} \]

6. **Conclusion:**
- The majority carrier density (holes) after doping is \(N_h = 10^{18} \, m^{-3}\).

### Final Answer:
The majority carrier density is \(10^{18} \, m^{-3}\).

---