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, we need to find the majority carrier density in a semiconductor that has been doped with a trivalent impurity. Here's a step-by-step solution:

### Step 1: Understand the given data
- The intrinsic electron density \( n_i \) of the semiconductor at room temperature is given as:
\[ n_i = 10^{16} \, \text{m}^{-3} \]
- After doping with a trivalent impurity, the electron density \( n_e \) decreases to:
\[ n_e = 10^{14} \, \text{m}^{-3} \]

### Step 2: Identify the type of doping
- Doping with a trivalent impurity (such as Boron) creates "holes" in the semiconductor, making holes the majority carriers and electrons the minority carriers.

### Step 3: Use the law of mass action
- According to the law of mass action, the product of the electron density and hole density in a semiconductor is constant and equal to the square of the intrinsic carrier concentration:
\[ n_i^2 = n_e \cdot n_p \]
where \( n_p \) is the hole density (majority carrier density).

### Step 4: Rearrange the equation to find \( n_p \)
- We can rearrange the equation to solve for \( n_p \):
\[ n_p = \frac{n_i^2}{n_e} \]

### Step 5: Substitute the values
- Substitute the known values into the equation:
\[ n_p = \frac{(10^{16})^2}{10^{14}} \]
- Calculate \( n_i^2 \):
\[ (10^{16})^2 = 10^{32} \]
- Now substitute this back into the equation:
\[ n_p = \frac{10^{32}}{10^{14}} = 10^{32 - 14} = 10^{18} \, \text{m}^{-3} \]

### Step 6: Conclusion
- The majority carrier density \( n_p \) is:
\[ n_p = 10^{18} \, \text{m}^{-3} \]

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

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