A semiconductor has equal electron and hole concentration of `2xx10^(8)m^(-3)`. On doping with a certain impurity, the electron concentration increases to `4xx10^(10)m^(-3),` then the new hole concentration of the semiconductor is

Options

A. `10^(6)m^(-3)`
B. `10^(8)m^(-3)`
C. `10^(10)m^(-3)`
D. `10^(12)m^(-3)`

Answer

To solve the problem, we need to find the new hole concentration after doping a semiconductor that initially has equal electron and hole concentrations. Let's break down the steps:

### Step 1: Understand the Initial Conditions
Initially, the electron concentration \( n_0 \) and hole concentration \( p_0 \) are both given as:
\[ n_0 = p_0 = 2 \times 10^8 \, \text{m}^{-3} \]

### Step 2: Determine the New Electron Concentration
After doping, the electron concentration \( n \) increases to:
\[ n = 4 \times 10^{10} \, \text{m}^{-3} \]

### Step 3: Use the Mass Action Law
According to the mass action law for semiconductors, the product of the electron concentration and hole concentration remains constant at thermal equilibrium:
\[ n \cdot p = n_0 \cdot p_0 \]
Substituting the known values:
\[ (4 \times 10^{10}) \cdot p = (2 \times 10^8) \cdot (2 \times 10^8) \]

### Step 4: Calculate the Initial Product of Concentrations
Calculating the right side:
\[ (2 \times 10^8) \cdot (2 \times 10^8) = 4 \times 10^{16} \]

### Step 5: Substitute and Solve for Hole Concentration
Now, substituting back into the equation:
\[ (4 \times 10^{10}) \cdot p = 4 \times 10^{16} \]
To find \( p \), divide both sides by \( 4 \times 10^{10} \):
\[ p = \frac{4 \times 10^{16}}{4 \times 10^{10}} = 10^{6} \, \text{m}^{-3} \]

### Conclusion
The new hole concentration \( p \) after doping is:
\[ p = 10^6 \, \text{m}^{-3} \]