The density of an electron-hole pair in a pure germanium is `3xx 10^(16) m^(-3)` at room temperature. On doping with aluminium, the hole density increases to `4.5 xx 10^(22) m^(-3)`. Now the electron density ( in `m^(-3))` in doped germanium will be

Options

A. `1xx10^(10)`
B. `2xx10^(10)`
C. `0.5xx10^(10)`
D. `4xx10^(10)`

Answer

To solve the problem, we need to find the electron density in doped germanium after doping with aluminum, given the initial conditions of pure germanium.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The density of electron-hole pairs in pure germanium at room temperature is given as:
\[ n_i = p_i = 3 \times 10^{16} \, \text{m}^{-3} \]
- Here, \( n_i \) is the initial electron density and \( p_i \) is the initial hole density.

2. **Identify Doping Conditions**:
- After doping with aluminum, the hole density increases to:
\[ p_f = 4.5 \times 10^{22} \, \text{m}^{-3} \]
- Here, \( p_f \) is the final hole density after doping.

3. **Use the Mass Action Law**:
- According to the mass action law, the product of the electron density (\( n \)) and hole density (\( p \)) remains constant and is approximately equal to the square of the intrinsic carrier concentration:
\[ n \cdot p \approx n_i^2 \]
- Therefore, we can write:
\[ n_f \cdot p_f \approx n_i^2 \]
- Where \( n_f \) is the electron density after doping.

4. **Substituting Known Values**:
- Substitute the known values into the equation:
\[ n_f \cdot (4.5 \times 10^{22}) \approx (3 \times 10^{16})^2 \]
- Calculate \( (3 \times 10^{16})^2 \):
\[ (3 \times 10^{16})^2 = 9 \times 10^{32} \]

5. **Rearranging the Equation**:
- Rearranging the equation to solve for \( n_f \):
\[ n_f = \frac{9 \times 10^{32}}{4.5 \times 10^{22}} \]

6. **Calculating \( n_f \)**:
- Simplifying the fraction:
\[ n_f = \frac{9}{4.5} \times 10^{32 - 22} = 2 \times 10^{10} \, \text{m}^{-3} \]

### Final Answer:
The electron density in doped germanium will be:
\[ n_f = 2 \times 10^{10} \, \text{m}^{-3} \]