A `Ge` specimen is dopped with `Al`. The concentration of acceptor atoms is `~10^(21) at oms//m^(3)`. Given that the intrinsic concentration of electron hole pairs is `~10^(19)//m^(3)`, the concentration of electron in the speciman is

Options

A. `10^(17)m^(-3)`
B. `10^(15)m^(-3)`
C. `10^(4)m^(-3)`
D. `10^(2)m^(-3)`

Answer

To solve the problem, we need to determine the concentration of electrons in a germanium (Ge) specimen that has been doped with aluminum (Al). The concentration of acceptor atoms (Al) is given as \( N_A \approx 10^{21} \, \text{atoms/m}^3 \) and the intrinsic concentration of electron-hole pairs is \( n_i \approx 10^{19} \, \text{m}^{-3} \).

### Step-by-Step Solution:

1. **Identify the relevant equations**:
The relationship between the concentration of electrons (\( n_e \)), holes (\( n_h \)), and intrinsic carrier concentration (\( n_i \)) in a semiconductor is given by:
\[ n_e \cdot n_h = n_i^2 \]
where \( n_h \) is the concentration of holes, \( n_e \) is the concentration of electrons, and \( n_i \) is the intrinsic carrier concentration.

2. **Determine the concentration of holes**:
In a p-type semiconductor, the concentration of holes (\( n_h \)) can be approximated as equal to the concentration of acceptor atoms since the majority carriers are holes:
\[ n_h \approx N_A = 10^{21} \, \text{m}^{-3} \]

3. **Substitute the known values into the equation**:
We know \( n_i \) and \( n_h \):
\[ n_i = 10^{19} \, \text{m}^{-3} \]
Now substituting into the equation:
\[ n_e \cdot n_h = n_i^2 \]
becomes:
\[ n_e \cdot (10^{21}) = (10^{19})^2 \]

4. **Calculate \( n_i^2 \)**:
\[ n_i^2 = (10^{19})^2 = 10^{38} \, \text{m}^{-6} \]

5. **Solve for \( n_e \)**:
Rearranging the equation gives:
\[ n_e = \frac{n_i^2}{n_h} = \frac{10^{38}}{10^{21}} = 10^{17} \, \text{m}^{-3} \]

6. **Final Result**:
Thus, the concentration of electrons in the specimen is:
\[ n_e = 10^{17} \, \text{m}^{-3} \]