Pure `Si` at `500K` has equal number of electron `(n_(e))` and hole `(n_(h))` concentration of `1.5xx10^(16)m^(-3)`. Dopping by indium. Increases `n_(h)` to `4.5xx10^(22) m^(-3)`. The doped semiconductor is of

Options

A. P - type having electron concentrations `n_(e)=5xx10^(9)m^(-3)`
B. n - Type with electron concentrations `n_(e)=5xx10^(22)m^(-3)`
C. P - Type with electron concentrations `n_(e)=2.5xx10^(10)m^(-3)`
D. n - Type with electron concentrations `n_(e)=2.5xx10^(23)m^(-3)`

Explanation

To solve the problem, we need to analyze the given information about the intrinsic and doped silicon semiconductor.

### Step-by-Step Solution:

1. **Identify the intrinsic carrier concentration**:
The intrinsic carrier concentration of pure silicon at 500 K is given as:
\[ n_i = n_e = n_h = 1.5 \times 10^{16} \, \text{m}^{-3} \]

2. **Determine the hole concentration after doping**:
When silicon is doped with indium, the hole concentration increases to:
\[ n_h = 4.5 \times 10^{22} \, \text{m}^{-3} \]

3. **Use the mass action law**:
According to the mass action law for semiconductors, we have:
\[ n_i^2 = n_e \cdot n_h \]
Where:
- \( n_i \) is the intrinsic carrier concentration,
- \( n_e \) is the electron concentration,
- \( n_h \) is the hole concentration.

4. **Substituting the values into the equation**:
We can substitute the known values into the equation:
\[ (1.5 \times 10^{16})^2 = n_e \cdot (4.5 \times 10^{22}) \]

5. **Calculate \( n_e \)**:
First, calculate \( (1.5 \times 10^{16})^2 \):
\[ (1.5 \times 10^{16})^2 = 2.25 \times 10^{32} \]
Now, substituting this back into the equation:
\[ 2.25 \times 10^{32} = n_e \cdot (4.5 \times 10^{22}) \]
To find \( n_e \), rearrange the equation:
\[ n_e = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}} = 5 \times 10^{9} \, \text{m}^{-3} \]

6. **Determine the type of semiconductor**:
Now we compare the concentrations of holes and electrons:
- Hole concentration \( n_h = 4.5 \times 10^{22} \, \text{m}^{-3} \)
- Electron concentration \( n_e = 5 \times 10^{9} \, \text{m}^{-3} \)

Since \( n_h \) is much greater than \( n_e \), the semiconductor is classified as a **P-type semiconductor**.

### Final Answer:
The doped semiconductor is a **P-type semiconductor** with an electron concentration of \( 5 \times 10^{9} \, \text{m}^{-3} \).

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