If a semi conductor has an intrinsic carrier concentration of` 1.41 xx 10^(16)// m^3`. When doped with `10^(21)// m^3` phosphorous atoms, then the concentration of holes / mat room temperature will be

Options

A. `2 xx 10^(21)`
B. `2 xx 10^(11)`
C. `1.41 xx 10^(10)`
D. `1.41 xx 10^(16)`

Answer

To solve the problem, we need to understand the behavior of intrinsic semiconductors and how doping affects the concentration of charge carriers, specifically holes.

### Step-by-Step Solution:

1. **Understand Intrinsic Carrier Concentration**:
The intrinsic carrier concentration (Ni) of the semiconductor is given as \(1.41 \times 10^{16} \, \text{m}^{-3}\). This value represents the concentration of both electrons (n) and holes (p) in an intrinsic (undoped) semiconductor at thermal equilibrium.

2. **Doping with Phosphorus**:
When the semiconductor is doped with phosphorus atoms, which are donor impurities, the concentration of electrons increases. The doping concentration is given as \(10^{21} \, \text{m}^{-3}\).

3. **Effect of Doping on Hole Concentration**:
Doping with donor atoms like phosphorus increases the number of electrons in the conduction band. However, it does not affect the concentration of holes directly. In a semiconductor, the relationship between electron concentration (n) and hole concentration (p) can be described by the mass action law:
\[ n \cdot p = N_i^2 \]
where \(N_i\) is the intrinsic carrier concentration.

4. **Calculate the New Electron Concentration**:
After doping, the electron concentration (n) can be approximated as:
\[ n \approx N_d + N_i \]
where \(N_d\) is the concentration of donor atoms. Since \(N_d\) is much larger than \(N_i\), we can simplify this to:
\[ n \approx N_d = 10^{21} \, \text{m}^{-3} \]

5. **Using Mass Action Law to Find Hole Concentration**:
Now, we can use the mass action law to find the hole concentration (p):
\[ p = \frac{N_i^2}{n} \]
Substituting the values:
\[ p = \frac{(1.41 \times 10^{16})^2}{10^{21}} \]
\[ p = \frac{1.9881 \times 10^{32}}{10^{21}} = 1.9881 \times 10^{11} \, \text{m}^{-3} \]

6. **Conclusion**:
Therefore, the concentration of holes at room temperature after doping with phosphorus is approximately \(1.9881 \times 10^{11} \, \text{m}^{-3}\).

### Final Answer:
The concentration of holes at room temperature will be approximately \(1.9881 \times 10^{11} \, \text{m}^{-3}\).