If a semiconductor has an intrinsic carrier concentration of `1.41 xx 10^(16)//m^(3)` when doped with `10^(21)//m^(3)` at 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)`

Explanation

To solve the problem, we need to analyze the effect of doping on the intrinsic carrier concentration of a semiconductor. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand Intrinsic Carrier Concentration
The intrinsic carrier concentration (n_i) of the semiconductor is given as:
\[ n_i = 1.41 \times 10^{16} \, \text{m}^{-3} \]

### Step 2: Understand Doping Concentration
The doping concentration (N_d) is given as:
\[ N_d = 10^{21} \, \text{m}^{-3} \]

### Step 3: Determine the Type of Doping
Since the doping concentration is significantly higher than the intrinsic carrier concentration, we can assume that the semiconductor is heavily doped. In this case, we can consider it as n-type doping, which means that the majority carriers will be electrons.

### Step 4: Calculate the Majority and Minority Carriers
For n-type semiconductors, the majority carrier concentration (n) can be approximated as the doping concentration (N_d) when it is much larger than the intrinsic concentration:
\[ n \approx N_d = 10^{21} \, \text{m}^{-3} \]

The minority carrier concentration (p) can be calculated using the mass action law:
\[ n \cdot p = n_i^2 \]
Where \( n_i \) is the intrinsic carrier concentration.

### Step 5: Calculate Minority Carrier Concentration
Using the values:
\[ n_i = 1.41 \times 10^{16} \, \text{m}^{-3} \]
We can find p:
\[ p = \frac{n_i^2}{n} = \frac{(1.41 \times 10^{16})^2}{10^{21}} \]

Calculating \( n_i^2 \):
\[ n_i^2 = (1.41 \times 10^{16})^2 = 1.9881 \times 10^{32} \, \text{m}^{-6} \]

Now substituting back to find p:
\[ p = \frac{1.9881 \times 10^{32}}{10^{21}} = 1.9881 \times 10^{11} \, \text{m}^{-3} \]

### Step 6: Conclusion
Thus, after doping, the majority carrier concentration (n) will be approximately \( 10^{21} \, \text{m}^{-3} \) and the minority carrier concentration (p) will be approximately \( 1.9881 \times 10^{11} \, \text{m}^{-3} \).

### Final Answer
- Majority carrier concentration (n) = \( 10^{21} \, \text{m}^{-3} \)
- Minority carrier concentration (p) = \( 1.9881 \times 10^{11} \, \text{m}^{-3} \)