In a pure silicon `(n_(i)=10^(16)//m^(3))` crystal at `300K, 10^(21)` atoms of phosphorus are added per cubic meter. The new hole concentration will be

Options

A. `10^(21) per m^(3)`
B. `10^(19) per m^(3)`
C. `10^(11) per m^(3)`
D. `10^(5) per m^(3)`

Explanation

To find the new hole concentration in a silicon crystal after doping with phosphorus, we can follow these steps:

### Step 1: Understand the given data
- Intrinsic carrier concentration of silicon, \( n_i = 10^{16} \, \text{m}^{-3} \)
- Concentration of phosphorus atoms added, \( N_D = 10^{21} \, \text{m}^{-3} \)
- Temperature, \( T = 300 \, \text{K} \)

### Step 2: Determine the electron concentration
When phosphorus is added to silicon, it acts as a donor, contributing free electrons to the conduction band. The electron concentration \( n \) in the n-type silicon can be approximated as:
\[ n \approx N_D \]
Thus, the electron concentration after doping is:
\[ n \approx 10^{21} \, \text{m}^{-3} \]

### Step 3: Use the mass action law
The mass action law states that the product of the electron concentration \( n \) and hole concentration \( p \) in intrinsic semiconductors is constant and equal to the square of the intrinsic carrier concentration:
\[ n \cdot p = n_i^2 \]
Where \( p \) is the hole concentration we want to find.

### Step 4: Rearrange the equation to find hole concentration
We can rearrange the equation to solve for \( p \):
\[ p = \frac{n_i^2}{n} \]

### Step 5: Substitute the known values
Now, substituting the known values into the equation:
\[ p = \frac{(10^{16})^2}{10^{21}} = \frac{10^{32}}{10^{21}} = 10^{32 - 21} = 10^{11} \, \text{m}^{-3} \]

### Step 6: Conclusion
The new hole concentration in the silicon crystal after doping with phosphorus is:
\[ p = 10^{11} \, \text{m}^{-3} \]