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`

Answer

To solve the problem of finding the new hole concentration in a silicon crystal after adding phosphorus atoms, 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: Use the mass action law
The mass action law states that:
\[ n_i^2 = n_e \cdot n_h \]
where:
- \( n_e \) is the electron concentration
- \( n_h \) is the hole concentration

### Step 3: Determine the electron concentration
When phosphorus (a donor impurity) is added to silicon, it increases the electron concentration. In n-type semiconductors, the concentration of electrons can be approximated as:
\[ n_e \approx N_D \]
Thus, we have:
\[ n_e = 10^{21} \, \text{m}^{-3} \]

### Step 4: Substitute into the mass action law
Now we can substitute \( n_i \) and \( n_e \) into the mass action law to find \( n_h \):
\[ n_h = \frac{n_i^2}{n_e} \]

### Step 5: Calculate the hole concentration
Substituting the values we have:
\[ n_h = \frac{(10^{16})^2}{10^{21}} = \frac{10^{32}}{10^{21}} = 10^{32 - 21} = 10^{11} \, \text{m}^{-3} \]

### Final Answer
The new hole concentration \( n_h \) is:
\[ n_h = 10^{11} \, \text{m}^{-3} \]

### Summary
The new hole concentration in the silicon crystal after adding \( 10^{21} \) atoms of phosphorus per cubic meter is \( 10^{11} \, \text{m}^{-3} \).

---