In a pure silicon `(n_(i)=10^(16)m^(-3))` crystal at 300 `K, 10^(23)` atoms of phosphorus are added per cubic meter. The new whole concentration will be

Options

A. `10^(23)m^(-3)`
B. `10^(10)m^(-3)`
C. `10^(11) m^(-3)`
D. `10^(9) m^(-3)`

Answer

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

### Step 1: Understand the intrinsic carrier concentration
In pure silicon at 300 K, the intrinsic carrier concentration (n_i) is given as:
\[ n_i = 10^{16} \, \text{m}^{-3} \]

### Step 2: Calculate the intrinsic hole concentration
In intrinsic semiconductors, the number of holes (N_H) is equal to the number of electrons (N_E). The relationship can be expressed as:
\[ N_H \cdot N_E = n_i^2 \]
Since \( N_H = N_E \) in intrinsic silicon, we can write:
\[ N_H^2 = n_i^2 \]
Thus, the hole concentration in intrinsic silicon is:
\[ N_H = N_E = n_i = 10^{16} \, \text{m}^{-3} \]

### Step 3: Add phosphorus atoms
When phosphorus (a donor impurity) is added, it introduces extra electrons into the silicon. The concentration of phosphorus atoms added is:
\[ N_D = 10^{23} \, \text{m}^{-3} \]

### Step 4: Calculate the new electron concentration
The new electron concentration (N_E') after doping can be approximated as:
\[ N_E' \approx N_D + n_i \]
Since \( N_D \) is much larger than \( n_i \), we can simplify this to:
\[ N_E' \approx N_D = 10^{23} \, \text{m}^{-3} \]

### Step 5: Calculate the new hole concentration
Using the relation from Step 2:
\[ N_H' = \frac{n_i^2}{N_E'} \]
Substituting the values we have:
\[ N_H' = \frac{(10^{16})^2}{10^{23}} \]
\[ N_H' = \frac{10^{32}}{10^{23}} \]
\[ N_H' = 10^{9} \, \text{m}^{-3} \]

### Conclusion
The new hole concentration after doping with phosphorus is:
\[ N_H' = 10^{9} \, \text{m}^{-3} \]