An intrinsic semi conductor has `10^(18) //m^3` free electron and is doped with pentavalent impurity of `10^(24) //m^3`. Then the free electrons density order increase by

Options

A. 4
B. 3
C. 5
D. 6

Answer

To solve the problem, we need to determine how much the density of free electrons in the intrinsic semiconductor increases when it is doped with a pentavalent impurity.

### Step-by-step Solution:

1. **Identify the initial density of free electrons**:
The intrinsic semiconductor has an initial free electron density of \( n_i = 10^{18} \, \text{m}^{-3} \).

2. **Identify the density of doped electrons**:
When the semiconductor is doped with a pentavalent impurity, the density of free electrons contributed by the dopant is \( n_d = 10^{24} \, \text{m}^{-3} \).

3. **Calculate the total density of free electrons after doping**:
The total density of free electrons after doping can be calculated as:
\[ n_{total} = n_i + n_d = 10^{18} + 10^{24} \]
Since \( n_d \) is much larger than \( n_i \), we can approximate:
\[ n_{total} \approx n_d = 10^{24} \, \text{m}^{-3} \]

4. **Determine the increase in electron density**:
The increase in the density of free electrons due to doping is given by:
\[ \Delta n = n_{total} - n_i = 10^{24} - 10^{18} \]
Again, since \( n_d \) is much larger than \( n_i \), we can approximate:
\[ \Delta n \approx 10^{24} \]

5. **Calculate the order of increase**:
To find the order of increase, we can take the ratio of the new density to the original density:
\[ \text{Order of increase} = \frac{n_{total}}{n_i} = \frac{10^{24}}{10^{18}} = 10^{24 - 18} = 10^{6} \]

6. **Conclusion**:
The order of increase in the free electron density is \( 6 \).

### Final Answer:
The free electron density increases by an order of \( 6 \).