The number density of electrons and holes in pure silicon at `27^@` C are equal and its value is `2.0 xx 10^16 m^(-3)`. On doping with indium the hole density increases to `4.5xx 10^22 m^(-3)`, the electron density in doped silicon is

Options

A. `10xx10^9 m^(-3)`
B. `8.89xx10^9 m^(-3)`
C. `11xx10^9 m^(-3)`
D. `16.78xx10^9 m^(-3)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
In pure silicon at \(27^\circ C\), the number density of electrons (\(n_i\)) and holes (\(p_i\)) are equal. Given:
\[ n_i = p_i = 2.0 \times 10^{16} \, \text{m}^{-3} \]

### Step 2: Identify the change after doping
After doping with indium, the hole density increases to:
\[ p' = 4.5 \times 10^{22} \, \text{m}^{-3} \]

### Step 3: Use the mass action law
According to the mass action law, the product of the electron density (\(n'\)) and hole density (\(p'\)) is constant at a given temperature:
\[ n' \cdot p' = n_i^2 \]
Where \(n_i\) is the intrinsic carrier concentration.

### Step 4: Substitute known values
We can rearrange the equation to find the new electron density (\(n'\)):
\[ n' = \frac{n_i^2}{p'} \]
Substituting the values:
\[ n' = \frac{(2.0 \times 10^{16})^2}{4.5 \times 10^{22}} \]

### Step 5: Calculate \(n'\)
Calculating \(n_i^2\):
\[ (2.0 \times 10^{16})^2 = 4.0 \times 10^{32} \]
Now substituting this into the equation:
\[ n' = \frac{4.0 \times 10^{32}}{4.5 \times 10^{22}} = \frac{4.0}{4.5} \times 10^{32 - 22} = \frac{4.0}{4.5} \times 10^{10} \]
Calculating the fraction:
\[ \frac{4.0}{4.5} \approx 0.8889 \]
Thus:
\[ n' \approx 0.8889 \times 10^{10} = 8.889 \times 10^{9} \, \text{m}^{-3} \]

### Final Answer
The electron density in the doped silicon is approximately:
\[ n' \approx 8.89 \times 10^{9} \, \text{m}^{-3} \]

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