Suppose a pure Si-crystal has `5xx10^(28) "atoms" m^(-3)`. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Give that `n_(i)=1.5xx10^(16)m^(-3)`.

Options

A. `4.5xx10^9 m^(-3)`
B. `4.5xx10^6 m^(-3)`
C. `2.5xx10^9 m^(-3)`
D. `2.5xx10^6 m^(-3)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of pentavalent atoms doped in the silicon crystal.

Given that the concentration of silicon atoms is \(5 \times 10^{28} \, \text{atoms/m}^3\) and the doping concentration is 1 ppm (parts per million), we can express 1 ppm as:

\[ 1 \, \text{ppm} = \frac{1}{10^6} \]

Now, we can calculate the number of pentavalent atoms doped in the silicon crystal:

\[ \text{Number of doped atoms} = \text{Total atoms} \times \text{Doping concentration} \]

\[ \text{Number of doped atoms} = 5 \times 10^{28} \, \text{atoms/m}^3 \times \frac{1}{10^6} = 5 \times 10^{22} \, \text{atoms/m}^3 \]

### Step 2: Determine the number of free electrons contributed by the doped atoms.

Since each pentavalent atom (like arsenic, As) donates one free electron to the silicon crystal, the number of free electrons (\(n_e\)) will be equal to the number of doped atoms:

\[ n_e = 5 \times 10^{22} \, \text{electrons/m}^3 \]

### Step 3: Calculate the hole concentration using the intrinsic carrier concentration.

We are given the intrinsic carrier concentration \(n_i = 1.5 \times 10^{16} \, \text{m}^{-3}\). The relationship between the electron concentration (\(n_e\)), hole concentration (\(n_h\)), and intrinsic carrier concentration (\(n_i\)) is given by:

\[ n_i^2 = n_e \cdot n_h \]

Rearranging this equation to find the hole concentration:

\[ n_h = \frac{n_i^2}{n_e} \]

Substituting the values:

\[ n_h = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} \]

Calculating \(n_i^2\):

\[ n_i^2 = (1.5 \times 10^{16})^2 = 2.25 \times 10^{32} \]

Now substituting back into the equation for \(n_h\):

\[ n_h = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 4.5 \times 10^{9} \, \text{holes/m}^3 \]

### Final Results:

- Number of free electrons (\(n_e\)): \(5 \times 10^{22} \, \text{electrons/m}^3\)
- Number of holes (\(n_h\)): \(4.5 \times 10^{9} \, \text{holes/m}^3\)