In sample of pure silicon `10^(13) "atom"//cm^(3)` is mixed of phosphorus. If all doner atoms are active then what will be resistivity at `20^(@)C` if mobility of electron is `1200 cm^(2)//"Volt"` sec :-

Options

A. `0.5209` ohm cm
B. `5.209` ohm cm
C. `52.09` ohm cm
D. `520.9` ohm cm

Explanation

To solve the problem, we need to calculate the resistivity (ρ) of a silicon sample doped with phosphorus. We are given the following data:

- Concentration of donor atoms (n) = \(10^{13} \, \text{atoms/cm}^3\)
- Mobility of electrons (μ) = \(1200 \, \text{cm}^2/\text{V}\cdot\text{s}\)
- Charge of an electron (e) = \(1.6 \times 10^{-19} \, \text{C}\)

The formula for resistivity (ρ) in terms of electron concentration (n), electron mobility (μ), and charge of the electron (e) is given by:

\[ \rho = \frac{1}{n \cdot e \cdot \mu} \]

### Step-by-step Solution:

1. **Identify the values**:
- \(n = 10^{13} \, \text{atoms/cm}^3\)
- \(e = 1.6 \times 10^{-19} \, \text{C}\)
- \(\mu = 1200 \, \text{cm}^2/\text{V}\cdot\text{s}\)

2. **Substitute the values into the resistivity formula**:
\[ \rho = \frac{1}{(10^{13}) \cdot (1.6 \times 10^{-19}) \cdot (1200)} \]

3. **Calculate the denominator**:
- First, calculate \(n \cdot e\):
\[ n \cdot e = (10^{13}) \cdot (1.6 \times 10^{-19}) = 1.6 \times 10^{-6} \]
- Then, calculate \(n \cdot e \cdot \mu\):
\[ n \cdot e \cdot \mu = (1.6 \times 10^{-6}) \cdot (1200) = 1.92 \times 10^{-3} \]

4. **Calculate resistivity**:
\[ \rho = \frac{1}{1.92 \times 10^{-3}} \approx 520.83 \, \Omega \cdot \text{cm} \]

5. **Round the answer**:
- The resistivity can be rounded to \(520.9 \, \Omega \cdot \text{cm}\).

### Final Answer:
The resistivity of the silicon sample at \(20^\circ C\) is approximately \(520.9 \, \Omega \cdot \text{cm}\).