The maximum electron density in the ionospherein the mornong is `10^(10)m^(-3)`. At noon time it increases to `2xx10^(10)m^(-3)`. Find the ratio of critical frequency at noon and the critical frequency in the morning.

Options

A. 2
B. 2.82
C. 4
D. 1.414

Answer

To solve the problem, we need to find the ratio of the critical frequency at noon to the critical frequency in the morning based on the maximum electron densities given.

### Step-by-Step Solution:

1. **Identify the maximum electron densities**:
- Maximum electron density in the morning, \( n_{morning} = 10^{10} \, m^{-3} \)
- Maximum electron density at noon, \( n_{noon} = 2 \times 10^{10} \, m^{-3} \)

2. **Use the formula for critical frequency**:
The critical frequency \( f \) is given by the formula:
\[ f = 9 \sqrt{n_{max}} \]
where \( n_{max} \) is the maximum electron density.

3. **Calculate the critical frequency in the morning**:
\[ f_{morning} = 9 \sqrt{n_{morning}} = 9 \sqrt{10^{10}} = 9 \times 10^{5} \, Hz \]

4. **Calculate the critical frequency at noon**:
\[ f_{noon} = 9 \sqrt{n_{noon}} = 9 \sqrt{2 \times 10^{10}} = 9 \sqrt{2} \times 10^{5} \, Hz \]

5. **Find the ratio of critical frequencies**:
\[ \text{Ratio} = \frac{f_{noon}}{f_{morning}} = \frac{9 \sqrt{2} \times 10^{5}}{9 \times 10^{5}} \]
The \( 9 \) and \( 10^{5} \) cancel out:
\[ \text{Ratio} = \sqrt{2} \]

6. **Calculate the numerical value of the ratio**:
\[ \sqrt{2} \approx 1.414 \]

### Final Answer:
The ratio of the critical frequency at noon to the critical frequency in the morning is \( \sqrt{2} \) or approximately \( 1.414 \).