On certain day, it is observed that the signals of higher than 5 MHz are not recived by reflection from `F_(1)` layer of ionosphere. The approximate maximum electron density of `F_(1)` layer ontheday is

Options

A. `7.9xx10^(11)//m^(3)`
B. `5xx10^(11)//m^(3)`
C. `4.2xx10^(11)//m^(3)`
D. `3xx10^(11)//m^(3)`

Answer

To solve the problem, we need to determine the maximum electron density of the F1 layer of the ionosphere based on the given frequency at which signals are not received.

### Step-by-Step Solution:

1. **Identify the given frequency**:
The problem states that signals higher than 5 MHz are not received. Therefore, we have:
\[ f_c = 5 \text{ MHz} = 5 \times 10^6 \text{ Hz} \]

2. **Use the formula for critical frequency**:
The critical frequency \( f_c \) is related to the maximum electron density \( n_{max} \) by the formula:
\[ f_c = 9 \sqrt{n_{max}} \]
Rearranging this formula to solve for \( n_{max} \):
\[ n_{max} = \left(\frac{f_c}{9}\right)^2 \]

3. **Substitute the value of \( f_c \)**:
Plugging in the value of \( f_c \):
\[ n_{max} = \left(\frac{5 \times 10^6}{9}\right)^2 \]

4. **Calculate \( \frac{5 \times 10^6}{9} \)**:
\[ \frac{5 \times 10^6}{9} \approx 5.5556 \times 10^5 \]

5. **Square the result**:
Now, we square this value:
\[ n_{max} \approx (5.5556 \times 10^5)^2 \approx 3.0864 \times 10^{11} \]

6. **Express in scientific notation**:
Rounding to significant figures, we can express this as:
\[ n_{max} \approx 3.1 \times 10^{11} \text{ m}^{-3} \]

7. **Select the closest option**:
Among the given options, \( 3 \times 10^{11} \text{ m}^{-3} \) is the closest to our calculated value.

### Final Answer:
The approximate maximum electron density of the F1 layer on that day is:
\[ \boxed{3 \times 10^{11} \text{ m}^{-3}} \]