For sky wave propagation of a `10 MHz` signal, what should be the minimum electron density in ionosphere?

Options

A. `1.2 xx 10^(8) m^(-3)`
B. `1.2 xx 10^(12) m^(-3)`
C. `1.2 xx 10^(14)m^(-3)`
D. `1.2 xx 10^(16) m^(-3)`

Answer

To find the minimum electron density in the ionosphere for sky wave propagation of a 10 MHz signal, we can follow these steps:

### Step 1: Convert Frequency to Hertz
The given frequency is 10 MHz. We can convert this to hertz:
\[ f = 10 \text{ MHz} = 10 \times 10^6 \text{ Hz} = 10^7 \text{ Hz} \]

### Step 2: Use the Relationship Between Frequency and Electron Density
The critical frequency \( f_c \) is related to the electron density \( n \) by the formula:
\[ f_c = 9 \sqrt{n} \]
where \( f_c \) is the critical frequency in hertz and \( n \) is the electron density in electrons per cubic meter.

### Step 3: Rearranging the Formula
We can rearrange the formula to solve for \( n \):
\[ n = \left(\frac{f_c}{9}\right)^2 \]

### Step 4: Substitute the Frequency into the Formula
Now we substitute \( f_c = 10^7 \text{ Hz} \) into the equation:
\[ n = \left(\frac{10^7}{9}\right)^2 \]

### Step 5: Calculate the Value
Calculating the right-hand side:
\[ n = \left(\frac{10^7}{9}\right)^2 = \frac{(10^7)^2}{9^2} = \frac{10^{14}}{81} \]

### Step 6: Simplify the Expression
To simplify \( \frac{10^{14}}{81} \), we can express it in a more manageable form:
\[ n = \frac{100}{81} \times 10^{12} \]

### Step 7: Final Calculation
Calculating \( \frac{100}{81} \):
\[ \frac{100}{81} \approx 1.23457 \approx 1.23 \]
Thus, we have:
\[ n \approx 1.23 \times 10^{12} \text{ electrons/m}^3 \]

### Final Answer
The minimum electron density in the ionosphere for sky wave propagation of a 10 MHz signal is approximately:
\[ n \approx 1.23 \times 10^{12} \text{ electrons/m}^3 \]

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