Question

The maximum electron density of a layer of the ionosphere is $9 \times 10^{12} \mathrm{~m}^{-3}$. The critical frequency of this layer is (A) \(9 \mathrm{MHz}\) (B) \(6 \mathrm{MHz}\) (C) \(27 \mathrm{MHz}\) (D) \(3 \mathrm{GHz}\)

Short answer

The critical frequency, \(f_c\), can be calculated using the formula \(f_c = \dfrac{1}{2\pi} \sqrt{\dfrac{e^2N}{\varepsilon_0 m_e}}\), where the maximum electron density, N, is given as \(9 \times 10^{12} \mathrm{~m}^{-3}\). Plugging in the values and constants into the formula, we find that the critical frequency \(f_c \approx 9 \mathrm{MHz}\). Thus, the correct answer is (A) \(9 \mathrm{MHz}\).

Step-by-step solution

Plug in the values

Plug the given values and constants into the formula: \(f_c = \dfrac{1}{2\pi} \sqrt{\dfrac{(1.6 \times 10^{-19})^2(9 \times 10^{12})}{(8.85 \times 10^{-12})(9.11 \times 10^{-31})}}\)

Calculate the critical frequency

Now, perform the calculation to find the critical frequency: \(f_c = \dfrac{1}{2\pi} \sqrt{\dfrac{(2.56 \times 10^{-38})(9 \times 10^{12})}{(8.85 \times 10^{-12})(9.11 \times 10^{-31})}}\) \(f_c = \dfrac{1}{2\pi} \sqrt{\dfrac{2.304 \times 10^{-25}}{8.06 \times 10^{-42}}}\) \(f_c = \dfrac{1}{2\pi} \sqrt{2.857 \times 10^{16}}\) \(f_c \approx 9 \times 10^{6}\) Thus, the critical frequency \(f_c \approx 9 \mathrm{MHz}\). The correct answer is (A) \(9 \mathrm{MHz}\).