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Chapter 20: Problem 2711

On a particular day, the maximum frequency reflected from the ionosphere is $8 \mathrm{MHz}\(. On another day it was found to increase to \)9 \mathrm{MHz}$. The ratio of the maximum electron densities of ionosphere on the two days is (A) \(\sqrt{(8 / 9)}\) (B) \((8 / 9)\) (C) \((8 / 9)^{(3 / 2)}\) (D) \((8 / 9)^{2}\)

Short Answer

Expert verified
The ratio of the maximum electron densities of ionosphere on the two days is \((\frac{8}{9})^2\).

Step by step solution

01

Critical frequency formula

The critical frequency (\(f_{c}\)), which is the highest frequency that can be reflected by the ionosphere for a given layer, is directly related to the electron density (\(N\)) of the ionosphere. This is given by the formula: \(f_{c} = 9 \sqrt{N}\) where \(f_{c}\) is in MHz and \(N\) is in electrons per cubic meter (el/m³).
02

Determine the electron densities

Using the given maximum frequencies, we can find the electron densities for each day using the critical frequency formula from Step 1. For day 1 (\(f_{c1} = 8\mathrm{MHz}\)): \(N_1 = \left(\frac{f_{c1}}{9}\right)^2\) For day 2 (\(f_{c2} = 9\mathrm{MHz}\)): \(N_2 = \left(\frac{f_{c2}}{9}\right)^2\)
03

Find the ratio of electron densities

Divide \(N_2\) by \(N_1\) to find the ratio of the maximum electron densities on the two days: \(\frac{N_2}{N_1} = \frac{(\frac{f_{c2}}{9})^2}{(\frac{f_{c1}}{9})^2} = \frac{f_{c2}^2}{f_{c1}^2}\)
04

Plug in given values and match the answer

Insert the given maximum frequencies into the ratio: \(\frac{N_2}{N_1} = \frac{(9 \ \mathrm{MHz})^2}{(8 \ \mathrm{MHz})^2} = \frac{81}{64} = \left(\frac{9}{8}\right)^2\) This matches option (D): \(\left(\frac{8}{9}\right)^2\). So, the ratio of the maximum electron densities of ionosphere on the two days is \((\frac{8}{9})^2\).

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