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The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be $\mathrm{f}_{\max }=9$ $\left(\mathrm{N}_{\operatorname{mux}}\right)^{1 / 2}$, where $\mathrm{N}_{\max }$ is the maximum electron density at that layer of the ionosphere on a certain day it is observed that signals of frequencies higher than $5 \mathrm{MHz}$ are not received by reflection from the $\mathrm{F}_1$ layer of the ionosphere while signals of frequencies higher than $8 \mathrm{MHz}$ are not received by reflection from the $\mathrm{F}_2$ layer of the ionosphere. Estimate the maximum electron densities of the $\mathrm{F}_1$ and $\mathrm{F}_2$ layers on that day.
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Verified Answer
As given that
The maximum frequency for reflection of sky waves
$$
\mathrm{f}_{\max }=9\left(\mathrm{~N}_{\max }\right)^{1 / 2}
$$
where, $\mathrm{N}_{\max }$ is a maximum electron density
For $F_1$ layer,
$$
\mathrm{f}_{\max }=5 \mathrm{MHz}=5 \times 10^6 \mathrm{~Hz}
$$
By equating (i) and (ii),
$$
5 \times 10^6=9\left(\mathrm{~N}_{\text {max }}\right)^{1 / 2}
$$
Maximum electron density
from (i) equation, $\mathrm{N}_{\max }=\left(\frac{f_{\max }}{9}\right)^2$
$$
\mathrm{N}_{\max }=\left(\frac{5}{9} \times 10^6\right)^2=3.086 \times 10^{11} / \mathrm{m}^3
$$
For $F_2$ layer,
$$
\mathrm{f}_{\text {max }}=8 \mathrm{MHz}=8 \times 10^6 \mathrm{~Hz}
$$
Byequating (i) and (iii),
$$
8 \times 10^6=9\left(\mathrm{~N}_{\max }\right)^{1 / 2}
$$
Maximum electron density
Byequation (i), $\mathrm{N}_{\max }=\left(\frac{\mathrm{f}_{\max }}{9}\right)^2$
$$
\mathrm{N}_{\max }=\left(\frac{8 \times 10^6}{9}\right)^2=7.9 \times 10^{11} / \mathrm{m}^3
$$
So maximum electron density for layer $\mathrm{F}_2$ is $\mathrm{N}_{\max }$ $=7.9 \times 10^{11} \mathrm{e} / \mathrm{m}^3$.
The maximum frequency for reflection of sky waves
$$
\mathrm{f}_{\max }=9\left(\mathrm{~N}_{\max }\right)^{1 / 2}
$$
where, $\mathrm{N}_{\max }$ is a maximum electron density
For $F_1$ layer,
$$
\mathrm{f}_{\max }=5 \mathrm{MHz}=5 \times 10^6 \mathrm{~Hz}
$$
By equating (i) and (ii),
$$
5 \times 10^6=9\left(\mathrm{~N}_{\text {max }}\right)^{1 / 2}
$$
Maximum electron density
from (i) equation, $\mathrm{N}_{\max }=\left(\frac{f_{\max }}{9}\right)^2$
$$
\mathrm{N}_{\max }=\left(\frac{5}{9} \times 10^6\right)^2=3.086 \times 10^{11} / \mathrm{m}^3
$$
For $F_2$ layer,
$$
\mathrm{f}_{\text {max }}=8 \mathrm{MHz}=8 \times 10^6 \mathrm{~Hz}
$$
Byequating (i) and (iii),
$$
8 \times 10^6=9\left(\mathrm{~N}_{\max }\right)^{1 / 2}
$$
Maximum electron density
Byequation (i), $\mathrm{N}_{\max }=\left(\frac{\mathrm{f}_{\max }}{9}\right)^2$
$$
\mathrm{N}_{\max }=\left(\frac{8 \times 10^6}{9}\right)^2=7.9 \times 10^{11} / \mathrm{m}^3
$$
So maximum electron density for layer $\mathrm{F}_2$ is $\mathrm{N}_{\max }$ $=7.9 \times 10^{11} \mathrm{e} / \mathrm{m}^3$.
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