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For sky wave propagation of a ` 15 MHz` signal what should be the minimum electron density in ionosphere ? |
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Answer» Correct Answer - `2.78 xx 10^(912) m^(-3)` Here , ` v_(c ) = 15 MHz = 15 xx 10^(6) Hz , N_(max) = ?` As ` v_( c ) = 9 (N_(max))^(1//2)` or `N_(max) = ((v_(c ))/( 9))^(2) = (v_( c)^(2))/( 81) = ((15 xx 10^(6))^(2))/(81)` ` = 2.78 xx 10^(12) m^(-3)` |
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