For skywave propagation of a $10 M H z$ signal, what should be the minimum electron density in ionosphere?

1.2 \times 10^{12} m^{- 3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.2 \times 10^{16} m^{- 3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.2 \times 10^{22} m^{- 3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.2 \times 10^{6} m^{- 3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $1.2 \times 10^{12} m^{- 3}$
The minimum electron density for skywave propogation is given by
$f_{c} = 9 \times \sqrt{N}$
Here $f_{c} = 10 M H z$
i.e $10^{7} = 9 \times \sqrt{N}$
$∴ N = 1.2 \times 10^{12} m^{- 3}$

Suggest Corrections

Similar questions

Q. For sky wave propagation of

10 M H z

signal, what should be the minimum electron density in ionosphere?

A pure semiconductor has equal electron and hole concentration of $10^{16} m^{- 3}$ . Doping by indium increases $n_{h}$ to $4.5 \times 10^{22} m^{- 3} . W h a t i s t h e v a l u e o f$ n_e$ in the doped semiconductor?

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

\frac{1.2 \times 1.2 + 1.2 \times 0.3 + 0.3 \times 0.3}{1.2 \times 1.2 \times 1.2 - 0.3 \times 0.3 \times 0.3} = \frac{m}{1.8}, F i n d m

.Write upto 3 decimal

The density of a liquid (mol. wt. = 70) is 1.2 g $m L^{- 1}$ . If 2 mL of liquid contains 35 drops, the number of molecules of liquid in one drop are :

Join BYJU'S Learning Program

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

The correct option is A 1.2×1012 m−3The minimum electron density for skywave propogation is given by fc=9×√N Here fc=10 MHz i.e 107=9×√N ∴N=1.2×1012 m−3

For skywave propagation of a $10 M H z$ signal, what should be the minimum electron density in ionosphere?

The correct option is A $1.2 \times 10^{12} m^{- 3}$
The minimum electron density for skywave propogation is given by
$f_{c} = 9 \times \sqrt{N}$
Here $f_{c} = 10 M H z$
i.e $10^{7} = 9 \times \sqrt{N}$
$∴ N = 1.2 \times 10^{12} m^{- 3}$