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I saw this graph about the electrons density in different altitudes and difference between night and day, the difference between the 2 electron densities (day and night) decreases till 300 Km (F2 layer) and then the difference increases again.

So, I wanted to know why is the recombination rate in F2 layer very low.

  • 1
    $\begingroup$ May not be necessary for getting a good answer, but if that graph is online, linking it (or even including it directly if copyright permits) could be helpful, especially for those of us not intimately familiar with ionospheric physics. $\endgroup$
    – user10851
    Commented Jan 8, 2013 at 15:26
  • $\begingroup$ The F-region is the point where most UV-light is absorbed, which continuously produces free electrons there. It is also the place where heavy ions (e.g., oxygen) become much less abundant than the lighter ones (e.g., hydrogen). The gas there is dominated by charged, not neutral particles which means it acts like a plasma. I will have to look into why the recombination rate is lower but I am guessing it's due to a higher temperature there. $\endgroup$ Commented Oct 18, 2016 at 15:56

1 Answer 1



The ionosphere is the boundary layer between the Earth's neutral atmosphere and the ionized, magnetized gas surrounding it called the magnetosphere (technically, the ionosphere is part of Earth's thermosphere). At low altitudes, the charge--neutral interactions can play a dominant role and even control the dynamics of the gas (e.g., neutral winds can push charged particles around affecting radio communications in some cases). At higher altitudes (e.g., F layer and above), the gas starts to behave more like a quasi-neutral gas exhibiting a collective behavior based upon long-range forces, called a plasma.

Though I should note that even at lower altitudes, the gas can be characterized as a plasma too. The distinction largely results from the region where radio waves are affected by variations in the free electron density, which determine the local plasma frequency ($\omega_{pe}^{2} = \tfrac{n_{e} \ e^{2}}{\varepsilon_{o} \ m_{e}}$).


The volumetric recombination rate is defined as (for an electron-proton plasma): $$ R = n_{e} \ n_{p} \ \alpha\left( H^{0}, T \right) $$ where $n_{s}$ is the number density of species $s$, $\alpha\left( H^{0}, T \right)$ is the recombination rate coefficient, $H^{0}$ is the first ionization potential of hydrogen (i.e., ~13.59 eV), and $T$ is the temperature of the electrons.

The recombination rate coefficient depends upon temperature roughly as $\alpha \propto T^{-1/2}$, thus at higher temperatures the rate drops. The recombination time -- duration necessary to neutralize a plasma in a given volume -- inversely depends upon the free electron density and $\alpha$, or $t_{rec} \approx \left( n_{e} \ \alpha \right)^{-1} \sim \sqrt{ \tfrac{T}{n_{e}^{2}} }$.

The total ionization state of the gas is generally defined by the Saha equation.

So, I wanted to know why is the recombination rate in F2 layer very low.

This is due to the increase in density and temperature, which causes $t_{rec} \sim \sqrt{ \tfrac{T}{n_{e}^{2}} }$ to increase. Meaning, the time necessary to neutralize the gas gets much larger in this region due to the increasing density and temperature. There is also the issue of there being fewer heavy ions (e.g., oxygen) which reduces the total ionization rate.

Interesting Examples

A Strömgren sphere describes the boundary between mostly neutral and mostly ionized gas. In other words, it is a sharp boundary in the optical depth of the medium to ionizing radiation.


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