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High velocity particles in the thermosphere

Physics Asked by Cesare on May 30, 2021

I do not understand why XUV absorption in the thermosphere leads to high velocities.

When I do the calculations on a sheet of paper (momentum and energy) I find that the velocity of (for example) atomic hydrogen before and after ionization is almost the same. To illustrate my point, please consider the following example,

let us imagine that there is a hydrogen ion traveling almost parallel to an electron. There is a very small deviation in the perpendicular velocity which causes the proton and electron to slowly approach each other over time. After a given time, the two will meet, combine, and a high-energy photon will be released. The energy released corresponds to the increased potential, so I cannot imagine the overall velocity changing much.

Now let us imagine the opposite, a hydrogen atom gets hit by a high-energy photon and it gets ionized. How come its velocity would change drastically resulting in the high "temperatures" of the thermosphere?

Is anyone aware of a simple calculation that can illustrate why the velocities are expected to change much?

One Answer

The classical answer is that there are 3 (ionizing) mechanisms for the interaction of light with matter: Photoelectric effect, Compton effect and pair production.

I made some rather simple calculations for the first two: Assuming a photon of 10 nm wavelength, the photon energy is $E_{photon}=frac{hc}{lambda}= 1.98 cdot 10^{17} J approx 123 eV$. Minus the 13.6 eV needed for hydrogen ionization, there is plenty of energy left going into kinetic energy of electron and proton. Neglecting the electron for simplification (I know that's wrong...) and assuming all this energy goes into the proton velocity $v_{proton} = sqrt{(2 cdot (E_{photon} - 13.6) / m_{proton})} approx 148000 m/s$.

Interestingly, when calculating the Compton effect for $180^circ$ backscattering of a 10 nm photon on an already free electron, I get a quite similar velocity of the electron...

Is that more or less comparable to your results (I made this calculations rather quick...)?

Correct answer by Charles Tucker 3 on May 30, 2021

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