Solar chromosphere: inter-conversion of Ca I, Ca II, role of photons

[jordankonisky]

TL;DR Summary

Does the mechanism of the formation of solar Ca I to Ca II require photons? What is the level of 393.37 nm photons in the solar chromosphere? Using Ca II filter to image solar Ca.

My question relates to the physics of the emission line of Ca II which originates in the solar chromosphere. My understanding is that ionized Ca which lacks a single electron in its valence orbital interacts with free electron in the chromosphere milieu to form Ca I (nonionized). The captured electron then moves from its capture orbital (4P 3/2) to the 4S 1/2 orbital and in the process generates a photon of lambda = 393.37 nm. This photon then travels to my telescope with a Ca II-K filter and then to my imaging camera.

My question is what is the mechanism by which that nonionized Ca I is converted to ionized Ca II. I have read that at the temperature of the chromosphere there is sufficient energy produced through atomic collisions to knock an electron out of the Ca I valence orbit to produce Ca II and a free electron. My question: does a photon of 393.37 nm play any role in the production of Ca II? Is it required that the leaving electron be kicked up to its capture orbital (4P 3/2) via absorption of a 393.37 nm photon before a collision can knock the electron out of Ca I to produce Ca II? Or can the collision free an electron from Ca I directly out of 4S ½ without the intervention of a 393.37 nm photon?

Finally, if there were no Ca in the chromosphere, wouldn’t there be some background of some 393.37 nm photons via the Black Body radiation profile? So, what I image are Ca II photons over a much lower concentration of free 393.37 nm photons. Hope I’m thinking about this right.

 

[jordankonisky]

Summary:: Does the mechanism of the formation of solar Ca I to Ca II require photons? What is the level of 393.37 nm photons in the solar chromosphere? Using Ca II filter to image solar Ca.

My question relates to the physics of the emission line of Ca II which originates in the solar chromosphere. My understanding is that ionized Ca which lacks a single electron in its valence orbital interacts with free electron in the chromosphere milieu to form Ca I (nonionized). The captured electron then moves from its capture orbital (4P 3/2) to the 4S 1/2 orbital and in the process generates a photon of lambda = 393.37 nm. This photon then travels to my telescope with a Ca II-K filter and then to my imaging camera.

My question is what is the mechanism by which that nonionized Ca I is converted to ionized Ca II. I have read that at the temperature of the chromosphere there is sufficient energy produced through atomic collisions to knock an electron out of the Ca I valence orbit to produce Ca II and a free electron. My question: does a photon of 393.37 nm play any role in the production of Ca II? Is it required that the leaving electron be kicked up to its capture orbital (4P 3/2) via absorption of a 393.37 nm photon before a collision can knock the electron out of Ca I to produce Ca II? Or can the collision free an electron from Ca I directly out of 4S ½ without the intervention of a 393.37 nm photon?

Finally, if there were no Ca in the chromosphere, wouldn’t there be some background of some 393.37 nm photons via the Black Body radiation profile? So, what I image are Ca II photons over a much lower concentration of free 393.37 nm photons. Hope I’m thinking about this right.

 

[jordankonisky]

I figured this out myself It is not the transition of the electron per se that is knocked out of the valence shell that generates the Ca K emission line.