Does ionization process release energy?
In the case of helium atoms, to ionize the first electron, energy is injected, which is 24.59 eV of the experimental results.
Will helium emit energy in the process? After the first electron is released, the second electron will automatically change its orbit, retreat from a high-energy orbit to the lowest-energy orbit (-54.42eV), and release energy, right? Is there such an ionization experiment that confirms or negates this conclusion?
Initial helium has 2 electrons in the same shell, it's full of electrons at first layer so it's the most stable atom. After 1st electron is ejected to infinity with 24.59eV energy, the 2nd relocates to the lowest energy level -54.42eV, about 30eV energy is emitted. Is this correct? Net energy is plus value in ionization process? What's wrong with this scenario?
I guess there is a problem with the design of experiments to measure helium ionization energy. It does not take into account that helium atoms emit high-frequency (far ultraviolet, near long-wave X-rays) photons at the first ionization. Because it was not taken into account, there would not be such a measurement result. But helium does emit energy at the first ionization.
The problem here arises from the electron orbital theory of the current mainstream physics. Mainstream theory now holds that electrons in the same orbit, such as 1S, have the same constant energy. This applies to hydrogen atom of a single electron, not to the helium atom. Helium atoms have two repulsive and interacting electrons, which can not be solved by Schrodinger equation and can not be explained by the solution of hydrogen atoms.
The two electrons of helium atom have the same energy in 1S orbit. When one is knocked out, the other's energy decreases, as evidenced by the fact that the second ionization energy is 54.4eV.
After losing one electron, the other reduces energy by releasing photons. This photon release has not been thought of and not been discovered by physical experiments. But it should exist.
So the 1S orbit is not ground state orbit, there is "underground state" orbit.
To zwol , by hydrogen-like I mean the ionized helium atom with only one electron. There is another word I can't remember.
According to the orbital theory of quantum mechanics, an orbital is a container that can hold one or two electrons. Whether one or two electrons are there, the parameters of the container remain the same. This is obviously incorrect.
Let’s say hydrogen atom. The 1S orbital radius of hydrogen atom derived from Schrodinger equation is 52.9pm, which is the same as that of Bohr model. Assuming that we manually force another electron into the 1S orbit, it will certainly affect the radius of the orbit, which can no longer be equal to 52.9pm. This is obvious because of the repulsive force between electrons. If we can get the value by Schroedinger equation, it must be bigger than 52.9.
Although helium atoms cannot be solved by Schrodinger equation, the same conclusion can be drawn by analogy. When one electron or two electrons are in orbital, the state of that orbit will be different. It is impossible to remain the same.
Back to helium atom ionization topic. In the process of knocking out the first electron (that is, when the first electron is moving from original place to infinity), the orbit of the second electron changes. Its energy level is reduced from some value to -54.4eV. These two processes are running concurrently, it's not true that after the first electron is knocked out, the orbital of the second electron begins to change. So, my original assumption that 29.8 eV would be emitted is incorrect. There is no such radiation.
The first ionization energy 24.6 is the result of the common contribution of both nucleus and another electron. If there is no other electron, or if the state of another electron is forcibly fixed, the first ionization energy will not be 24.6, but a higher number.
From this analysis, we can also see that the transition of energy level is an ordinary movement of electrons, there is time for the process. The ionization is the upward transition. Electron is knocked to infinity from original lower energy level.
Best Answer
As far as I know the effect you describe isn't seen in helium. That's because the reorganisation takes place while the photon is ejecting the electron. The experimentally observed ionisation energy of helium is the energy difference between the unionised atom and the ion after reorganisation.
However this effect is certainly seen when heavier atoms are ionised by high energy photons such as X-rays. This is the basis of X-ray fluorescence. An X-ray photon ejects a $1s$ electron from the atom and the other electrons then reorganise and emit X-rays in the process.
This effect is alse seen when atoms are ionised by electron beams an in fact it is the mechanism by which X-ray tubes work. In an X-ray tube an incoming electron ionises a tungsten atom by ejecting one of the inner electrons, then the rest of the electrons reorganise and emit an X-ray photon in the process.