I sort of understand what has happened with the Experiment with turning Lead in to gold.

https://www.symmetrymagazine.org/article/lhc-near-miss-collisions-turn-lead-into-gold

So given the energy required to remove electrons is called Ionization energy and this is detailed as 1st 2nd, 3rd Expressed in Joules (or KiloJoules / mol / Electron Volts

Periodic Table Guide

Ionization Energy Chart of all Elements (Full Chart Inside)

Ionization energy chart of all the elements is given below.First ionization energy, second ionization energy as well as third ionization energy of the

So what energy is required to strip all the electronic from heavier elements such as Lead, which as this has a mass of 82 then there would be 82 electronics.

I know the LHC works at Mega and Giga electron volts, so if they are removing all or most of the electrons, is this a value that equates to all the different ionization energies added together.

I am kinda guessing we don't need values for 4th,5th and higher values, as at least within Chemistry we deal with reactions that i think rely on the outer shell (valance) electrons.

Furthermore, I probably don't quite understand some of this, but they must know how much energy is required for each element to lose all its electrons (if that is what happens)

Thanks

Paul

Transforming lead into gold (as found out in the LHC) is not about re-arranging electrons, but about re-arranging nucleons (protons and neutrons).

BTW, it's an extremely low-efficiency process (requiring billions of years just to produce enough gold to make a pair of earrings).

Edited May 17May 17 by joigus
minor re-wording

6 minutes ago, joigus said:

BTW, it's an extremely low-efficiency process (requiring billions of years just to produce enough gold to make a pair of earrings).

Yup. I saw a report that said 89,000 atoms a second. A billion years gets you a around a gram