In the Frank-Hertz experiment, electrons (inelastic ) hit atoms and excite their energy levels. What happens to the electron after it has hit (what I believe is) the nucleus? I imagine that if the kinetic energy of the incoming electron is just enough to make one of the atom's electrons jump from one level to another, then the incoming electron ends up at near 0 speed near/on the nucleus, at this point we'd have an electron stuck to the nucleus, but that isnt permitted. So what's going on?

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    $\begingroup$ When it says the electron hits the atom, it means the electron interacts with an electron of the atom, it does not mean hitting the nucleus of the atom $\endgroup$ – Hugo V Nov 6 '18 at 11:28
  • $\begingroup$ @HugoV , why so ? The nucleus is much bigger and has opposite charge on top of that. There are many more chance of colision $\endgroup$ – Manu de Hanoi Nov 6 '18 at 13:49
  • $\begingroup$ Thats not correct, an atom is ~100,000 times bigger than its nucleus. I'll write an answer to try to make it clear $\endgroup$ – Hugo V Nov 6 '18 at 14:01

An atom is $\sim 10^5$ bigger than its nucleus, meaning that the electron cloud surrounding the atom extends much further out than the nucleus itself. When an incoming electron approaches the atom, it will first encounter this electron cloud and interact with it by a repulsive force. The nucleus will be much further away, so its force on the electron is much smaller. When you also consider that the interaction is actually hapening with the atoms outter most electron, while the other electrons are shielding most of the positive charge of the nucleus, as can be seen on the image below, then its even clearer that the electron is not interacting with the nucleus.

Nucleus shielding

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  • $\begingroup$ although at "higher energies" the electron can "hit" the nucleus: hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/elescat.html $\endgroup$ – Manu de Hanoi Nov 6 '18 at 15:38
  • $\begingroup$ Of course, it is not impossible for an electron to scatter from a nucleus, it's just a completely different process than the one described in the question $\endgroup$ – Hugo V Nov 6 '18 at 15:41

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