I have this question for a long time that in LASER when we do the pumping by heating how does molecular collision causes the atoms(molecules) to reach in excited state. E.g. in $\ce{He-Ne}$ laser when the excited $\ce{He}$ atom collides with the $\ce{Ne}$ atom the $\ce{Ne}$ atom gets excited and $\ce{He}$ atom reaches its ground state.

My question are

  1. What happens at the molecular level that causes $\ce{Ne}$'s electron to jump into higher state? Does Ne atom absorb some kind of photon by some mechanism?
  2. Another thing is if we take $\ce{He}$ atom as the frame of reference then the high energy $\ce{Ne}$ atom is striking $\ce{He}$ so the high enrgy $\ce{Ne}$ must give its energy to $\ce{He}$ not the other way round.Where did I go wrong here?

1 Answer 1


The general rule is that you do not need to absorb the photon to excite an electron. What you need is to provide some portion of energy. It can be done by means of a photon which is absorbed by an atom, of course, but it is not the only way.

Alternatively, you could collide an atom with some other object and upon the collision some of the kinetic energy is converted into internal energy which is then redistributed by its components. One of the components of the internal energy is the electronic energy, so if enough kinetic energy is converted into this particular part of internal energy, an electron gets excited. This process is usually referred to as the collisional excitation. This is the process by which an electron in $\ce{He}$ atom is excited: $\ce{He}$ atom in its ground electronic state collides with an energetic electron which causes electron excitation.

A collision between $\ce{He}$ atom in an excited electronic state and ground state $\ce{Ne}$ atom also involves an energy transfer. But this time it is the excitation energy (possibly together with some amount of kinetic energy as well) which is transferred from $\ce{He}$ atom to $\ce{Ne}$ atom. Due to a near coincidence between the $\mathrm{2\,^3S_1}$ and $\mathrm{2\,^1S_0}$ energy levels of the $\ce{He}$ electronic states, and the $\mathrm{3s^2}$ and $\mathrm{2s^2}$ energy levels of $\ce{Ne}$, the transfer of excitation energy from $\ce{He}$ atom to $\ce{Ne}$ atom is quite efficient.

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Image courtesy of Wikipedia.

Disclaimer: I'm a professional chemist, not a physicist, so I might be wrong in what follows...

... but as far as I understand the matter at the microscopic level the energy is transferred by means of electromagnetic radiation the quantum of which is indeed photon. But do not think about this process as a photon being somehow "produced" in a form of a "billiard ball" by one atom, then transferred to another atom and finally magically "disappearing" here. This classical way of thinking is the root of confusion.

Before the energy transfer we have two systems each described by its own wave function. In order for the energy to be transferred the systems should start to interact with each other, but once they do so they become entangled and we do not even longer have separate systems. Rather, we have a big composite system described by its own wave function. After the interaction we again can speak about the individual systems each of which is again in own quantum state, though, the final state of each system is different from the starting one due to energy transfer. And in order to predict the final state of both systems (to be more precise, the probabilities of all possible final states) one could use the rules of quantum mechanics.

  • $\begingroup$ Why do we need excited $\ce{He}$? Why don't unexcited high speed $\ce{He}$ atom collides and causes $\ce{Ne}$ to excite? How does this happen technically? When $\ce{He}$ approach $\ce{Ne}$ the outer electrons must repel each other then their must be some mechanism which transfer momentum to the outermost electron of Ne. I'm sure the mechanism has something to do with the excitation of He otherwise if we look from He's frame of reference then He itself must get excited, creating a paradox. Perhaps He's electron goes to ground state there by creating a photon which is absorbed by $\ce{Ne}$. $\endgroup$
    – user31782
    Commented Jul 23, 2015 at 14:34
  • $\begingroup$ @user31782, now I see that you have two questions out there. The first step of the process is inelastic collision of energetic electrons with ground state helium atoms. These collisions produce excited helium atoms as described in the answer. But when excited helium atoms collide with ground state neon atoms the story is a bit different: it is still about energy transfer, but this time conceptually an electron in a helium atoms is deexcited and this energy is transferred to an electron in a neon atom which is excited. $\endgroup$
    – Wildcat
    Commented Jul 23, 2015 at 15:30
  • $\begingroup$ @user31782, unexcited high speed $\ce{He}$ atom will also cause an electron in $\ce{Ne}$ atom to excite, but no so efficiently. The excitation energy transfer is much more efficient then the kinetic energy transfer in this case due to near coincidence between the energy levels of species in question. $\endgroup$
    – Wildcat
    Commented Jul 23, 2015 at 15:40
  • $\begingroup$ @user31782, well, if you are interested in the actual mechanism of the energy transfer, I would say Chem.SE is the wrong place to ask such question (Physics.SE is the right place) since there is absolutely no chemistry in the mechanism of the energy transfer. $\endgroup$
    – Wildcat
    Commented Jul 23, 2015 at 15:54
  • $\begingroup$ Again, I don't think that unexited $\ce{He}$ atom would be able to excite $\ce{Ne}$ atom, because if we see from $\ce{He}$'s frame then $\ce{Ne}$ is striking with a higher energy and hence must excite $\ce{He}$ atom instead -- paradox. Secondly ,yes I want to know the actual scenario of what exactly happens with the inelastic collision. How is the kinetic energy given to the outermost electron. I can't ask question on Phys.SE for some reason. I guess my question can be on topic here because I am asking about atomic chemistry. And why does the near coincidence of energy levels [cont].. $\endgroup$
    – user31782
    Commented Jul 23, 2015 at 18:09

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