When an electron ($e^-$) is added to $\ce{Cl}$ ,i.e., $$\ce{Cl + e^- -> Cl^- + ∆_{gh}H}$$ We can see that energy is released. My question is where is this energy coming from? What happens within the atom such that it loses energy and releases it to the surrounding?
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$\begingroup$ Could you please explain why you downvoted it so that I can correct it? $\endgroup$– InquisitiveOct 16, 2019 at 13:54
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$\begingroup$ I think that the energy is not released by atom. It is released by the system, consisting of atom, electron and the entity the electron came from, e.g. other atom. $\endgroup$– SimonEOct 16, 2019 at 19:57
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1$\begingroup$ chemistry.stackexchange.com/questions/78838/… chemistry.stackexchange.com/questions/98723/… $\endgroup$– MithoronOct 17, 2019 at 0:15
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1$\begingroup$ I voted to close this question because of the many other duplicates, but I realized a little late that those might not actually address your question, which is why I added a brief answer. Your question seems to be twofold: "where does the emitted energy come from" and "how does the electron get rid of it". $\endgroup$– Buck Thorn ♦Oct 17, 2019 at 20:14
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The main question here is "why is it exothermic". The following diagram illustrates the conundrum:
This portrays a ball rolling over a track. Thanks to energy conservation, in frame 2 the ball has gained kinetic energy KE after descending into a well, KE diminishing as the ball reemerges again in frame 3. You can think of an electron as the ball and of the track as the potential generated by the atom. In this classical picture the electron will retain a fixed elliptical orbit or escape trajectory as it zooms past the nucleus.
If there was no way for an electron (or ion) to shed energy then there would be no bound state. One way for the electron to shed energy is by emitting a photon, which causes it to relax into a lower energy bound state (in the cartoon this would be a tighter oscillation within the well).
The cartoon above is admittedly a gross oversimplification, particularly inaccurate in suggesting that the system is classical and 1-dimensional. In reality binding of the electron to the atom with emission of a photon or some nonradiative process is best treated quantum-mechanically, but the essential principle of conservation of energy is retained.
Further reading
A deeper explanation of electron affinity involves some heavy-duty physics, but this presentation on the Breit-Wigner cross section might be useful. It explains:
Broad states which can be formed by collisions between the particles into which they decay are referred to as resonances. They are described by the Breit-Wigner formula.
Not the most recent reference, but the following article which mentions different ways of measuring and computing electron affinities may also be useful: Meunier et al, Molecular Simulation, 1999, Vol. 23. pp. 109-125
answered Oct 17, 2019 at 10:23
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$\begingroup$ Is it the energy in the electron + nucleus system that is being shed when the outer electron enters the region where electrostatic force of nucleus is experienced? If yes, why does the interaction result is shedding of energy? $\endgroup$ Oct 17, 2019 at 20:19
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$\begingroup$ I cannot give you a detailed answer, an explanation involves pretty advanced quantum mechanics, but I added some refs to my answer. $\endgroup$– Buck Thorn ♦Oct 17, 2019 at 20:29