I'm a little bit confused about the concept of an ionisation energy and the difference between the ground-state energies of ionic and neutral particles. Until now, I thought that they were the same, and no book or article made me question it. However, after some reflection, I'm convinced that I was wrong, and should have thought more about the two concepts. Consider for instance a particle $\mathrm{A}$ in its ground-state and at rest which suffers ionisation, being transformed into its ion $\mathrm{A}^+$, also in its ground-state, and a free electron $\mathrm{e}^-$, both at rest: $$\mathrm{A}\to \mathrm{A}^++\mathrm{e}^-.$$ The ionisation energy of $\mathrm{A}$, let it be $I_\mathrm{A}$, is defined as the energy which necessary adding to $\mathrm{A}$ in order to make such process happen, i.e. $$I_\text{A}=\epsilon_{\mathrm{A}^+,0}+\epsilon_{\mathrm{e}^-}-\epsilon_{\mathrm{A},0},$$ being $\epsilon_{\mathrm{A},0}$ and $\epsilon_{\mathrm{A}^+,0}$ the ground-state energies of $\mathrm{A}$ and $\mathrm{A}^+$, respectively, and $\epsilon_{\mathrm{e}^-}$ the energy of a free electron (which is also a ground-state energy since the electron as only one energy level associated to its possible two states). Therefore, the ionisation energy is only equal to the difference between the ground-state energies of ionic and neutral particles, if the energy of the electron is negligible when compared to it. But, is this energy really negligible?
3 Answers
Therefore, the ionisation energy is only equal to the difference between the ground-state energies of ionic and neutral particles, if the energy of the electron is negligible when compared to it. But, is this energy really negligible?
Assume we carry out the ionisation by means of a photon of energy $h\nu$:
$$\mathrm{A}+h\nu \to \mathrm{A}^++\mathrm{e}^-$$
Then:
$$I_\text{A}=\epsilon_{\mathrm{A}^+,0}+\epsilon_{\mathrm{e}^-}-\epsilon_{\mathrm{A},0}-h\nu$$
If:
$$\epsilon_{\mathrm{e}^-}=h\nu$$
Then:
$$I_\text{A}=\epsilon_{\mathrm{A}^+,0}-\epsilon_{\mathrm{A},0}$$
So with the right type of photon the electron energy is cancelled out.
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$\begingroup$ Shouldn't the ionisation energy be the difference in the internal energies between the products and reactants? If the photon ionises the particle without transferring energy to the translational modes of the products, the quantity $h\nu$ would be the ionisation energy. $\endgroup$ Commented Feb 24, 2021 at 15:24
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$\begingroup$ That's not how this works. It works the way the other answer explains. $\endgroup$– GertCommented Feb 24, 2021 at 15:31
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$\begingroup$ This is worth reading: en.wikipedia.org/wiki/… $\endgroup$– GertCommented Feb 24, 2021 at 15:48
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When talking about ionization I think usually a non-relativistic framework is considered. In this case, the minimum energy required to ionize the atom is the one that leaves the electron with no momentum. Since we are in a non-relativistic framework, no momentum means that the electron's energy will also be null, if the electron is far enough and so considerable free.
Ionization is the process to create an ion and a free electron, both at rest and separated hypothetically by an infinite spatial distance. The separation process is presumed to be reversible.
When you want to obtain the ground state energy of the ion by itself from the first ionization energy and a ground state energy of a neutral atom, you must include the work of separating an electron from the product ion.
For reference, we use ionization energies in conjunction with electron affinities to build Born-Haber cycles for chemical reactions that predict such otherwise generally unmeasurable energies such as the lattice energy for an ionic crystal.
