The Hydrogen atom fundamental energy is -13.6 eV.
Is there an atom that has an energy level lower to -13.6 eV ?
if no, then why, in semiconductor physics, the integral on energy start at $-\infty$ instead of $-13.6\ eV$ ?
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The integrals in semiconductor physics usually have a factor of density of states $\rho(E)$ which goes to zero outside certain energy limits. So you can integrate to infinity, but the density of states will only be non-zero for certain ranges.
Mathematically then, if you want to integrate a function $f(E)$, you are replacing a sum over energy states by an integral times the density of states.
$$\int_{-\infty}^{\infty} \rho(E) f(E) dE \leftrightarrow\sum_{E_i} f(E_i)$$
You can see the density of states below from this link. For silicon the relevant bands span about 20 eV.
KF Gauss
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Yes. Neglecting effects of other electrons, the ground-state energy scales like $Z^2$. So probably all other elements have more negative ground-state energies than hydrogen does.
I recommend reviewing either the Bohr or Schrodinger models for a hydrogen-like atom that has a nucleus that has $Z$ protons.
G. Smith
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"neglecting the effects of other electrons" is about as accurate an approximation as neglecting the kinetic energy or the electrostatic attraction to the nucleus. (To be blunt: it is completely useless.) The ionization energy of hydrogen, as it turns out, is quite high (Wikipedia has a good graph), with only about six other elements having higher ionization energies (i.e. lower ground-state energies w.r.t. the single-free-electron ion). – Emilio Pisanty Mar 14 '20 at 00:45
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1@EmilioPisanty I’d like to understand whether we are interpreting the question differently or whether I am totally confused about multi-electron atoms, which is certainly possible. You are talking about ionization energies, but the OP asked about “an energy level”. I was under the impression that multi-electron atoms can be approximated in a way that allows one to assign an energy level to each electron based on an approximate single-electron orbital, and that for the electrons closest to the nucleus the outer electrons have little effect on this energy level. Is this totally wrong? – G. Smith Mar 14 '20 at 02:02
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@EmilioPisanty The effect of screening by outer electrons on core energy levels is limited so 1s core levels should roughly scale with $(Z-1)^2 \approx Z^2$. – my2cts Mar 14 '20 at 07:26
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@G. Smith, it depends on whether we are interested in the first ionization energy (aka work function), then E. Pisanty is right. If the deepest core level is relevant then you are right and you get $Z^2$ scaling. As far as OP's question, I think both are irrelevant, it is really the bandwidth and density of states that matters. Depending on what he wants, this bandwidth may be as low as 2eV. – KF Gauss Mar 14 '20 at 12:15
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@G.Smith that is indeed how the HF framework works, but that is precisely the point: HF is wrong -- or, more precisely, HF insists on seeing multi-electron dynamics as a set of independent single-electron problems in a way that really does not hold up when seen in the light of the quantum mechanics of fully-correlated many body systems. Electron orbitals are not physical entities, and the energies associated with them are essentially a fiction (they're eigenvalues of the self-consistent Hartree-Fock single-electron operator, and not of the system's Hamiltonian). – Emilio Pisanty Mar 26 '20 at 14:59
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In other words, in multi-electron systems, there is only one single physical energy, which is assigned to the system as a whole; there are no physical per-electron energies. Single electrons don't have energy levels or states; there are only global multi-electron states. Thus, it is meaningless to say "the core $1s$ orbital has a Hartree-Fock energy of xxx" (say, using @my2cts's ultimately-meaningless shielding arguments) unless that can be assigned a full multi-electron eigenstate of the Hamiltonian. – Emilio Pisanty Mar 26 '20 at 15:17
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... and, for states with nominal HF energies above the ionization potential, this is impossible. Yes, in the restricted world of HF those states exist, but they are at the same energies as the ionized continuum and they do couple with it (via configuration-interaction terms that lie outside the scope that HF can incorporate), so they will mix with that continuum; in other words, they are autoionizing states and are better described as Fano resonances in that continuum than states at all. – Emilio Pisanty Mar 26 '20 at 15:20
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1In any case, to summarize (CC @MathieuKrisztian), and to state things as plainly as they need to be stated: this answer is dead wrong. There are no physical bound states above the ionization threshold. Neglecting the effects of other electrons is like neglecting the effect of the Sun's gravity when calculating the Earth's orbit around the Sun. The only way for ground-state energies to scale with $Z^2$ would be to live in a world where electrons are fermions and don't interact electrostatically with each other. – Emilio Pisanty Mar 26 '20 at 15:23
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@EmilioPisanty Perhaps you are not familiar with core ionisations then. HF is not as wrong as you believe. Of course when a core electron is ejected the final state is only approximately given by the HF approximation. Such states have a certain life time and a high ionisation energy, not so far from their HF orbital energy value, which is my point. E.g. if you remove a 1s electron from a C in diamond , the ionised atom will be have like N for a short while. – my2cts Mar 26 '20 at 17:05
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Your arguments do not hold. I computed these core excitations myself with Hartree-Fock, @EmilioPisanty. – my2cts Mar 26 '20 at 17:15
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@my2cts "Such states have a certain life time" -- which means that they are not states or levels, they are resonances. This is standard material in atomic physics beyond undergraduate level. HF is useful scaffolding but it is just scaffolding. Whether you calculated the approximate positions of these resonances yourself makes no difference to anything. – Emilio Pisanty Mar 27 '20 at 11:52
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Just to clarify your patronising comments, I am perfectly familiar with core excitations; the Fano resonances in helium are bread and butter material for attosecond transient absorption spectroscopy and related frameworks. The keywords there are resonance and transient. These are not states. If you don't want to agree with that, pick up some modern literature or a suitable textbook. – Emilio Pisanty Mar 27 '20 at 11:55
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1@EmilioPisanty Who is patronising who? Your assessment of core excitations is meaningless . These states exist for quite a long time. I would call them metastable. With your logic the only states are ground states . Why so pedantic? – my2cts Mar 27 '20 at 13:25
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@my2cts There is a clear difference between autoionizing resonances and the radiative decay of bound states. You're obviously free to extend your personal definition of "state" to include resonances, but that's entirely at odds with all of the modern atomic-physics literature. (Or perhaps you have specific examples of your personal definition being used in recent literature that you can point to? It's pointless to continue this conversation unless you can do so.) – Emilio Pisanty Mar 27 '20 at 14:54
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@EmilioPisanty In my book there are also excited states. This includes core excitations, positronium, electron-hole pairs, ... – my2cts Mar 27 '20 at 16:01
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The most important point here, regardless of whether one considers core excitations as resonances or metastable, is that this answer is actually irrelevant to OPs question which is fundamentally about the meaning of density of states in semiconductors. I do agree that the critique by Emilio is patronizing and a bit over the top. – KF Gauss Mar 27 '20 at 19:54
He Ifor neutral helium). The energy levels are calibrated with zero at the ground state, with the zero in your convention corresponding to the first ionization limit (markedHe IIin the table forHe I, indicating that helium becomes ionized). For helium, the difference between the ground state and the ionization threshold is 24eV. – Emilio Pisanty Mar 13 '20 at 23:20