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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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  • $\begingroup$ See for yourself: the NIST Atomic Spectra Database has all the known energy levels. Set the units to eV and choose your atom (say, He I for 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 (marked He II in the table for He I, indicating that helium becomes ionized). For helium, the difference between the ground state and the ionization threshold is 24eV. $\endgroup$ – Emilio Pisanty Mar 13 '20 at 23:20
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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.

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  • $\begingroup$ "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). $\endgroup$ – Emilio Pisanty Mar 14 '20 at 0:45
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    $\begingroup$ @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? $\endgroup$ – G. Smith Mar 14 '20 at 2:02
  • $\begingroup$ Isn’t that how the Hartree-Fock approximation works? $\endgroup$ – G. Smith Mar 14 '20 at 2:04
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    $\begingroup$ In 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. $\endgroup$ – Emilio Pisanty Mar 26 '20 at 15:23
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    $\begingroup$ @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? $\endgroup$ – my2cts Mar 27 '20 at 13:25
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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.

Image from https://plot.ly/python/v3/ipython-notebooks/density-of-states/

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