2

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 $-13.6\ eV$ ?

2 Answers2

3

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/

KF Gauss
  • 8,314
1

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
  • 52,489