# Is it true, that for all atoms the second checkmark is valid?

Is the second checkmark true for all atoms or only one-electron type of atoms?

Photoelectric Effect

The photon theory:

$$\checkmark$$ Building on Planck's ideas, Einstein proposed that light itself is particulate, quantized into tiny "bundles" of energy, later called photons.

$$\checkmark$$ Each atom changes its energy by an amount $$ΔE_{atom}$$ when it absorbs or emits one photon, one "particle" of light, whose energy is related to its frequency, not its amplitude: $$E_{photon}=hν=ΔE_{atom}$$

$$\checkmark$$ A photon of a certain minimum energy must be absorbed to free an electron from the surface

It's the Photoelectric Effect in the context of Quantum Theory and Bohr's Model.

The reason why I doubt it's for the one electron atoms (species) is that this same formula shown in the second checkmark is used in Bohr's Model and Bohr's Model only works for hydrogen-like atoms.

It's in the last line

The Bohr Model of the Hydrogen Atom

$$\checkmark$$ Two years after the nuclear model was proposed, Niels Bohr (1885–1962), a young Danish physicist working in Rutherford's laboratory, suggested a model for the H atom that did predict the existence of line spectra.

$$\checkmark$$ Postulates of the Model in his model, Bohr used Planck's and Einstein's ideas about quantized energy and proposed three postulates:

1. The H atom has only certain energy levels, which Bohr called stationary states. Each state is associated with a fixed circular orbit of the electron around the nucleus. The higher the energy level, the farther the orbit is from the nucleus.

2. The atom does not radiate energy while in one of its stationary states. Even though it violates principles of classical physics, the atom does not change its energy while the electron moves within an orbit.

3. The atom changes to another stationary state (the electron moves to another orbit) only by absorbing or emitting a photon. The energy of the photon ($$hν$$) equals the difference in the energies of the two states:

$$E_{photon}=ΔE_{atom}=E_{final}-E_{initial}=hν$$

• Well, is the "atom" that is too specific. It holds true for isolated atom - independently of their atomic number. But the photoelectric effect can takes place from a metal surface, or from a molecule. What is general is that the Energy of the photon does correspond to the energy levels interval of quantisised states. Oct 5, 2021 at 9:53
• so... Energy of photon = ΔEatom = hv for all atoms or just one electron atoms? Oct 5, 2021 at 10:06
• Welcome to Physics Stack Exchange! Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. Oct 5, 2021 at 10:45
• Emilio Pisanty-okay it's okay thanks anyway Oct 5, 2021 at 10:53

The Bohr model was a useful step to the construction of a more sophisticated theory - quantum mechanics - which was able to model multi-electron atoms (and more complex systems like molecules), something the Bohr model cannot do properly. In a Bohr-type model of an atom you can speak of an electron changing from one energy level to another. This is not really the case in quantum mechanics where you must (ideally) treat the all the electrons as being in one connected state.

However the energy level of an emmitted photon is going to the difference between the energy of the initial state and the final state of the atom.

The key thing to remember is that energy has to be conserved. The final system's energy (atom+photon) must have the same energy as the inital state (just an atom).

• what if the atom absorbs n photons then it should be ΔE = nhv? Oct 5, 2021 at 10:36
• @ElieMakdissi The photoelectric effect is seen in conductors where the energy levels of the electrons are with the lattice of the conducto, not with individual atoms. Individual atom interactions are just that, interactions . Oct 5, 2021 at 11:04
• @annav The photoelectric effect is present in both conductors and isolated atoms. The core of the effect (removal of electrons dependent on the frequency of the light and not its intensity) is the same in both cases, independently of whether you call the minimum frequency a "work function" or an "ionization potential". Oct 6, 2021 at 16:52
• @EmilioPisanty sure,it is the same underlying physics , but historically the effect is defined in conductors, and the n in nhν is irrelevant to the effect. Oct 6, 2021 at 17:13
• @annav Perhaps historical literature from the early 20th century draws a distinction between the two. Current primary literature doesn't, and hasn't done so for a long time. Oct 7, 2021 at 12:07

Actually your question isn't related much to the photoelectric effect. The only connection is that the photoelectric effect was among the first evidences suggesting that photons are discrete and carry a discrete amount of energy, according to the equation shown.

The third member ( delta E atom) of the equation says that absorption (or emission) of a photon occurs when the energy of the latter matches the energy difference between quantisised levels of the entity undergoing such phenomenon.

This happens in isolated atoms, independently of their atomic number, ions, and even molecules or lattices.

• thanks a lot ^^ Oct 6, 2021 at 13:23