1
$\begingroup$

Professor taught us that an electron gains or looses only those energies which are equal to difference in two energy levels. That is $E_1 + \Delta E = E_2$ or $E_1 + \Delta E = E_3.$

What if we give, say, $\pu{11.3 eV}$ and the remainder is $\pu{-2.2 eV},$ so the electron jumps to $n = 2$ where the energy is $\pu{-3.4 eV}$ and the remainder $\pu{-1.2 eV}$ is released?

I'm really confused with the working here.

$\endgroup$
11
  • 2
    $\begingroup$ Have you thought about why atomic absorption spectra contain just narrow lines and not wide bands ? $\endgroup$ – Poutnik Apr 21 at 8:01
  • $\begingroup$ Yes, that's because we have specific 'energy levels' and not continuous i.e. only for n=1,2,3.... and not n=1,1.1,1.2..... and because the energy levels aren't continuous the spectra isn't continuous but discrete? Am I right? $\endgroup$ – Vishwas Sharma Apr 21 at 9:17
  • $\begingroup$ Yes, and also atoms absorb photons of just specific energy levels, equal to differences between atom discrete allowed energy levels. $\endgroup$ – Poutnik Apr 21 at 9:20
  • 1
    $\begingroup$ It can't reject the excess energy because it can't absorb it in the first place. You need to understand that photons can provide all energy they have or none. There's no absorb half reject other half. $\endgroup$ – Nisarg Bhavsar Apr 21 at 14:35
  • 1
    $\begingroup$ @NisargBhavsar So let me get this straight. So it's just like how Bohr explained it. We have a ladder that the electron can climb up to. It can either make its way up by stepping on the next or can go down by stepping down, but only on the steps and not in between. So either give me the required amount of energy or don't. Right?! $\endgroup$ – Vishwas Sharma Apr 24 at 6:24
3
$\begingroup$

Quantum mechanics received its name from the concept of the "quantum" which is Latin for "how much" or "amount" and is used to mean "small discrete piece or fragment". Classical physics prior to the development of QM described the energy of bodies as a continuum of possibilities. The novelty introduced by QM was the idea that within the continuum only a subset of the possible energies are in fact allowed, corresponding to allowed states. This means by extension that the energies of a body in different possible states differ by discrete rather than continuous amounts. This discreteness becomes increasingly evident for small and confined systems, where the wave-like properties of matter become more evident, as demonstrated for instance by the self-interference of the electron wavefunction in the double-slit scattering experiment.

Electrons in atoms fulfill all of the conditions for observation of QM phenomena. The particles involved are subatomic (very small) and are confined (by a Coulombic potential). The result is that electrons in atoms exist in a discrete number of allowed states, each state defined by a fixed energy and wavefunction that can be interpreted as the probability distribution of the electron.

When an atom is excited by a photon (a discrete light particle with a fixed amount of energy) it absorbs the entire photon. All of the photon's energy is acquired by the atom. However, such an excitation implies a transition between possible states of the atom. It is necessary that the energy of the photon match the difference in energy of the atomic states. In the case where the two states are bound (if an electron is not excited into the continuum of ionized free states) the result is that only a subset of photon energies can be absorbed by the atom.


As noted by Shoubhik R Maiti, photons can also be scattered by atoms and in the process transfer only part of their energy to the atoms. However the principle and meaning of quantization of electronic states remains the same. Electronic states are quantized (discrete) and so only particular amounts of energy can be absorbed (quanta corresponding to differences in energies of discrete levels).

$\endgroup$
3
  • 1
    $\begingroup$ +1 This is a good answer. I would also add that there are some cases where it's not necessary to have a metastable excited state, for example Raman scattering. A molecule can absorb photons to go into a virtual electronic state (that does not exist as a measurable state—uncertainty principle) and then releases a photon back, going into an actual excited state which exists for some time. $\endgroup$ – Shoubhik R Maiti Apr 23 at 22:07
  • 1
    $\begingroup$ @ShoubhikRMaiti Thanks, that would invalidate some of what I wrote in the last paragraph, although I was thinking of absorption, not scattering. You might want to write an answer - if not I will add a caveat to my own to incorporate your comment. $\endgroup$ – Buck Thorn Apr 24 at 3:45
  • $\begingroup$ No I think what you have added to your answer is fine. Yes scattering is a two-photon process so it's quite unusual and different from the normal one-photon absorption or emission etc. $\endgroup$ – Shoubhik R Maiti Apr 24 at 9:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.