If an electron absorbs a photon to get exited to a higher energy level, it should either come back to same state or any other lower state by emitting the required photon. How then can there be a net transfer of energy to the atom? Heating up means increase in kinetic (vibrational) energy of the atoms. If the energy absorbed by a photon is re-emitted as a photon then how do atoms extract energy from incident photons?
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2$\begingroup$ You might find it instructive to read about the Mössbauer effect. $\endgroup$– dmckee --- ex-moderator kittenCommented Apr 25, 2013 at 16:07
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$\begingroup$ @dmckee---ex-moderatorkitten I really wish you'd post that as answer. Also, more generally, en.wikipedia.org/wiki/Nuclear_spectroscopy (I don't know if that Wiki page existed 7 years ago, though. Hello from the future! Erm...it's a bit scary here right now.) $\endgroup$– Tanner StrunkCommented Apr 1, 2020 at 20:35
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2 Answers
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Let us clear some misunderstandings:
if an electron absorbs a photon to get exited to a higher energy level,
It is not the electron that absorbs the photon to go to a higher energy level. It is the whole atom, which is represented by a potential well with energy levels filled by electrons up to a point. A photon with the correct energy, i.e. an energy that covers the difference of the level where the electron is and a higher energy empty level, will be absorbed by the whole atom. The electron will decay with a characteristic decay time from the higher energy level to a lower one and a photon will take up the energy again. It can happen that cascades of photons may take up the energy.
it should either come back to same state or any other lower state by emitting the required photon....
Again, it is the system nucleus+electrons that absorbs and emits quantized photons.
In this case, ( of absorption and emission), little kinetic energy is transferred to the atom, from momentm conservation within the Heisenberg Uncertainty Principle.
*BUT HOW CAN IT TRANSFER ITS ENERGY TO NEARBY ATOMS *
I hope it is clear that it does not. It could happen that, for example, the dipole field of an atom interacts with the field of another atom and transfers kinetic energy it has received, but it will be from an almost continuum scatterings, where the photons are off mass shell (virtual). Black body radiation agrees for large wavelengths between classical and quantum forms. The need of quantization appears at higher energies to avoid the ultraviolet catastrophy.
THE only way is through electron- atom collisions (absurd).....
No, photons from the continuum can interact with the dipole and quadrupole moments of the atoms and transfer energy there. Infrared photons of the continuum can scatter off the left over fields of the atoms/molecules transferring momentum /kinetic energy and increasing temperature. It is a continuum process, not a quantized one. Relaxation of the vibrational levels of the lattice will release quantized photons.
heating up means increase in vibrational energy of the atoms,
right but also of kinetic energy
then how exactly do they gain it from excited electrons?
They do not, they gain it from the continuum of infrared photons interactions with the left over fields of the atoms and molecules.
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Essentially what you are asking about is the "photoelectric effect". The intensity of the light does not contribute to energy needed for the material to expel a photon.
The Minimum energy needed to release a photon is given by the work function $\Phi=hf$ where h is Planck's constant and f is frequency.
Here is a flash simulation of what is going on.
What the affect the intensity has is that it will change the number of electrons that are excited by the incident light which is different.
