# How do molecules gain kinetic energy from light?

I know that molecules can absorb light through electronic and vibrational excitations, which certainly increase the internal energy of a molecule. This idea is always connected to the quantum theory in my head (transition between discreet levels by absorbing a photon with a certain energy etc.) Now, in the most basic classical picture, temperature of let's say a liquid is basically average kinetic energy of all molecules or their average velocity.

What I cannot see is how a photon may give a molecule an actual momentum to increase its kinetic energy and consequently increase the temperature? I mean it can excite an electron in the molecule or make it vibrate, but as a whole the molecule does not really move faster. Or is it the acoustic vibrational modes that give the molecule an actual kick? I mean they should still be vibrations, but at least the vibrations which involve moving molecule as a whole.

• I may be wrong, but this may be best understood by Classical Momentum Conservation Law. If photon is absorbed by an electron, then it may go into a higher orbit at some point in time. In that precise moment of time nucleus is dragged towards receding electron. Then molecules other atoms are dragged towards that atom with momentum. – Agnius Vasiliauskas Aug 20 '19 at 18:24

A photon has momentum $$\mathbf{p} = \hbar\mathbf{k}$$. This is a vector, so by conservation of momentum, the molecule has to gain momentum $$m\mathbf{v} = \hbar\mathbf{k}$$ in the proper direction.

During absorption, the energy $$hf$$ of the incident photon is split in two (or more if there's rotation and vibrations, but let's keep it simple): $$hf = \frac{1}{2}mv^2 + \text{levdiff}$$. Molecule speed $$v$$ is defined through $$m\mathbf{v} = \hbar\mathbf{k}$$, so there is still only one incident energy $$hf$$ that corresponds to an energy level difference $$\text{levdiff}$$ (neglecting the spread due to uncertainty, etc.).

• Thank you for your response! But isn't it in contradiction with the rule that a photon energy has to exactly match (in theory of course) the energy difference between a molecule's electronic levels with an exception of processes with a phonon excitation as in Raman spectroscopy? Otherwise I can imagine that any excitation is possible if a photon has sufficient energy, and what is left from the excitation just goes to the molecule's speed up. – CHILLQQ Aug 20 '19 at 20:03
• There is always narrow range of energies that are acceptable (line width). This is due to temperature (initial movements of source and absorber), and Heisenberg uncertainties (photon energy, electron state). At room temperature and above, the added translational kinetic energy is negligible compared to inter-molecule collisions that cause volume expansion. – Notabot Aug 20 '19 at 20:11
• @CHILLQQ Radiation does not have to have exactly frequency $(E_2-E_1)/h$ to be absorbed. There is some leeway, because every transition has non-zero spectral width. Part of energy goes to electronic excitation, part to kinetic energy of the molecule (much smaller part, due to large mass of molecule, so this is often neglected). If frequency of radiation is too high however, the sensitivity of the transition is very small and the transition effective cross section is very small, so the absorption is so small it is ignored - hence the rule you mention. – Ján Lalinský Aug 20 '19 at 22:10
• @CHILLQQ You are absolutely right. The kinetic energy acquired by the molecule’s change in momentum plus the energy difference has to equal the total photon energy. This means that the incoming photon needs to have a slightly larger energy than the energy difference to be absorbed. – Gilbert Aug 21 '19 at 0:01