# How can atoms be influenced by light? [closed]

I was watching a SciShow video "Most dangerous Chemicals". It said that a compound called Azidozide azide explodes on being exposed to light. My question is: how does the the atom feel light. Heat can be felt (let us not go into the philosophical question why do we feel warm while standing below the sun) and can excite molecules. How can light excite molecules?

## closed as too broad by Nilay Ghosh, airhuff, Jon Custer, Todd Minehardt, pentavalentcarbonDec 31 '17 at 23:19

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Light can be ‘felt’ because light is transmitted by photons. Depending on how you choose to observe light, it will either manifest itself as waves — for example in interference experiments — or as particles — for example when describing the photoelectric effect.

When described as a wave, the energy of a certain source of light will be determined by its wavelength by equation $(1)$ in which $h$ is the Planck constant and $c$ is the speed of light. $$E = \frac{hc}\lambda\tag{1}$$ It may not immediately be obvious how this translates to the energy of a photon, but the key is the de Broglie wavelength as given in equation $(2)$.

$$\lambda = \frac hp\tag{2}$$

This equation connects all particle and wave properties of types of matter. (Indeed, due to the wave-particle dualism all particles can theoretically behave as waves and vice-versa; the wave nature of particles as large as entire molecules has been shown in interference experiments.)

When a photon with a certain energy travels through an atom or molecule, it may interact with the electrons present; often this interaction causes the photon to be absorbed and the electron to be excited to a higher energy level. As chemists, we describe this as an electronic transition from one orbital to another. The entire process can also be reversed, i.e. if an electron is excited to a higher energy level by any means, it can relax following certain quantum-mechanical selection rules back down to a lower energy level and emit a photon of a definite wavelength. The latter process can be observed in the flame colour given by sodium atoms: their 3s electron is excited to a 4p level thermally due to the heat of the flame. Upon relaxation, the electron emits a photon of $\pu{589nm}$ wavelength which we perceive as yellow. We could write that in an equation: $$\ce{4p ->[][\lambda = \pu{589nm}] 3s}\tag{3}$$

If the target of a photon happens to be a molecule rather than an atom, we need to consider what type of orbitals we are exciting from and exciting into. Typically, the occupied orbitals will be bonding orbitals while unoccupied orbitals will be antibonding. Thus with the correct photon wavelength, an electron can be excited from a bonding into an antibonding orbital which we can, in layman’s terms, call ‘breaking a bond’. Whether or not the bond actually breaks depends on whether or not the atoms can dissociate apart.

A beautiful reaction to exemplify this is the reaction of chlorine and hydrogen in equimolar mixtures. Under ambient conditions, these two are metastable. If the activation energy is supplied, they will react to form hydrogen chloride.

$$\ce{H2 + Cl2 -> 2 HCl}\tag{4}$$

Activation energy can be supplied not only by a flame but also by light as long as its wavelength is short enough. A flash of blue light will work, a flash of red light will not. What happens? First, a photon excites an electron in a chlorine molecule. It will reach an antibonding, unoccupied orbital, the $4\sigma$ orbital. This causes the $\ce{Cl-Cl}$ bond to break liberating two chlorine radicals.

$$\ce{Cl2 + h\nu -> [Cl2^*]^\ddagger -> 2 Cl^.}\tag{5}$$

These chlorine radicals are then energetic enough to react with hydrogen molecules to give hydrogen radicals and so on — a radical chain reaction.

\begin{align}\ce{Cl^. + H2 &-> HCl + H^.}\tag{6}\\ \ce{H^. + Cl2 &-> HCl + Cl^.}\tag{7}\end{align}

And this goes on until all the reactive species have been consumed and the radicals recombine. The reaction releases $\pu{92.3kJ/mol}$ and is therefore strongly exothermic, even explosive.

This mechanism of course also works for infrared light which we perceive as heat. However, since the energy of infrared light is very low, few reactions can be initiated by it.

Photons in the UV or visible ranges of the EM spectrum can have sufficient energy to excite electrons. In simplified terms, this usually involves an electron entering an anti-bonding orbital, which may be sufficient to start a reaction. If that reaction yields enough energy to start the next one, you have a chain reaction. Alternatively, those electrons relax back to their ground states, photons will be emitted, and the atom or molecule will give off visible light of specific frequencies. This is called fluorescence. So in a nutshell light gives off energy and that is what will excite your molecules along with the energy from anything else that makes contact with the chemical. Ask if you need clarification.

• I think this answer could be improved if it addressed how the absorption of light could result in an explosion per the OP's example, not just the emission of a photon. – airhuff Dec 15 '17 at 1:47
• Thanks for the suggestion. I'm not so sure what it is but simple warmth of light could be enough to cause it to explode perhaps because of the burning point of the chemical is so low the heat overcomes it and it ignites. And when i mentioned how the energy from contact of outside sources ignites it the friction would be that cause of explosion due to the heat generated. Feel free to correct my thinking. – Nathaniel Barnhill Dec 15 '17 at 17:03