1
$\begingroup$

When we examine IR spectra, we see troughs corresponding to absorption at exactly a specific frequency that corresponds to the energy needed to stretch certain bonds (although translational motions and intermolcular forces can broaden the accepted frequency).

Coordination compounds absorb light at exactly the frequency corresponding to the crystal-field splitting energy.

However, when we talk about energy from photons required to break a bond, we say "at least" the frequency instead of "exactly" in the examples above. Take $\ce{Cl-Cl}$ for example, we say that the photon must have at least a frequency of 607 THz (or wavelength of no more than 496 nm). There is still a definite energy involved here: 242 kJ/mol to promote an electron from a bonding MO in the molecule to an anti-bonding MO.

As another example, when we talk about ionizing radiation, we say that there is at least a certain amount of energy needed to ionize a compound. For example, my textbook says that a photon needs "at least 1216 kJ/mol" to ionize water. There is still a definite energy level involved: bringing the bonding electron from its negative potential energy MO to 0.

In all of these examples, definite energies were involved. Why is it that sometimes we say that that a photon needs to have exactly the energy needed, and other times the minimum energy needed?

$\endgroup$
2
$\begingroup$

When you ionize or break something, the resulting particles (an electron and an ion, or maybe two atoms, or whatever) would fly away from each other with arbitrary speed. Their kinetic energy would absorb any excess energy of the photon. This is the "at least" case.

On the other hand, when you excite some particle (molecule or whatever) in such a way that it remains in one piece, then you can't send it flying, because the momentum conservation forbids that. This is the "exactly equal" case.

Surely this is a simplification; in fact, a photon may excite a molecule using only a part of its energy, while keeping the rest of it to itself (Raman scattering), but that's another story.

$\endgroup$

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.