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Topic: Vibronic spectrum of iodine  (Read 5833 times)

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jasusu

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Vibronic spectrum of iodine
« on: July 30, 2024, 08:52:58 AM »
Could someone explain to me what exactly is a convergence limit in the vibronic spectrum of iodine? I know that when the v' increases, the vibrational energy spacing decreases. This is from a source I found but I don't quite understand:

"At a point called the convergence limit, the spacing between bands decreases to zero. Beyond this convergence limit, the spectrum is continuous because the excited state of the I2 molecule is not bound."

Why the excited state isn't "bound" anymore? And does the quantization of energy affect the fact that the spectrum becomes continuous at 500 nm?

Thank you!

Offline Corribus

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Re: Vibronic spectrum of iodine
« Reply #1 on: July 30, 2024, 09:54:05 AM »
Are you familiar with the Leonard Jones or other anharmonic potential surface?

In the harmonic approximation, you can dump an infinite amount of energy into a molecular system and it never flies apart because the restoring force is always proportional (and opposite direction to) the displacement (i.e., Hooke's Law). In this case, the vibrational states are evenly spaced. The vibrational transitions would therefore never converge to a limit, and the system is always bound - that is, the atoms will never fly apart.

In the anharmonic system, which is more realistic, you CAN dump enough kinetic energy into the system that the atoms can overcome the restoring force and fly apart - that is, become unbound from each other. The vibrational states get closer and closer together as the molecule approaches this dissociation energy. So, if you were to plot the energy gap between states as a function of the energy of the lower (origin) state, they will eventually converge at a limit where the gaps between states become infinitesimally small. This is the point where you can no longer put more kinetic energy into the system and have it remain bound as a molecule - the atoms will fly apart because they are no longer trapped in a potential well.

From a quantum mechanical point of view, "bound" means that the particles are confined by a potential boundary, and in this case they exhibit quantized energy levels. Think: particle in a box. When you confine particles to a region, they take on quantized energy states in which the particles can only take on certain energies. In an unbound system, particles are not confined and they may take on any energy states - the states are continuous. The potential surface (defined by electrostatic attraction) between two particles (two iodine atoms, say) confines them to a region of space when they form a molecule, meaning that their allowed vibrational states are quantized. This is true only when the particles do not have enough kinetic energy to escape their mutual attraction. If the two atoms have enough potential energy such that they are no longer confined by the potential surface, then they are no longer bound by that surface and their energy states are continuous/unquantized. As noted, this only happens with an anharmonic surface. In a harmonic surface, the states are always quantized because a harmonic system is always bound by the potential surface, which extends to infinity - it is literally impossible to give the particles enough energy to escape their mutual attraction.
 
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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