Photodissociated iodine laser and population inversion

Iodine molecules ($$\ce{I2}$$) can absorb in the visible region and dissociate into $$\ce{2 I^.}$$ radicals. One of the I atoms is in ground electronic state $$\mathrm{^2P_{3/2}}$$ and the other I atom is in the excited $$\mathrm{^2P_{1/2}}$$. So, photodissociation of $$\ce{I2}$$ produces 50% of the species in the excited state and 50% in ground state.

The $$\mathrm{^2P_{1/2}\rightarrow^2P_{3/2}}$$ of iodine has been extensively used in laser production, in chemical iodine laser or methyl iodide photodissociation laser for example. But in those cases, most of the $$\ce{I}$$ atoms are produced in the excited state. The criteria for laser production is population inversion, where >50% of the species are in the excited state.

Here I am not sure if the photodissociation of iodine molecule satisfies the criteria of population inversion, as exactly 50% of the species is in excited state. So my question is—can we consider 50% excited species as population inversion and if so can photodissociated $$\ce{I_2}$$ generate laser?

• You are making a chemical laser. A laser depend on keeping a difference in population between the two levels involved, and of course feedback for a laser oscillator. It does not matter how many I* you make as long as the lasing transition does not go the the lowest ground state level. You should make the transition end at a higher level in the ground state I atoms that has no thermal population, (i.e. a make a 3 level laser) so that the laser radiation cannot pump this level to the higher level from which lasing occurs. You must of course keep producing I* atoms. – porphyrin Jan 9 at 19:46