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 &gt;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 positive difference in population between the two levels involved (upper &gt;lower), 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 must make the transition end at a higher level in the ground state I atoms that has no (i.e. insignificant) thermal population, thus a make a 3 level laser meaning that the laser radiation cannot pump  this lower level to the higher level from which lasing occurs.  You must of course keep producing I* atoms.