The exciton binding energy in methyl ammonium lead bromide is reported to be around 40 meV but for iodide, it is even less than room temperature energy. As a result of this, the charge carriers are free electrons and holes in iodide. Is there anything to do with the ionic size of Bromide? The dielectric constant has a part to play in this. But why the dielectric constant is reduced in the case of Bromide perovskite?

More Bibek S Dhami's questions See All
Similar questions and discussions