Question

Suppose a simple dye molecule can be modeled as having a hydrogen-like energy states with 6 electrons. Initially, the electrons settle to the lowest total energy state. Imagine that we illuminate this molecule with just right energy to excite the electrons from the n = 3 level to the n = 5 level. When such excited electrons decay back to their lower available energy states, photons of the same energy as the incoming photons are emitted. In addition, photons with different wavelengths can be emitted. This is known as fluorescence. (a) Draw the energy-level diagram (for n = 1 to n = 6) and show how electrons are filling the energy levels before light is illuminated. (b) After the illumination, the electron’s arrangement changes in the energy diagram. Draw the energy-level diagram again (for n = 1 to n = 6) and show how the electrons are filling the energy levels after the light illumination. (c) The excited electrons in (b) decays to the lower energy states. Identify two transitions producing photons with longer wavelengths than the incoming photons. Write your answer in this format: Eph (6 →1), if the photon is generated by the electron transitioning from n = 6 to n = 1 (this is not the correct answer). (d) Explain why other transitions (such as from n = 5 to n = 1) are not possible.

Answer 1