a. Using the information provided below, derive an expression to show that the energy of an electronic transition measured from UV-VIS spectroscopy is a function of the wavelength of the light absorbed using the equations below. Make sure you indicate your final answer! γλ=c AE = hv where Eis the energy for the transition, v is the frequency of light, A is the wavelength, c and hare the speed of light and Plank's constant respectively (both constants). Insert an image (.jpeg, png, .heic) showing all of your work for deriving the expression in the box below.
Electronic Transitions and Spectroscopy
The term “electronic” connotes electron, and the term “transition” implies transformation. In a molecule, the electrons move from a lower to a higher energy state due to excitation. The two energy states, the ground state and the excited state are the lowest and the highest energy states, respectively. An energy change is observed with this transition, which depicts the various data related to the molecule.
Photoelectron Spectroscopy
Photoelectron spectroscopy (PES) is a part of experimental chemistry. It is a technique used in laboratories that involves projecting intense beams of radiation on a sample element. In response, the element ejects electrons for which the relative energies are measured.

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