Part 1: 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. ν λ = c ΔE = hν where E is the energy for the transition, ν is the frequency of light, λ is the wavelength, c and h are the speed of light and Plank's constant respectively (both constants). Part 2: From the derived equation, as the absorption maximum (λmax) for a molecule increases to a longer wavelength (for example from 300 to 700 nm), how does this affect the energy for the electronic transition?
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.
Part 1:
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
ν λ = c
ΔE = hν
where E is the energy for the transition, ν is the
Part 2:
From the derived equation, as the absorption maximum (λmax) for a molecule increases to a longer wavelength (for example from 300 to 700 nm), how does this affect the energy for the electronic transition?
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