Download presentation
1
Spectroscopy – study the interaction of matter
and electromagnetic radiation. irradiate sample measure scattering, absorption, or emission in terms of measured parameters. interpret results.
2
= c/ E = h Visible light ~ 400-700 nm UV light ~ 200-400 nm
where h is Planck’s constant (6.63x10-34 J/s) and E is given as Joules. A mole of photons therefore has an energy of ENA, where NA is Avagadro’s number (6.022x1023). Visible light ~ nm UV light ~ nm
4
Quantum Mechanical Description of Matter
Quantum mechanics can be used to describe the spatial distribution of all matter (or at least for relatively simple systems like single electrons) in terms of discrete, quantized energy levels using solutions to the Schrödinger equation: Hn = Enn where (x, y, z) is a wave function that describes the position of a particle such as an electron as a distribution in space, H is the Hamiltonian operator that operates on the wave function to describe an observable property such as energy (E), and n is the quantum number that describes each of the quantized energy states for the system.
5
Quantum Mechanical Description of Matter
Soultions of the time-dependent Schrödinger equation can be used to describe the various energy levels of a system, i.e. for a particle in a one-dimensional box of width a, the energy levels can be described by While the energy levels of the hydrogen atom are more complex, they too can be described by quantum mechanics: En (kcal/mol) = -(22e4)/((40)2h2n2) = -(1312)/n2 n = 1 n = 2 n = 3 Note that each system is unique. The spacing of energy levels on the left is increasing as n increases, while the spacing decreases as n increases for the hydrogen atom above.
6
Energy levels of matter
Etot = Etrans + Erot + Evib + Eelec + Ee-spin + Enuc
7
Molecular Orbitals
8
Boltzmann Distribution
Describes the distribution of atoms among different energy states:
9
Selection Rules E = hv net displacement of charge
e- spins stay opposite p s
10
Absorption of light – Beer-Lambert Law
11
Photoselection
12
Fluorescence Polarization
Anisotropy: Polarization: r = (I|| - I)/(I|| + 2I) P = (I|| - I)/(I|| + I) = 2P/(3 - P) = 3r/(2 + r) For a single fluorophore: I|| cos2 I sin2sin2 = ½sin2 Therefore, since cos2 + sin2 = 1 r = ½(3cos2 - 1)
13
Jablonski Diagram
14
Morse Diagram
15
Franck-Codon Principle
16
Excitation vs. Emission Spectra
Stoke’s Shift Invariance of emission spectra to excitation wavelength
17
Mirror Image Rule
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.