Spectroscopy – study the interaction of matter and electromagnetic radiation.

• irradiate sample• measure scattering, absorption, or

emission in terms of measured parameters.

• interpret results.

= c/E = h

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 ~ 400-700 nmUV light ~ 200-400 nm

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.

Quantum Mechanical Description of Matter

En hman

2 2

28

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

En (kcal/mol) = -(22e4)/((40)2h2n2) = -(1312)/n2

Quantum Mechanical Description of Matter

n = 1

n = 2

n = 3

While the energy levels of the hydrogen atom are more complex, they too can be described by quantum mechanics:

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.

Etot = Etrans + Erot + Evib + Eelec + Ee-spin + Enuc

Energy levels of matter

Molecular Orbitals

nn

EkT

ex

gr

exp

Boltzmann Distribution

Describes the distribution of atoms among different energy states:

Selection Rules

• E = hv

• net displacement of charge

• e- spins stay opposite

s

p

AII

ct

log

10

0

Absorption of light – Beer-Lambert Law

z

y

x

d

d

cos

sin

sindd

sinsin

Photoselection

Anisotropy: Polarization:r = (I|| - I)/(I|| + 2I) P = (I|| - I)/(I|| + I) = 2P/(3 - P) = 3r/(2 + r)

For a single fluorophore:I|| cos2I sin2sin2 = ½sin2

Therefore, since cos2 + sin2 = 1r = ½(3cos2 - 1)

Fluorescence Polarization

Jablonski Diagram

Morse Diagram

Franck-Codon Principle

•Excitation vs. Emission Spectra

•Stoke’s Shift

•Invariance of emission spectra to excitation wavelength

Mirror Image Rule