Fluorescent molecules can be used to track biomolecular events one molecule at a
ID: 1289367 • Letter: F
Question
Fluorescent molecules can be used to track biomolecular events one molecule at a time. Fluorescence is very similar to the atomic transitions studied in lecture. Consider a fluores- cent molecule where the electron that participates in fluorescence has ground state energy of ?4.6 eV. When the molecule is illuminated with light this electron is excited up to a state with energy ?2.3 eV.
1) (3 pt) When the electron drops back to the ground state, what wavelength photon is given off?
2) (1 pt) What color is this?
3) (1 pt) When the electron is in the excited state excess charge will accumulate at points in the molecule. If the fluorescent molecule is in a polar solvent (like water) the solvent can help stabilize the excited state. If the excited state has an energy of ?2.6 eV when the fluorescent molecule is in water, what is the wavelength of the emitted photons?
4) (2 pt) Now consider the protein-ligand system shown in the figure. When the molecules are separated the fluorescent molecule is exposed to water, but it gets surrounded by non- polar protein when the molecules bind together. If you are looking at these two molecules under a microscope, what will you see if the molecules are separated? Bound? (Hint: the proteins are both much too small to see under the microscope, but you can see the emitted photons.)
5) (3 pt) The average time the electron stays in the excited state before dropping back into the ground state is called its mean lifetime. The fact that the lifetime is not infinite introduces some uncertainty in the energy of the photon, and thus the wavelength. If the lifetime of the excited state in this problem is 3 ns, what is the spread of uncertainty in the wavelength?
Explanation / Answer
1) so dE = 4.6-2.3 = 2.3 eV
2) lambda = hc/E = 1240/2.3= 539 nm
which is green light
3) so now dE = 2.0 eV
lambda = 1240/3 = 620 nm
4) No figure is included
5) dE dt = hbar/2
dE*3.0E-9 = (4.14E-15/(4*pi))
dE = 1.10E-7 eV
E = hc/lambda
dE = - hc/lambda^2 dlambda
dE/E = dlambda/lambda
dlambda = 1.10E-7*620/2.0 = 3.41E-5 nm
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