Since pioneering work in the early 1990s, the areas of single molecule detection and microscopy have — Physical Chemistry — Kinetics Chemistry Question
Observing Single Molecules
Since pioneering work in the early 1990s, the areas of single molecule detection and microscopy have exploded and expanded from chemistry and physics into life sciences. Great progress came about with the demonstration of room–temperature imaging (with a near–field scanning optical microscope) of the carbocyanine dye 1,1'–didodecyl–3,3,3',3'–tetramethylindo–carbocyanine perchlorate (diIC12). In this experiment, dye molecules are spread on a sample surface and localized according to their fluorescence signals. The structure of diIC12 is shown below.
[VISUAL]
Indicate which part of the diIC12 molecule is responsible for its fluorescence. Mark the correct answer.
(1) The benzene rings
(2) The dodecyl side chains
(3) The four methyl groups at the heterocyclic rings
(4) The C–N chain connecting the two benzene rings
(5) The perchlorate ion
Model Answer
Molecular fluorescence in the visible region is due to delocalized electrons in extended π–systems, so the correct answer is:
(4) i. e. : The C–N chain connecting the two benzene rings.
The surface densities of the molecules have to be sufficiently low, if you want to observe them as individual fluorescent spots under a microscope. No more than 10 molecules per μm 2 on the sample surface is a good value.
10 dm 3 of a solution of diIC12 in methanol are deposited on a very clean glass cover slide. The drop covers a circular area having a diameter of 4 mm.
Calculate the molar concentration of the solution necessary to obtain the value of 10 molecules per μm 2 . (For this calculation we assume that the transfer of the dye molecules from solution to the sample surface by evaporation of the solvent is homogeneous on the whole wetted area.)
Model Answer
A circle with a diameter of 4 mm covers a surface area of
S = r 2 with r = 2×10 –3 m, so S = 1.26×10 –5 m 2
The number of molecules in this area is:
10 / (10 –6 m) 2 × 1.26×10 –5 m 2 = 126×10 6 molecules
They are transferred onto the surface by the evaporation of 10 dm 3 of solution, so the concentration has to be
126×10 6 / (10×10 –6 dm 3 ) = 1.26×10 13 molecules per dm 3
which corresponds to a molar concentration of
c = 1.26×10 13 dm –3 / (6.022×10 23 mol –1 ) = 2.1×10 –11 mol dm –3
The sample is illuminated with the 543.5 nm–line of a green He–Ne laser. The excitation power is adjusted so that the illuminated area (100 nm in diameter) is hit by 3×10 10 photons per second.
What is the excitation power that has been used?
Model Answer
When E = hc / , the energy per photon is:
E = hc / = 6.626×10 –34 J s × 3.00×10 8 m s –1 / 543.5 × 10 –9 m = 3.66×10 –19 J
3×10 10 photons per second amount to an excitation power of
P = 3.65×10 –19 J × 3×10 10 s –1 = 1.1×10 –8 J s –1 = 11 nW
The absorption cross section is an important parameter for the calculation of the expected fluorescence signal from a single molecule. It may be regarded as the effective area of the molecule that captures all incoming photons. At room temperature, this value corresponds approximately to the size of the dye molecule.
An illuminated diIC12 molecule absorbs 2.3×10 5 photons per second under the described conditions. Calculate the absorption cross section of the diIC12 molecule in Å 2 (It can be assumed that the 100 nm diameter area is uniformly illuminated).
Model Answer
On average, there are 10 molecules per μm 2 , so that one molecule occupies statistically an area of Smol = (1×10 –6 m) 2 / 10 = 1×10 –13 m 2 .
The total illuminated area of × (50×10 –9 m) 2 = 7.85×10 –15 m 2 receives 3×10 10 photons per second, and the area occupied by a single molecule receives
3×10 10 s –1 × 1×10 –13 m 2 / (7.85×10 –15 m 2 ) = 3.82×10 11 photons per second. Only 2.3×10 5 photons are absorbed every second, so the area which is capturing photons is:
σ = 1×10 –13 m 2 × 2.3×10 5 s –1 / (3.82×10 11 s –1 ) = 6×10 –20 m 2 or 6 Å 2
(or σ = (7.85×10 –15 m 2 / 3×10 10 s –1 ) × 2.3×10 5 s –1 = 6×10 –20 m 2 )
The fluorescence quantum yield, i.e. the average number of fluorescence photons created for each absorbed photon, is 0.7 for diIC12 (7 fluorescence photons are created for every 10 absorbed photons). The collection efficiency of the generated fluorescence photons by the experimental setup (including filters to suppress remaining excitation light) is 20 %, and the photon detection efficiency of the highly sensitive photodetector is 55 % over the range of the molecular fluorescence.
How many fluorescence photons will actually be detected on average by the photodetector during a 10 ms acquisition window if one diIC12 molecule is located in the illuminated area?
Model Answer
A dilC12 molecule that absorbes 2.3×10 5 photons per second emits
Nfluo = 0.7 × 2.3×10 5 s –1 = 161×10 3 fluorescence photons per second. Due to the detection efficiency, this results in Ndet = 161×10 3 s –1 × 0.2 × 0.55 = 17 710 detected photons per second. In a time interval of 10 ms, the number of detected photons is:
17 710 s –1 × 10×10 –3 s = 177 photons.
The fluorescence image is constructed by raster scanning the illuminated area across the sample surface.
What diameter do you expect for the fluorescence spot corresponding to one single dye molecule? Mark the correct answer.
(1) One pixel
(2) 543.5 nm
(3) 100 nm
(4) 200 nm
(5) Approximately 1 m
Model Answer
Each point in the illuminated sample area is hit by the same number of photons per second (uniform illumination). A molecule that is located in the spot’s center is emitting as many fluorescence photons as if it was sitting anywhere else in the illuminated spot. As the illuminated area is raster–scanned across the sample surface, the molecule will be visible as long as it is inside the illuminated area. This is the reason why the fluorescence spot of one molecule will have a size equal to the illuminated area, i.e. 100 nm in diameter (14.3).