User:Allison K. Alix/Notebook/Thesis Research/2013/10/29: Difference between revisions
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|style="background-color: #EEE"|[[Image:owwnotebook_icon.png|128px]]<span style="font-size:22px;"> | |style="background-color: #EEE"|[[Image:owwnotebook_icon.png|128px]]<span style="font-size:22px;"> October 29th 2013</span> | ||
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Revision as of 13:09, 5 November 2013
October 29th 2013 | <html><img src="/images/9/94/Report.png" border="0" /></html> Main project page <html><img src="/images/c/c3/Resultset_previous.png" border="0" /></html>Previous entry<html> </html>Next entry<html><img src="/images/5/5c/Resultset_next.png" border="0" /></html> | |||||||||||||||||||
To Do
Calculating %MB openExample: N = 7.9 particles 7.9 particles x 1 mol/6.02x1023 particles x 1/6.93 x 10-16L = 1.89x10-8M NOTE: 6.93x10-16 is the focal volume of the spot, previous calculated 1.89x10-8M x 1nM/10-9M = 18.9nM Actual concentration of this sample was 22.4nM (According to dilution calculations) %MB open = 18.9nM/22.4nM x 100% = 84.4nM
DataThe above image is an FCS curve showing the average diffusion time of 9 10000X diluted green bead samples. τD was calculated to be 58.7ms Calculationsradius of the spot τD = (ω023ηπd)/4kT where τD is the diffusion time of the sample ω0 is the radius of the spot η is the viscosity of the solvent (water), 0.001028kg/m s d is the diameter of the particle k is Boltzmann's constant = 1.3806488 × 10-23 m2 kg s-2-2 K-1 T is the temperature of the room (18.7 °C, 291.7K) ω0 = sqrt([(0.05874s(4)(1.3806488x10-23m2kg/s2K)(291.7K)]/[3(0.001028kg/m s)(π)(2x10-7m)]) ω0 = 0.698μm volume of the spot V =4/3(π)r3 = 4/3(π)(6.98x10-7)3 = 1.42x10-18 x 1000L/1m3 = 1.42x10-15L
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