User:Allison K. Alix/Notebook/Thesis Research/2013/10/29
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V =4/3(π)r<sup>3</sup>  V =4/3(π)r<sup>3</sup>  
  +  
  +  = 4/3(π)(6.98x10<sup>7</sup>)<sup>3</sup>  
  +  
+  = 1.42x10<sup>18</sup> x 1000L/1m<sup>3</sup>  
+  
+  = 1.42x10<sup>15</sup>L  
Revision as of 12:13, 30 October 2013
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To Do
Calculating %MB openExample: N = 7.9 particles 7.9 particles x 1 mol/6.02x10^{23} particles x 1/6.93 x 10^{16}L = 1.89x10^{8}M NOTE: 6.93x10^{16} is the focal volume of the spot, previous calculated 1.89x10^{8}M x 1nM/10^{9}M = 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. τ_{D} was calculated to be 58.7ms Calculationsradius of the spot τ_{D} = (ω_{0}^{2}3ηπd)/4kT where τ_{D} is the diffusion time of the sample ω_{0} 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 × 1023 m^{2} kg s^{2}2 K^{1} T is the temperature of the room (18.7 °C, 291.7K) ω_{0} = sqrt([(0.05874s(4)(1.3806488x10^{23}m^{2}kg/s^{2}K)(291.7K)]/[3(0.001028kg/m s)(π)(2x10^{7}m)]) ω_{0} = 0.698μm volume of the spot V =4/3(π)r^{3} = 4/3(π)(6.98x10^{7})^{3} = 1.42x10^{18} x 1000L/1m^{3} = 1.42x10^{15}L
