User:Allison K. Alix/Notebook/Thesis Research/2013/06/12
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(→Calculating AuNP diameter and concentration) 
(→Calculating AuNP concentration in g/L) 

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5.7668x10<sup>16</sup> g/1 AuNP x 5.219x10<sup>12</sup>AuNP x 1/20mL x 1000mL/1L = '''0.1505g/L'''  5.7668x10<sup>16</sup> g/1 AuNP x 5.219x10<sup>12</sup>AuNP x 1/20mL x 1000mL/1L = '''0.1505g/L'''  
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+  ==Dilution of AuNP==  
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+  According to Reference 1, for the silica coating procedure, AuNP with a radius of ~19nM must be in a concentration of 0.058g/L and 37.4mL is needed.  
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+  M<sub>1</sub>V<sub>1</sub>=M<sub>2</sub>V<sub>2</sub>  
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+  (0.1505g/L)(x)=(0.058g/L)(37.4mL)  
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+  x = 13.37mL in 21.33mL H<sub>2</sub>O  
Revision as of 14:05, 12 June 2013
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AuNP synthesis
Note: These AuNPs will be used in the silica coating procedure. See calculations below. Calculating AuNP diameter and concentration (nM)λ_{spr}= 527nm d= [ln(λ_{spr}λ_{0}/L_{1})]/L_{2} where λ_{0}= 512nm, L_{1}= 6.53, and L_{2}=0.0216 (These are fit parameters from theoretical calculations^{1}) d= [ln(527nm512nm/6.53)]/0.0216 d= 38.5nm c = A_{450}/ε_{450} c = 1.89/(4.36x10^{9}) c = 0.43nM Calculating AuNP concentration in g/L1) Calculate the volume of 1 AuNP Volume of 1 AuNP = 4/3πr^{3}, where r = d/2 = 19.25nm V= 4/3π(19.25x10^{9}m)^{3} V= 2.988x10^{23}m^{3} V= 2.988x10^{17}cm^{3} 2) Calculate the mass of 1 AuNP density of Au = 19.3g/cm^{3} 19.3g/cm^{3} x 2.988x10^{17}cm^{3} = 5.7668x10^{16} g per 1 AuNP 3) Calculate # of AuNP in 20mL solution 0.43x10^{9}mol/L x 20mL x 1L/1000mL x 6.02x10^{23}AuNP/1 mol = 5.219x10^{12} AuNP 4) Calculate mass of all AuNP in 20mL solution 5.7668x10^{16} g/1 AuNP x 5.219x10^{12}AuNP x 1/20mL x 1000mL/1L = 0.1505g/L Dilution of AuNPAccording to Reference 1, for the silica coating procedure, AuNP with a radius of ~19nM must be in a concentration of 0.058g/L and 37.4mL is needed. M_{1}V_{1}=M_{2}V_{2} (0.1505g/L)(x)=(0.058g/L)(37.4mL) x = 13.37mL in 21.33mL H_{2}O
