User:Allison K. Alix/Notebook/Thesis Research/2013/06/12: Difference between revisions
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==Dilution of AuNP== | ==Dilution of AuNP== | ||
According to Reference 1, for the silica coating procedure, AuNP with a radius of ~19nM must | According to Reference 1, for the silica coating procedure, 34.7mL AuNP with a radius of ~19nM must have a concentration of 0.058g/L | ||
M<sub>1</sub>V<sub>1</sub>=M<sub>2</sub>V<sub>2</sub> | M<sub>1</sub>V<sub>1</sub>=M<sub>2</sub>V<sub>2</sub> | ||
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x = 13.37mL in 21.33mL H<sub>2</sub>O | x = 13.37mL in 21.33mL H<sub>2</sub>O | ||
==Calculating the amount of PVP to add to AuNP== | |||
Revision as of 11:06, 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)λspr= 527nm d= [ln(λspr-λ0/L1)]/L2 where λ0= 512nm, L1= 6.53, and L2=0.0216 (These are fit parameters from theoretical calculations1) d= [ln(527nm-512nm/6.53)]/0.0216 d= 38.5nm c = A450/ε450 c = 1.89/(4.36x109) c = 0.43nM Calculating AuNP concentration in g/L1) Calculate the volume of 1 AuNP Volume of 1 AuNP = 4/3πr3, where r = d/2 = 19.25nm V= 4/3π(19.25x10-9m)3 V= 2.988x1023m3 V= 2.988x1017cm3 2) Calculate the mass of 1 AuNP density of Au = 19.3g/cm3 19.3g/cm3 x 2.988x1017cm3 = 5.7668x10-16 g per 1 AuNP 3) Calculate # of AuNP in 20mL solution 0.43x10-9mol/L x 20mL x 1L/1000mL x 6.02x1023AuNP/1 mol = 5.219x1012 AuNP 4) Calculate mass of all AuNP in 20mL solution 5.7668x10-16 g/1 AuNP x 5.219x1012AuNP x 1/20mL x 1000mL/1L = 0.1505g/L Dilution of AuNPAccording to Reference 1, for the silica coating procedure, 34.7mL AuNP with a radius of ~19nM must have a concentration of 0.058g/L M1V1=M2V2 (0.1505g/L)(x)=(0.058g/L)(37.4mL) x = 13.37mL in 21.33mL H2O Calculating the amount of PVP to add to AuNP | |
