# User:Allison K. Alix/Notebook/Thesis Research/2013/06/12

(Difference between revisions)
 Revision as of 14:01, 12 June 2013 (view source) (→Calculating AuNP diameter and concentration)← Previous diff Revision as of 14:05, 12 June 2013 (view source) (→Calculating AuNP concentration in g/L)Next diff → Line 56: Line 56: 5.7668x10-16 g/1 AuNP x 5.219x1012AuNP x 1/20mL x 1000mL/1L = '''0.1505g/L''' 5.7668x10-16 g/1 AuNP x 5.219x1012AuNP x 1/20mL x 1000mL/1L = '''0.1505g/L''' + + ==Dilution of AuNP== + + 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. + + M1V1=M2V2 + + (0.1505g/L)(x)=(0.058g/L)(37.4mL) + + x = 13.37mL in 21.33mL H2O

## 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)

λspr= 527nm

d= [ln(λspr0/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 = A450450

c = 1.89/(4.36x109)

c = 0.43nM

## Calculating AuNP concentration in g/L

1) 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 AuNP

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.

M1V1=M2V2

(0.1505g/L)(x)=(0.058g/L)(37.4mL)

x = 13.37mL in 21.33mL H2O