How to estimate the molar concentration of quantum dots?
For
CdTe, CdSe and CdS quantum dots, the molar concentrations can be
calculated according to Peng et al. [1] It is important to noted that
there is a correction of this paper. [2] It can be summarized as
following steps:
Step 1. Determine the particle size of QDs.
1.1 An optical method:
1.1.1 Measure the absorption spectrum of the QDs sample by a UV-vis spectroscopy. Write down the wavelength λ(nm) and the absorbance A(O.D. value) at the peak position of the first exciton absorption peak.
1.1.2 Determine the particle size D(nm)
of the QDs sample by Figure 2 in Peng’s paper. Or calculate the size
using the fitting functions, as below, of the curves in Figure 2.
CdTe: D = (9.8127×10-7)λ3 – (1.7147×10-3)λ2 + 1.0064 λ – 194.84
CdSe: D = (1.6122×10-9)λ4 – (2.6575×10-6)λ3 + (1.6242×10-3)λ2 – 0.4277 λ + 41.57
CdS: D = (-6.6521×10-8)λ3 + (1.9557×10-4)λ2 – (9.2352×10-2)λ + 13.29
1.2 Alternatively, we can measure the particle size D(nm) directly from a TEM image.
Step 2. Determine the extinction coefficient of QDs.
There are also two ways to calculate the extinction coefficient(ε) in unit of L/(mol•cm) or cm-1M-1 of QDs:
2.1 Considering the transition energy (ΔE)
CdTe: ε = 3450ΔE(D)2.4
CdSe: ε = 1600ΔE(D)3
CdS: ε = 5500ΔE(D)2.5
Here, the transition energy ΔE is corresponding to the first absorption peak, in unit of eV.
Relationship between ΔE and λ :
eΔE=hγ=hc/λ
where, e is charge of an electron, which is 1.6×10-19 C; h is Planck constant, which is 6.626×10-34 Js; c is light speed, which is 3×1017 nm/s. So that
ΔE(eV)•λ(nm)=1242(eV•nm)
The fitting functions above were according to the Brus[3] and Wang et al.[4]
2.2 Emporical function (without considering the transition energy)
CdTe: ε = 10043(D)2.12
CdSe: ε = 5857(D)2.65
CdS: ε = 21536(D)2.3
The difference between the
results of these two groups of functions is small and neglectable,
within the particle size rang of 4 to 7 nm for CdTe, 2.5 to 6 nm for
CdSe, and 2 to 5.5 nm for CdS.
Step 3. Determine the molar concentration of QDs.
The relationship between the absorbance and molar concentration is called the Lambert-Beer’s Law, which is also known as the Beer-Lambert Law or the Beer’s Law.
A = εCL
Where A is the absorbance, ε is extinction coefficient as mentioned above. C is the molar concentration (mol/L or M)of QDs, L is the path length(cm) of the radiation beam used for recording the absorption spectrum.
PS: I thought up until now that A was proportionate to lg(L)!!! How silly I was!
Till now, we got the values of the molar concentration of our QD samples. Congratulations!
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BUT~~~~~~~
What about quantum dots other than CdTe, CdSe, CdS, e.g. PbS, or QDs with a core/shell structure, e.g. CdSe/ZnS?
Actually the molar concentration of QDs samples can be figured out using standard atomic absorption (AA) measurement alone.
PS: No guarantee for this method. I’ve never tried it, since we don't have any AA spectrometer in our lab.
Step1. Determine the molar concentration of each metal element in the QDs sample by AA spectrometer.
For example, we can determine the molar concentrations of Cd, Se, Zn and S in a CdSe/ZnS QDs solution sample by AA measurement.
Let us say, [Cd]=[Se]=1 mM, [Zn]=[S]=0.2 mM.
Step2. Calculate the total volume of QDs.
We assume that the nano-scaled materials have the same density as
the according bulk materials. We can find the densities and molar
masses of the bulk materials. For example,
Density of CdSe: 5.816 g/cm3;
Density of ZnS: 4.090 g/cm3;
Molar mass of CdSe: 191.37 g/mol;
Molar mass of ZnS: 97.474 g/mol.
So that 1 mM CdSe refers to
CdSe per liter solution.
Likewise, 0.2 mM ZnS refers to
ZnS per liter solution.
Therefore, the total volumn of QDs per liter is 0.033+0.0048 = 0.038 cm3.
Step3. Calculate the molar concentration of QDs.
According to TEM or DSL(dynamic light scattering) measure, we can
obtain the size of a single QD. Divide the total volume per liter of
QDs, which obtained in Step 2, by the volume of a single QD, we get the
molar concentration of QDs.
Let’s say the a single QD has a diameter of 4 nm. The volumn will be 0.03351 nm3, according to the formula for sphere volume 4/3πR3.
The Number of particles in one liter solution is
Hence, the molar concentration of QDs solution is 1.8 μM.
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Remarks:
1. The
results are acceptable, though there are some errors due to the
estimations and assumptions we made during the calculations above.
2. The
UV-vis method is simple and convenience. But it requires the extinction
coefficients of the nano-materials, which are not known generally. So
far, this method is limited to CdSe, CdTe and CdS QDs.
3. The
AA method can be used for all kinds of nanoparticles in principle.
However, AA spectrometer is not such common-used equipment as UV-vis
spectrometer.
References
1. Yu, W.W., et al., Experimental determination of the extinction coefficient of CdTe, CdSe, and CdS nanocrystals. Chemistry of Materials, 2003. 15(14): p. 2854-2860.
2. Yu, W.W., et al., Experimental determination of the extinction coefficient of CdTe, CdSe and CdS nanocrystals (vol 15, pg 2854, 2003). Chemistry of Materials, 2004. 16(3): p. 560-560.
3. Brus, L., Electronic Wave-Functions in Semiconductor Clusters - Experiment and Theory. Journal of Physical Chemistry, 1986. 90(12): p. 2555-2560.
4. Wang, Y. and N. Herron, Nanometer-Sized Semiconductor Clusters - Materials Synthesis, Quantum Size Effects, and Photophysical Properties. Journal of Physical Chemistry, 1991. 95(2): p. 525-532.
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