The visible absorption and photoluminescence of CdSe nanoparticles depend on the size of the particle. A sudden injection of room temperature selenium solution into hot cadmium solution produces quickly-growing seed crystals. Samples are withdrawn from the hot solution and quenched at room temperature to produce a series of increasing particle sizes and observable change in color.
- Wear eye protection
- Chemical gloves recommended
- Fume hood recommended
Step 1. Add 13 mg of CdO to a 25 mL round bottom flask clamped in a heating mantle. Warning: CdO is an inhalation hazard and this operation should be done in a fume hood. Avoid plastic containers to reduce static problems.
Step 2. To the same flask, add by pipet 0.6 mL oleic acid and 10 mL octadecene. Swirl the flask to mix the liquids. Insert a thermometer capable of measuring 225 °C.
Step 3. Heat the cadmium solution. (For a thermowell on high this takes about 20 minutes.) When the temperature reaches 225 °C, use a clean and dry pipet to quickly transfer 1 mL of the room temperature selenium solution to the 225 °C cadmium solution and start timing. The injection should happen as quickly as possible.
Step 4. Remove approximately 1 mL samples at frequent intervals using a 9 inch glass Pasteur pipet as the CdSe particles grow in size. Have a partner record the times for each withdrawal, starting from the time the selenium was added. In the video the first five samples were removed at 10 second intervals.
Step 5. Continue removing samples at longer intervals so there is a noticeable color change. In the video ten samples are removed within 3 minutes of the initial injection.
Step 6. Record the absorbance spectra of the solutions to find the maximum wavelength peak. (Particle sizes can be estimated using the x-intercept or the peak wavelength; see the calculation methods shown below.)
Also graph the absorbance wavelength as a function of growth time.
Step 7. Record the emission spectra of the solutions to find the maximum wavelength peak. In the movie, sequential samples are placed in a beam of 400 nm wavelength light. What is the evidence for band gap excitation rather than molecular absorbance?
– Option: Samples show narrower peaks in the absorption spectrum when they are quenched more quickly. Instead of collecting in test tubes, they can be collected in flasks containing liquid nitrogen.
The quantum dots can be isolated from octadecene and resuspended in an organic solvent.
Selenium Stock Solution
Add 30 mg of Se and 5 mL octadecene to a 10 mL round bottom flask over a stirrer hot plate.
– Warning: Se is an inhalation hazard and this operation should be done in a fume hood.
Measure by syringe 0.4 mL trioctylphosphine from its Sure-Seal bottle and add to the flask.
Add a magnetic stir bar. Stir and warm the solution as necessary to completely dissolve the selenium. Cool to room temperature. This stock solution may be prepared ahead of time, has enough Se precursor for five preparations, and can be stored in a sealed container for at least several months.
CAUTION: Avoid physical contact with cadmium oxide and cadmium selenide as both are carcinogens.
- CdO, Aldrich 202894
- Oleic Acid, technical grade, Aldrich 364525
- Selenium, Aldrich 209651
- trioctylphosphine, technical grade, Aldrich 117854
- Octadecene, technical grade, Aldrich O806
- analytical balance (0.0001 gram) in a fume hood
- 10 mL round bottom flask and stopper
- stirrer hotplate
- 1 mL syringe with needle
- 1/4″ magnetic stir bar
- 25 mL round bottom flask
- 25 mL heating mantle and controller
- 1 mL graduated pipet
- 10 mL pipet
- 250 °C thermometer
- Pasteur pipets and bulbs
- test tubes
- absorbance spectrometer
- emission spectrometer
Calculations and Graphs
Graph the absorbance wavelength as a function of growth time.
The x-intercept of the linear portion of the absorbance as a function of wavelength graph is a measure of Eg.
Eg = h c / λ
h =6.626×10-34 J s
c = 2.998×108 m/s
e = 1.602×10-19 C
ε0 = 8.854×10-12 C2/N/m2
m0 = 9.110×10-31 kg
λbulk = 512 nm
ε = 5.7
me* = 0.19
mh* = 0.80
λbulk = 709 nm
ε = 10.6
me* = 0.13
mh* = 0.45
λbulk = 365 nm
ε = 8.66
me* = 0.24
mh* = 0.59
The effective mass model suggests
E = hc/λ + (1/me* + 1/mh*)h2/(8r) – 1.8e2/(εr)
where r is the radius of the nanoparticle. The second term is the particle-in-a-box confinement energy for an electron-hole pair in a spherical quantum dot and the third term is the Coulomb attraction between an electron and hole modified by the screening of charges by the crystal.
After multiplying by r2, rearranging, and using the quadratic formula,
r = (1.8e2/ε ± √((1.8e2/ε)2 – (1/me* + 1/mh*)h3c/(2λ)))/(2hc/λ)
What is the diameter of the nanoparticles?
An alternative method of finding the nanoparticle diameter (W. William Yu, Lianhua Qu, Wenzhuo Guo, and Xiaogang Peng, Chem. Mater., 2003, 15 (14), pp 2854-2860) is based on an empirical fit of the observed wavelength of the peak, λ, for CdSe particles with known sizes measured by TEM.
diameter = 1.6122×10-9 λ4 − 2.6575×10-6 λ3 + 1.6242×10-3 λ2 − 0.4277 λ + 41.57
What is the diameter of the nanoparticles using this relationship? Do the two methods give similar sizes?