Chapter7

Chapter7 - Chapter 7: Quantum Theory Chem 6A, Section D Oct...

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Unformatted text preview: Chapter 7: Quantum Theory Chem 6A, Section D Oct 13, 2011 1 Chem 6A Michael J. Sailor, UC San Diego Problem: Diffraction of light from a CD grating A CD is held 720 cm from a wall. A 530 nm laser is diffracted from the surface, and the diffracted spot appears 257 cm from the specular beam. What is the track spacing on the CD? Chem 6A Michael J. Sailor, UC San Diego diffracted beam (n=2) diffracted beam (n=1) specular beam y tanθ = x/y 2 3 3 Chem 6A Michael J. Sailor, UC San Diego Problem: Fraunhofer Diffraction Incident laser beam beams in phase CD tracks CD surface Track spacing = d Chem 6A Michael J. Sailor, UC San Diego 4 Problem: Fraunhofer Diffraction ff di Additional path traveled ra ct ed be am θ beams in phase CD surface θ Track spacing = d 5 Chem 6A Michael J. Sailor, UC San Diego Problem: Fraunhofer Diffraction 2) n= 1) (= n m ea m ( ba ed be ct ed eam ra ct b ff di fra lar f di ecu sp d sinθ = nλ d = nλ/sinθ tanθ = (257 cm / 720 cm) θ = 0.3428 d = (1 x 530 nm) / sin(0.3428) = 1577 nm, or 1.577 micrometers y Additional path traveled tanθ = x/y θ Track spacing = d Chem 6A Michael J. Sailor, UC San Diego 6 Solution: Diffraction of light from a CD grating diffracted beam (n=2) diffracted beam (n=1) specular beam dsinθ = nλ d = nλ/sinθ tanθ = (257 cm / 720 cm) θ = 0.3428 d = (1 x 530 nm) / sin(0.3428) = 1577 nm, or 1.577 micrometers y tanθ = x/y 7 Chem 6A Michael J. Sailor, UC San Diego Discrete vs Continuous Spectra White Light Spectrum 400 Ne Line Spectrum 450 500 550 Wavelength, nm Chem 6A Michael J. Sailor, UC San Diego 600 8 Frequency Spectra-Ne gas vs violin Intensity 1 Violin G audible spectrum 0.8 0.6 0.4 0.2 0 0 500 1000 1500 2000 2500 3000 3500 4000 Frequency (Hz) Intensity 1 Ne gas visible spectrum 0.8 0.6 0.4 0.2 0 3 10 14 3.5 1014 4 1014 4.5 1014 5 1014 5.5 1014 6 1014 Frequency (Hz) Chem 6A Michael J. Sailor, UC San Diego 9 Chapter 7 (cont) Chem 6A, Section D Oct 18, 2011 Chem 6A Michael J. Sailor, UC San Diego 10 Announcements: • Thurs Oct 20 quiz (#4) will be the periodic table quiz – Practice quiz available online at: http://sailorgroup.ucsd.edu/Chem6A_sailor/BlankPeriodicTable.pdf (You will also need to name 20 of the elements in that table, given the element symbol) – Bring student ID • Thurs Oct 27 quiz (#5) will be on Chapter 6 Chem 6A Michael J. Sailor, UC San Diego 11 Quiz 3 score histogram Chem 6A Michael J. Sailor, UC San Diego 12 Grades so far (after quizzes 1-3) 90 number of students 80 70 60 50 40 30 20 10 0 F D C- C C+ B- B B+ A- A A+ Chem 6A Michael J. Sailor, UC San Diego 13 Advice on studying (1) Read the chapters (2) Do the homework (3) Do related homework problems (in the book or in the online ARIS resource) (4) Be sure you understand how to do all the example problems worked out in class (5) Visit helproom, section, or my office hours to answer questions or problems you can’t solve Chem 6A Michael J. Sailor, UC San Diego 14 Visible Light Wavelengths and Energies: Color Infrared Red Orange Yellow Green Blue Indigo Violet Ultraviolet R O Y G B I V Wavelength >800 nm 630 nm 590 nm 560 nm 510 nm 440 nm 420 nm 400 nm <350 nm Energy 2.0 eV 2.1 eV 2.2 eV 2.4 eV 2.8 eV 3.0 eV 3.1 eV eV = electron Volts 15 Chem 6A Michael J. Sailor, UC San Diego Comparison of light wavelengths to hair thickness: Name Hair thickness, Hair thickness, micrometers (µm) nanometers (nm) Sam (m) 50 50,000 Max 60 60,000 Garrett 44 44,000 Quinlan 66 66,000 Sam (f) 34 34,000 Tim 45 45,000 Mio 62 62,000 Chem 6A Michael J. Sailor, UC San Diego 16 UVA and UVB UVA = 400 nm–315 nm UVB = 315 nm–280 nm UVC = 280 nm–100 nm Chem 6A Michael J. Sailor, UC San Diego 17 Problem: Energy-Wavelength conversion When a copper ion is heated to 1200 °C in a fireworks explosion, it emits blue light at 450 nm. To what energy does this correspond, in electron volts? Set up but do not solve. Chem 6A Michael J. Sailor, UC San Diego 18 Solution: Energy-Wavelength conversion E= hc = λ 6.6261 x 10-34 J.s 3 x 108 m s 109 nm 450 nm eV m 1.6022 x 10-19 J = 2.76 eV Chem 6A Michael J. Sailor, UC San Diego 19 Problem: Energy-Wavelength conversion When a copper ion is heated to 1200 °C in a fireworks explosion, it emits blue light at 450 nm. To what energy does this correspond, in Joules? Set up but do not solve. Chem 6A Michael J. Sailor, UC San Diego 20 Solution: Energy-Wavelength conversion E= hc = λ 6.6261 x 10-34 J.s 3 x 108 m s 450 nm 109 nm m = 4.42 x 10-19 J Chem 6A Michael J. Sailor, UC San Diego 21 Problem: Particle-wave duality of matter What is the wavelength of an electron (mass = 9.11 x 10-31 kg) and a baseball (mass = 0.1 kg) traveling at the same speed of 35 m/s? Chem 6A Michael J. Sailor, UC San Diego 22 Solution: Particle-wave duality of matter h λ= for an electron mv 6.626 × 10−34 kg ⋅ m2 / s 109 nm λ= × = 21,000nm −31 9.11 × 10 kg ⋅ 35m / s 1m h for a baseball mv 6.626 × 10−34 kg ⋅ m2 / s 109 nm λ= × = 1.9 × 10−25 nm 0.1kg ⋅ 35m / s 1m λ= Chem 6A Michael J. Sailor, UC San Diego 23 Problem: Photochemistry The energy of the O-O bond in O2 is 496 kJ/ mol. What is the maximum wavelength of a photon of light that can split the oxygen bond? a) 241 nm b) 330 nm c) 410 nm d) 6.0 x 10-10 nm e) none of the above Chem 6A Michael J. Sailor, UC San Diego 24 Solution: photochemistry Energy in a bond: 496 kJ mol mol 6.02x1023 molecule = 8.24x10-19 J/molecule E= hc/λ λ = hc/E = molecule 8.24x10-19 J 6.626x10-34 J sec 1000J kJ 3x108 m sec = 2.41 x 10-7 m = 241 nm 25 Chem 6A Michael J. Sailor, UC San Diego Discrete vs Continuous Spectra White Light Spectrum Ne Line Spectrum Why do atoms give off such complicated spectra when they get hot? 400 450 500 550 Wavelength, nm Chem 6A Michael J. Sailor, UC San Diego 600 26 Problem: Bohr’s model and the energy of atomic transitions Calculate the wavelength of a photon emitted when an electron falls from the n = 3 state to the n = 2 state in the hydrogen atom. 27 Chem 6A Michael J. Sailor, UC San Diego Solution: Wavelength of atomic transitions The Rydberg equation (7.3, pg 221 in text) ⎛ྎ 1 1 1 ⎞ྏ = RH ⎜ྎ 2 − 2 ⎟ྏ λ ⎝ྎ n1 n2 ⎠ྏ RH = 1.097 x 10-2 nm-1, n1 = 2 and n2 = 3 ⎛ྎ 1 ⎞ྏ 1 ⎞ྏ 1 −2 1 −2 ⎛ྎ 1 = 1.097 × 10 ⎜ྎ 2 − 2 ⎟ྏ = 1.097 × 10 ⎜ྎ − ⎟ྏ ⎜ྎ 2 3 ⎟ྏ € ⎝ྎ 4 9 ⎠ྏ λ ⎝ྎ ⎠ྏ λ = 656 nm Chem 6A Michael J. Sailor, UC San Diego 28 Problem: Energy of atomic transitions One of the emission lines from the star σ Ori AB in the constellation Orion occurs at a wavelength of 486.3 nm. The line arises from atomic hydrogen contained in that star. This line probably corresponds to the following electronic transition: a) n=5 to n=1 b) n=4 to n=1 c) n=3 to n=1 d) n=4 to n=2 e) n=5 to n=2 Chem 6A Michael J. Sailor, UC San Diego 29 Chem 6A Michael J. Sailor, UC San Diego 30 Solution: Energy of Atomic Transitions The Rydberg expression tells you the relationship between wavelength (λ) and the transition between two levels: ⎛ྎ 1 1 1 ⎞ྏ = R H ⎜ྎ 2 − 2 ⎟ྏ λ ⎝ྎ n1 n 2 ⎠ྏ RH = Rydberg constant n1 = starting atomic level n2 = ending atomic level € 31 Chem 6A Michael J. Sailor, UC San Diego Note: Your book also gives this expression in terms of energy instead of wavelength. The equation for an individual energy level (chapter 7, p225): ⎛ྎ Z 2 hcRH ⎞ྏ E = ⎜ྎ ⎟ྏ 2 ⎝ྎ n1 ⎠ྏ Z = nuclear charge h = Planck’s constant c = speed of light RH = Rydberg constant To calculate the energy difference between two levels: € Note h.c.R = (6.626x10-34J.s)(3x108m/s)(1.0967x107m-1) = 2.18x10-18J Chem 6A Michael J. Sailor, UC San Diego 32 Solution: Energy of atomic transitions Applying the Rydberg equation to answer (a): n1 = 5, n2 = 1 = -0.01053 nm-1 (negative value indicates light is emitted, not absorbed) So λ = 95 nm for the transition from n1 = 5 to n2 = 1 Chem 6A Michael J. Sailor, UC San Diego 33 Solution: Energy of atomic transitions calculate the wavelength of each transition: transition a) n=5 to n=1 b) n=4 to n=1 c) n=3 to n=1 d) n=4 to n=2 e) n=5 to n=2 λ, nm 95 97 102 486 434 So the answer is (d) n=4 to n=2 Chem 6A Michael J. Sailor, UC San Diego 34 Problem: Energy of atomic transitions One of the emission lines from the star σ Ori AB in the constellation Orion occurs at a wavelength of 486.3 nm. The line arises from atomic hydrogen contained in that star. This line probably corresponds to the following electronic transition: a) n=5 to n=1 b) n=4 to n=1 c) n=3 to n=1 d) n=4 to n=2 e) n=5 to n=2 Chem 6A Michael J. Sailor, UC San Diego 35 Solution: Energy of electronic transitions Another way to solve this problem: Assume n1 = 1, calculate the value of n2 given that λ= 486.3 nm ⎛ྎ 1 1 1 ⎞ྏ = R H ⎜ྎ 2 − 2 ⎟ྏ λ ⎝ྎ n1 n 2 ⎠ྏ ⎛ྎ 1 1 ⎞ྏ ⎛ྎ 1 1 ⎞ྏ = R H ⎜ྎ − 2 ⎟ྏ = R H ⎜ྎ1 − 2 ⎟ྏ λ ⎝ྎ 1 n 2 ⎠ྏ ⎝ྎ n 2 ⎠ྏ € Rearranging, € 1 1 1− =2 λ ⋅ RH n2 Chem 6A Michael J. Sailor, UC San Diego 36 Solution: Energy of electronic transitions 1 1 1− =2 λ ⋅ RH n2 Solve for n2: ⎛ྎ ⎞ྏ−1/ 2 1 1 n2 = = 1− ⎛ྎ ⎞ྏ ⎜ྎ λ ⋅ R H ⎟ྏ 1 ⎝ྎ ⎠ྏ € ⎜ྎ1 − ⎟ྏ ⎝ྎ λ ⋅ R H ⎠ྏ ⎞ྏ−1/ 2 ⎛ྎ 1 = 1.1 n 2 = ⎜ྎ1 − −2 ⎟ྏ ⎝ྎ 486.3 ⋅ 1.097 × 10 ⎠ྏ 37 Chem 6A Michael J. Sailor, UC San Diego Solution: Energy of electronic transitions 1.1 is not a valid quantum number! Assume that n1 = 2 ⎛ྎ 1 1 ⎞ྏ ⎛ྎ 1 1 ⎞ྏ 1 = R H ⎜ྎ 2 − 2 ⎟ྏ = R H ⎜ྎ − 2 ⎟ྏ λ ⎝ྎ 2 n 2 ⎠ྏ ⎝ྎ 4 n 2 ⎠ྏ Solve for n2: −1/ 2 ⎛ྎ 1 1 1 ⎞ྏ =− n2 = ⎛ྎ 1 ⎞ྏ ⎜ྎ 4 λ ⋅ R H ⎟ྏ 1 ⎝ྎ ⎠ྏ ⎜ྎ − ⎟ྏ ⎝ྎ 4 λ ⋅ R H ⎠ྏ ⎞ྏ−1/ 2 ⎛ྎ 1 = 3.998 = 4 n 2 = ⎜ྎ0.25 − −2 ⎟ྏ ⎝ྎ 486.3 ⋅ 1.097 × 10 ⎠ྏ € Chem 6A Michael J. Sailor, UC San Diego 38 Solution: Energy of electronic transitions So the transition is n = 4 to n = 2; the answer is (d): a) n=5 to n=1 b) n=4 to n=1 c) n=3 to n=1 d) n=4 to n=2 e) n=5 to n=2 Chem 6A Michael J. Sailor, UC San Diego 39 Problem: Ionization energy and electron screening (a) Using the Schroedinger expression for energy of a 1-electron atom, calculate the ionization energy of an electron in the 3s orbital of a sodium atom, in kJ/mol. Chem 6A Michael J. Sailor, UC San Diego 40 The Periodic Table of the Elements 1 18 1 H 2 2 1.0079 3 13 14 15 6.941 9.01218 11 12 10.811 13 12.011 14.0067 15.9994 18.9984 20.1797 14 15 16 17 18 Al Si Na Mg 22.9898 19 K Rb Cs 4 21 Ra 7 24 V 8 25 Cr Y Zr Nb Mo 91.224 92.9064 72 73 La Hf Ta Co 55.847 58.9332 44 45 Tc W 10 27 Fe Ru 95.94 98.9063 74 75 178.49 180.948 104 105 9 26 Mn 47.88 50.9415 51.9961 54.9381 40 41 42 43 132.905 137.327 138.906 87 88 89 Fr 6 23 Ti 87.62 88.9059 56 57 Ba 5 22 Sc 40.078 44.9559 38 39 Sr 85.4578 55 3 24.305 20 Ca 39.0983 37 B Rh 101.07 102.906 76 77 Re Os 183.85 186.207 106 107 Ir 190.2 108 192.22 109 28 Ni 58.69 46 Pd 11 63.546 47 Ag P 26.9815 28.0855 30.9738 30 31 32 33 29 Cu N Zn 65.39 48 Cd Ga 69.723 49 Ge As 72.61 74.9216 50 51 In Sn Sb 106.42 107.868 112.411 78 79 80 114.82 81 118.71 82 121.75 83 Pt Tl Pb Au Hg 195.08 196.967 200.59 204.383 207.2 Bi O S 9 F Cl Ne Ar 32.066 35.4527 34 35 39.948 36 Se Kr 78.96 52 Te Br 79.904 53 I Xe 127.6 126.905 84 85 Po 83.8 54 At 131.29 86 Rn 208.98 208.982 209.987 222.018 Ac Unq Unp Unh Uns Uno Une 223.02 226.025 227.028 - 58 59 60 61 62 63 64 65 Ce Pr Nd Pm Sm Eu Gd Tb 140.12 90 Lanthanides Actinides 12 C 8 4.0026 10 5 Be 7 He 17 4 Li 6 16 140.91 91 144.24 92 157.25 96 158.92 97 Th 232.038 Pa 231.04 U 238.03 146.92 93 150.35 94 151.96 95 66 67 Dy Ho 162.5 98 71 164.93 99 167.26 100 168.93 101 174.97 103 Fm Md No Pu Am Cm Bk Cf Es 239.05 251.08 254.09 249.08 70 Lu 237.05 247.07 69 Tm Yb Np 241.06 68 Er 257.1 258.1 173.04 102 255 Chem 6A Michael J. Sailor, UC San Diego Lr 262.1 41 Solution: E = -1310(Z2/n2) kJ/mol ⎛ྎ Z 2 Z 2 ⎞ྏ E = -1310 ⎜ྎ 2 − 2 ⎟ྏ ⎝ྎ n final ninitial ⎠ྏ ⎛ྎ 112 112 ⎞ྏ E = - 1310 ⎜ྎ 2 − 2 ⎟ྏ 3 ⎠ྏ ⎝ྎ ∞ ⎛ྎ 121 ⎞ྏ E = - 1310 ⎜ྎ − = 17, 612 kJ/mol ⎝ྎ 9 ⎠ྏ Chem 6A Michael J. Sailor, UC San Diego 42 Problem: Ionization energy (cont) (b) The measured ionization energy (IE) of an electron in the 3s orbital of a sodium atom is 495 kJ/mol. Why is this so far off from the value we just calculated? Chem 6A Michael J. Sailor, UC San Diego 43 Solution-Why is the answer so far off from the actual number? Calculate what would be the charge on the nucleus for an Ionization Energy of 495 kJ/mol: Electron only feels an “effective charge” of 1.84 (not 11) because of charge screening by the inner electrons Chem 6A Michael J. Sailor, UC San Diego 44 Failures of the Bohr model of the atom (1) Bohr model only works for hydrogen. It fails to predict any other element's gas phase spectrum. (2) Classical physics predicts that a charged particle undergoing acceleration radiates light. Electrons couldn't stay in fixed orbits (angular acceleration) according to classical physics. Chem 6A Michael J. Sailor, UC San Diego 45 Quantum mechanics mid 1920's: DeBroglie: an electron can be described as a wave Heisenberg, Schroedinger: used standing waves instead of orbits to describe electrons around atoms. Key features of quantum mechanics: (1) Electrons act like waves (2) They exist in specific modes, called wave functions (3) The wave functions are described using parameters called quantum numbers Chem 6A Michael J. Sailor, UC San Diego 46 Heisenberg’s Uncertainty Principle "You can't know both the exact position and the exact momentum of any particle at exactly the same time" h Δx ⋅ Δmv ≥ 4π so electrons randomly exist in some fuzzy probability haze around the nucleus € h = 6.6261 x 10-34 J.s Chem 6A Michael J. Sailor, UC San Diego 47 What is a wave function? Quantum mechanics treats the electron as a 3-dimensional standing wave Standing Waves: 1-d: Violin string y = sin(x) 2-d: the surface of a drum z = sin(x)cos(y) 3-d: real complicated to visualize z = sin(x)sin(y)sin(z) Chem 6A Michael J. Sailor, UC San Diego 48 Atomic wave functions (orbitals) Orbitals are probability maps, showing the surface that contains the electron 95% (or more) of the time. We also call these electron probability contours 3 s orbital 3 px orbital 3 dxz orbital www.rsc.org/chemsoc/visualelements/orbital/ 49 Chem 6A Michael J. Sailor, UC San Diego Quantum numbers Quantum Number Called Describes n l Principle quantum number SIZE and ENERGY Angular momentum (Azimuthal) quantum number SHAPE ml Magnetic quantum number ORIENTATION ms Electron spin quantum number INTRINSIC ANGULAR MOMENTUM OF THE ELECTRON Chem 6A Michael J. Sailor, UC San Diego 50 The Particle in a Box An example of a 2-dimensional standing wave for an electron 20 edelocalized wave Relative Energy elocalized particle n=4 15 10 n=3 5 confined wave n=2 n=1 What are the energies of these wavefunctions? 0 -0.2 0 0.2 0.4 x 0.6 0.8 1 1.2 L 51 Chem 6A Michael J. Sailor, UC San Diego Derivation of Particle in a Box Equation Classical physics: E = ½ mv2 Allowed wavelengths for electrons in the box: λ= 2L/n, where n = 1, 2, 3, … de Broglie relationship: λ = h/mv Relative Energy 20 n=4 15 10 n=3 5 n=2 n=1 0 -0.2 0 0.2 0.4 x 0.6 L Chem 6A Michael J. Sailor, UC San Diego 0.8 1 1.2 52 Derivation of Particle in a Box Equation Allowed wavelengths: λ= 2L/n, where n = 1, 2, 3, … Relative Energy de Broglie relationship: λ = h/mv so mv = h/λ Classical physics: E = ½ mv2 or E = ½ (mv)2/m substituting, E = ½ (h/λ)2/m = ½ h2/λ2m Since λ = 2L/n (from above) Then E = ½ h2n2/22L2m, or 2 E= 20 n=4 15 10 n=3 5 n=2 n=1 2 nh 8mL2 0 -0.2 0 0.2 0.4 x 0.6 0.8 1 1.2 L 53 Chem 6A Michael J. Sailor, UC San Diego € Stern-Gerlach Experiment: Ag atoms Screen OVEN Containing Ag Magnet The atoms split into two paths in a magnetic field This experiment tells us that each individual electron has a magnetic moment; there must be a 4th quantum number: Electron spin, or ms Chem 6A Michael J. Sailor, UC San Diego 54 Quantum numbers Quantum Number Called Describes n Principle quantum number SIZE and ENERGY l Angular momentum (Azimuthal) quantum number SHAPE ml Magnetic quantum number ORIENTATION ms Electron spin quantum number INTRINSIC ANGULAR MOMENTUM OF THE ELECTRON 55 Chem 6A Michael J. Sailor, UC San Diego Allowable values for quantum numbers Quantum Number values example n 1, 2, 3, …∞ 2 l 0…n-1 0, 1 ml -l …+l -1,0,1 ms +1/2, -1/2 +1/2 Chem 6A Michael J. Sailor, UC San Diego 56 Quantum numbers Chem 6A Michael J. Sailor, UC San Diego 57 Problem: Quantum numbers The set of quantum numbers n = 4, l = 2, ml = 0 and ms = +1/2 describes an electron in which orbital? a) 4f b) 4d c) 4p d) 4s e) none of the above Chem 6A Michael J. Sailor, UC San Diego 58 Answer: Quantum numbers n = 4 is principle QN (energy) l = 2 is the type of orbital: type s pdf So this is a d-type orbital value of l 0 1 2 3 ANSWER: 4d Not needed for this problem: ml = 0 is the orientation of the d-orbital ms =+1/2 is the spin on the electron Chem 6A Michael J. Sailor, UC San Diego 59 Question: How many orbitals in a shell? How many orbitals are in the n = 4 level? a) 16 b) 4 c) 32 d) 18 e) none of the above Chem 6A Michael J. Sailor, UC San Diego 60 Answer: How many orbitals in a shell? One way to think of this is that each time n increases by 1, it adds an additional l value, adding an additional set of orbitals: l = 0 l = 1 l = 2 l = 3 Total # of orbitals 1 n=1 1s n=2 2s 2p n=3 3s 3p 3d n=4 4s 4p 4d 1+3=4 1+3+5=9 4f 1+3+5+7=16 ANSWER: 16 61 Chem 6A Michael J. Sailor, UC San Diego Quantum dots: Artificial atoms http://www.invitrogen.com Chem 6A Michael J. Sailor, UC San Diego 62 Summary: The Particle in a Box elocalized particle edelocalized wave Relative Energy Confined electrons exist as standing waves, whose energies take on discrete values 20 n=4 15 10 n=3 5 confined wave 22 nh E= 8mL2 n=2 n=1 0 -0.2 0 0.2 0.4 x 0.6 0.8 1 1.2 L 63 Chem 6A Michael J. Sailor, UC San Diego Quantum dots: Artificial atoms O O OH O CH3 R = radius of the nanoparticle Chem 6A Michael J. Sailor, UC San Diego aspirin 64 “Artificial atoms” made from CdSe in solution The properties of a nanomaterial derive from its size— form determines function 2 nm 2.5 nm 3 nm 4 nm 5 nm 6 nm Felice Frankel Michael J. Sailor, UC San Diego Bawendi Chem 6A Michael J. Sailor, UC San Diego research group, MIT 65 Absorption and fluorescence spectra of quantum dots depend on the size of the dot Photoluminescence emission spectra Absorption spectra 61 Å 6 nm 34 hrs 12 hrs 80 min 3 nm 33 Å 35 min 3 min 2 nm 20 sec 22Å 10 sec 1.5 2.0 2.5 3.0 Energy (eV) 2.0 2.5 3.0 3.5 Energy (eV) Chem 6A Michael J. Sailor, UC San Diego 66 Biological Applications of Quantum Dots Advantages: Stable Many Distinct Colors 84 microns Applications: Biological Staining Drug Discovery Genomics Mouse 3T3 fibroblasts simultaneously stained with red and green quantum dots Silicon quantum dots imaging a tumor in a mouse "Semiconductor nanocrystals as fluorescent biological labels." Bruchez, M.; Moronne, M.; Gin, P.; Weiss, S.; Alivisatos, A. P. Science 1998, 281, 2013-2016. 67 Chem 6A Michael J. Sailor, UC San Diego Biodegradable silicon-based quantum dots Lift-off Etching in HF 210mA/cm2, 150s Free-standing film Porous Si Si substrate (P++) Ultrasonic fracture Activation H2O, 24h Filtering 200nm pores Luminescent Nanoparticles Nanoparticles Microparticles Park, J.-H. et al. Nature Mater. 2009, 8, 331-336. Chem 6A Michael J. Sailor, UC San Diego 68 Extras 69 Chem 6A Michael J. Sailor, UC San Diego 69 ...
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