chem 6a, section d oct 13, 2011 - sailor research group...
TRANSCRIPT
Chem 6A Michael J. Sailor, UC San Diego
Chapter 7:Quantum Theory
Chem 6A, Section D Oct 13, 2011
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?
diffracted beam (n=2)diffracted beam (n=1)specular beam
ytanθ = x/y
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Chem 6A Michael J. Sailor, UC San Diego
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Chem 6A Michael J. Sailor, UC San Diego
Problem: Fraunhofer Diffraction
Track spacing = d
Incident laser beam
beams in phase
CD surface
CD tracks
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Chem 6A Michael J. Sailor, UC San Diego
diffra
cted b
eam
Problem: Fraunhofer Diffraction
Track spacing = d
beams in phase
CD surface
θ
θ
Additional path traveled
Chem 6A Michael J. Sailor, UC San Diego
Problem: Fraunhofer Diffraction
Track spacing = d
θ
Additional path traveled
d sinθ = nλd = nλ/sinθtanθ = (257 cm / 720 cm)θ = 0.3428d = (1 x 530 nm) / sin(0.3428)= 1577 nm, or 1.577 micrometers
diffra
cted b
eam (n
=2)
diffra
cted b
eam (n
=1)
specul
ar be
am
ytanθ = x/y
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Chem 6A Michael J. Sailor, UC San Diego
Solution: Diffraction of light from a CD grating
dsinθ = nλd = nλ/sinθ
tanθ = (257 cm / 720 cm)θ = 0.3428d = (1 x 530 nm) / sin(0.3428)= 1577 nm, or 1.577 micrometers
diffracted beam (n=2)diffracted beam (n=1)specular beam
ytanθ = x/y
Chem 6A Michael J. Sailor, UC San Diego
Discrete vs Continuous Spectra
aa
600550500450400Wavelength, nm
White LightSpectrum
Ne LineSpectrum
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Chem 6A Michael J. Sailor, UC San Diego
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0 500 1000 1500 2000 2500 3000 3500 4000
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Frequency (Hz)
Frequency Spectra-Ne gas vs violinViolin G audible spectrum
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3 1014 3.5 1014 4 1014 4.5 1014 5 1014 5.5 1014 6 1014
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nsity
Frequency (Hz)
Ne gas visible spectrum
Chem 6A Michael J. Sailor, UC San Diego
Chapter 7 (cont)Chem 6A, Section D Oct 18, 2011
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Chem 6A Michael J. Sailor, UC San Diego
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
Quiz 3 score histogram
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Chem 6A Michael J. Sailor, UC San Diego
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Chem 6A Michael J. Sailor, UC San Diego
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
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Chem 6A Michael J. Sailor, UC San Diego
Color Wavelength EnergyInfrared >800 nm
R Red 630 nm 2.0 eVO Orange 590 nm 2.1 eVY Yellow 560 nm 2.2 eVG Green 510 nm 2.4 eVB Blue 440 nm 2.8 eVI Indigo 420 nm 3.0 eVV Violet 400 nm 3.1 eV
Ultraviolet <350 nm
eV = electron Volts
Visible Light Wavelengths and Energies:
Chem 6A Michael J. Sailor, UC San Diego
Comparison of light wavelengths to hair thickness:
Name Hair thickness,micrometers (µm)
Hair thickness,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
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Chem 6A Michael J. Sailor, UC San Diego
UVA and UVB
UVA = 400 nm–315 nmUVB = 315 nm–280 nmUVC = 280 nm–100 nm
Chem 6A Michael J. Sailor, UC San Diego
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.
Problem: Energy-Wavelength conversion
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Chem 6A Michael J. Sailor, UC San Diego
Solution: Energy-Wavelength conversion
hcλ
E = =
6.6261 x 10-34 J.s 3 x 108 m 109 nm eVs 450 nm m 1.6022 x 10-19 J
= 2.76 eV
Chem 6A Michael J. Sailor, UC San Diego
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.
Problem: Energy-Wavelength conversion
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Chem 6A Michael J. Sailor, UC San Diego
Solution: Energy-Wavelength conversion
hcλ
E = =
6.6261 x 10-34 J.s 3 x 108 m 109 nms 450 nm m
= 4.42 x 10-19 J
Chem 6A Michael J. Sailor, UC San Diego
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?
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Chem 6A Michael J. Sailor, UC San Diego
Solution: Particle-wave duality of matter
€
λ =hmv
for an electron
λ =6.626×10−34kg ⋅m2 / s9.11×10−31kg ⋅ 35m / s
×109nm
1m= 21,000nm
λ =hmv
for a baseball
λ =6.626×10−34kg ⋅m2 / s
0.1kg ⋅ 35m / s×
109nm1m
= 1.9×10−25nm
Chem 6A Michael J. Sailor, UC San Diego
Problem: PhotochemistryThe 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
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Chem 6A Michael J. Sailor, UC San Diego
Solution: photochemistryEnergy in a bond:
496 kJ mol 1000Jmol 6.02x1023 molecule kJ
= 8.24x10-19 J/moleculeE= hc/λλ = hc/E =
molecule 6.626x10-34 J sec 3x108 m8.24x10-19 J sec
= 2.41 x 10-7 m = 241 nm
Chem 6A Michael J. Sailor, UC San Diego
Discrete vs Continuous Spectra
aa
600550500450400Wavelength, nm
White LightSpectrum
Ne LineSpectrum
Why do atoms give off such complicated spectra when they get hot?
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Chem 6A Michael J. Sailor, UC San Diego
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.
Chem 6A Michael J. Sailor, UC San Diego
Solution: Wavelength of atomic transitions
€
1λ
= RH1n12 −
1n22
⎛
⎝ ⎜
⎞
⎠ ⎟
The Rydberg equation (7.3, pg 221 in text)
RH = 1.097 x 10-2 nm-1, n1 = 2 and n2 = 3
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1λ
=1.097 ×10−2 122−132
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ =1.097 ×10
−2 14−19
⎛
⎝ ⎜
⎞
⎠ ⎟
λ = 656 nm
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Chem 6A Michael J. Sailor, UC San Diego
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
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Chem 6A Michael J. Sailor, UC San Diego
Solution: Energy of Atomic TransitionsThe Rydberg expression tells you the relationship between wavelength (λ) and the transition between two levels:
€
1λ
= RH1n12 −
1n22
⎛
⎝ ⎜
⎞
⎠ ⎟
RH = Rydberg constantn1 = starting atomic leveln2 = ending atomic level
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):
Note h.c.R = (6.626x10-34J.s)(3x108m/s)(1.0967x107m-1) = 2.18x10-18J
To calculate the energy difference between two levels:
Z = nuclear chargeh = Planck’s constantc = speed of lightRH = Rydberg constant
€
E =Z 2hcRn12
⎛
⎝ ⎜
⎞
⎠ ⎟ H
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Chem 6A Michael J. Sailor, UC San Diego
Solution: Energy of atomic transitions
Applying the Rydberg equation to answer (a):
= -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
n1 = 5, n2 = 1
Chem 6A Michael J. Sailor, UC San Diego
Solution: Energy of atomic transitionscalculate the wavelength of each transition:
So the answer is (d) n=4 to n=2
transition λ, nma) n=5 to n=1 95b) n=4 to n=1 97c) n=3 to n=1 102d) n=4 to n=2 486e) n=5 to n=2 434
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Chem 6A Michael J. Sailor, UC San Diego
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
Solution: Energy of electronic transitions
€
1λ
= RH1n12 −
1n22
⎛
⎝ ⎜
⎞
⎠ ⎟
€
1− 1λ ⋅ RH
=1n22
Rearranging,
Another way to solve this problem: Assume n1 = 1, calculate the value of n2 given that λ= 486.3 nm
€
1λ
= RH11−1n22
⎛
⎝ ⎜
⎞
⎠ ⎟ = RH 1−
1n22
⎛
⎝ ⎜
⎞
⎠ ⎟
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Chem 6A Michael J. Sailor, UC San Diego
Solution: Energy of electronic transitions
Solve for n2:
€
1− 1λ ⋅ RH
=1n22
€
n2 =1
1− 1λ ⋅ RH
⎛
⎝ ⎜
⎞
⎠ ⎟
= 1− 1λ ⋅ RH
⎛
⎝ ⎜
⎞
⎠ ⎟
−1/ 2
€
n2 = 1− 1486.3 ⋅1.097 ×10−2
⎛
⎝ ⎜
⎞
⎠ ⎟ −1/ 2
=1.1
Chem 6A Michael J. Sailor, UC San Diego
Solution: Energy of electronic transitions
Solve for n2:
€
n2 =1
14−
1λ ⋅ RH
⎛
⎝ ⎜
⎞
⎠ ⎟
=14−
1λ ⋅ RH
⎛
⎝ ⎜
⎞
⎠ ⎟
−1/ 2
€
n2 = 0.25 − 1486.3 ⋅1.097 ×10−2
⎛
⎝ ⎜
⎞
⎠ ⎟ −1/ 2
= 3.998 = 4
1.1 is not a valid quantum number! Assume that n1 = 2
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1λ
= RH122−1n22
⎛
⎝ ⎜
⎞
⎠ ⎟ = RH
14−1n22
⎛
⎝ ⎜
⎞
⎠ ⎟
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Chem 6A Michael J. Sailor, UC San Diego
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
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.
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Chem 6A Michael J. Sailor, UC San Diego
The Periodic Table of the Elements1 18
1 2
H 2 13 14 15 16 17 He1.0079 4.0026
3 4 5 6 7 8 9 10
Li Be B C N O F Ne6.941 9.01218 10.811 12.011 14.0067 15.9994 18.9984 20.1797
11 12 13 14 15 16 17 18
Na Mg 3 4 5 6 7 8 9 10 11 12 Al Si P S Cl Ar22.9898 24.305 26.9815 28.0855 30.9738 32.066 35.4527 39.948
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr39.0983 40.078 44.9559 47.88 50.9415 51.9961 54.9381 55.847 58.9332 58.69 63.546 65.39 69.723 72.61 74.9216 78.96 79.904 83.8
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe85.4578 87.62 88.9059 91.224 92.9064 95.94 98.9063 101.07 102.906 106.42 107.868 112.411 114.82 118.71 121.75 127.6 126.905 131.29
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn132.905 137.327 138.906 178.49 180.948 183.85 186.207 190.2 192.22 195.08 196.967 200.59 204.383 207.2 208.98 208.982 209.987 222.018
87 88 89 104 105 106 107 108 109
Fr Ra Ac Unq Unp Unh Uns Uno Une223.02 226.025 227.028 - - - - - -
58 59 60 61 62 63 64 65 66 67 68 69 70 71
Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu140.12 140.91 144.24 146.92 150.35 151.96 157.25 158.92 162.5 164.93 167.26 168.93 173.04 174.97
90 91 92 93 94 95 96 97 98 99 100 101 102 103
Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr232.038 231.04 238.03 237.05 239.05 241.06 247.07 249.08 251.08 254.09 257.1 258.1 255 262.1
Chem 6A Michael J. Sailor, UC San Diego
Solution:
E = -1310 Z 2
nfinal2 −Z 2
ninitial2
⎛
⎝ ⎜
⎞
⎠ ⎟
E = -1310 112
∞2 −112
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⎛
⎝ ⎜
⎞
⎠ ⎟
E = -1310 −121
9⎛
⎝ ⎜ ⎞
⎠ =17, 612 kJ/mol
E = -1310(Z2/n2) kJ/mol
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Chem 6A Michael J. Sailor, UC San Diego
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
Solution-Why is the answer so far off from the actual number?
Electron only feels an “effective charge” of 1.84 (not 11) because of charge screening by the inner electrons
Calculate what would be the charge on the nucleus for an Ionization Energy of 495 kJ/mol:
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Chem 6A Michael J. Sailor, UC San Diego
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
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
mid 1920's: DeBroglie: an electron can be described as a waveHeisenberg, Schroedinger: used standing waves instead of orbits to describe electrons around atoms.
Key features of quantum mechanics:
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Chem 6A Michael J. Sailor, UC San Diego
Heisenberg’s Uncertainty Principle
"You can't know both the exact position and the exact momentum of any particle at exactly the same time"
so electrons randomly exist in some fuzzy probability haze around the nucleus
€
Δx ⋅ Δmv ≥ h4π
h = 6.6261 x 10-34 J.s
Chem 6A Michael J. Sailor, UC San Diego
What is a wave function?
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)
Quantum mechanics treats the electron as a 3-dimensional standing wave
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Chem 6A Michael J. Sailor, UC San Diego
Atomic wave functions (orbitals)
3 s orbital 3 px orbital 3 dxz orbital
www.rsc.org/chemsoc/visualelements/orbital/
Orbitals are probability maps, showing the surface that contains the electron 95% (or more) of the time. We also call these electron probability contours
Chem 6A Michael J. Sailor, UC San Diego
Quantum numbersQuantum Number
Called Describes
n Principle quantum number SIZE and ENERGY
l Angular momentum (Azimuthal) quantum number
SHAPE
mlMagnetic quantum number ORIENTATION
msElectron spin quantum number
INTRINSIC ANGULAR MOMENTUM OF THE ELECTRON
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Chem 6A Michael J. Sailor, UC San Diego
The Particle in a Box
e-
localized particle
delocalized wave
confined wave
e-
0
5
10
15
20
-0.2 0 0.2 0.4 0.6 0.8 1 1.2
Rela
tive
Ener
gy
x
n = 1
n = 2
n = 3
n = 4
L
An example of a 2-dimensional standing wave for an electron
What are the energies of these wavefunctions?
Chem 6A Michael J. Sailor, UC San Diego
Derivation of Particle in a Box Equation
0
5
10
15
20
-0.2 0 0.2 0.4 0.6 0.8 1 1.2
Rela
tive
Ener
gy
x
n = 1
n = 2
n = 3
n = 4
L
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
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Chem 6A Michael J. Sailor, UC San Diego
Derivation of Particle in a Box Equation
0
5
10
15
20
-0.2 0 0.2 0.4 0.6 0.8 1 1.2
Rela
tive
Ener
gy
x
n = 1
n = 2
n = 3
n = 4
L
Allowed wavelengths: λ= 2L/n, where n = 1, 2, 3, …
de Broglie relationship:λ = h/mv so mv = h/λ
Classical physics: E = ½ mv2 or E = ½ (mv)2/msubstituting,E = ½ (h/λ)2/m = ½ h2/λ2mSince λ = 2L/n (from above)Then E = ½ h2n2/22L2m, or
€
E =n2h2
8mL2
Chem 6A Michael J. Sailor, UC San Diego
Stern-Gerlach Experiment:
OVENContaining Ag
ScreenAg atoms
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
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Chem 6A Michael J. Sailor, UC San Diego
Quantum numbersQuantum Number
Called Describes
n Principle quantum number SIZE and ENERGY
l Angular momentum (Azimuthal) quantum number
SHAPE
mlMagnetic quantum number ORIENTATION
msElectron spin quantum number
INTRINSIC ANGULAR MOMENTUM OF THE ELECTRON
Chem 6A Michael J. Sailor, UC San Diego
Allowable values for quantum numbersQuantum 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
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Chem 6A Michael J. Sailor, UC San Diego
Quantum numbers
Chem 6A Michael J. Sailor, UC San Diego
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
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Chem 6A Michael J. Sailor, UC San Diego
Answer: Quantum numbersn = 4 is principle QN (energy)l = 2 is the type of orbital:type s p d fvalue of l 0 1 2 3
So this is a d-type orbital
Not needed for this problem:ml = 0 is the orientation of the d-orbitalms =+1/2 is the spin on the electron
ANSWER: 4d
Chem 6A Michael J. Sailor, UC San Diego
How many orbitals are in the n = 4 level? a) 16 b) 4 c) 32 d) 18 e) none of the above
Question: How many orbitals in a shell?
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Chem 6A Michael J. Sailor, UC San Diego
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
n = 1 1s 1
n = 2 2s 2p 1+3=4
n = 3 3s 3p 3d 1+3+5=9
n = 4 4s 4p 4d 4f 1+3+5+7=16
ANSWER: 16
Chem 6A Michael J. Sailor, UC San Diego
Quantum dots: Artificial atomshttp://www.invitrogen.com
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Chem 6A Michael J. Sailor, UC San Diego
Summary: The Particle in a Box
e-
localized particle
delocalized wave
confined wave
e-
€
E =n2h2
8mL20
5
10
15
20
-0.2 0 0.2 0.4 0.6 0.8 1 1.2
Rela
tive
Ener
gy
x
n = 1
n = 2
n = 3
n = 4
L
Confined electrons exist as standing waves, whose energies take on discrete values
Chem 6A Michael J. Sailor, UC San Diego
Quantum dots: Artificial atoms
O OH
OO
CH3
aspirinR = radius of the nanoparticle
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Chem 6A Michael J. Sailor, UC San DiegoMichael J. Sailor, UC San Diego
Felice Frankel
“Artificial atoms” made from CdSe in solution
Bawendi research group, MIT
6 nm2 nm 3 nm2.5 nm 5 nm4 nm
The properties of a nanomaterial derive from its size—form determines function
Chem 6A Michael J. Sailor, UC San Diego
Absorption and fluorescence spectra of quantum dots depend on the size of the dot
Energy (eV)
2.0 2.5 3.0 3.5
Energy (eV)1.5 2.0 2.5 3.0
34 hrs
12 hrs
80 min
35 min
3 min
20 sec
10 sec
33 Å
61 Å
22Å
Photoluminescence emission spectra
Absorption spectra
6 nm
2 nm
3 nm
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Chem 6A Michael J. Sailor, UC San Diego
Biological Applications of Quantum Dots
Advantages:StableMany Distinct Colors
"Semiconductor nanocrystals as fluorescent biological labels." Bruchez, M.; Moronne, M.; Gin, P.; Weiss, S.; Alivisatos, A. P. Science 1998, 281, 2013-2016.
84 microns
Mouse 3T3 fibroblasts simultaneously stained with red and green quantum dots
Applications:Biological StainingDrug DiscoveryGenomics
Silicon quantum dots imaging a tumor in a mouse
Chem 6A Michael J. Sailor, UC San Diego
Biodegradable silicon-based quantum dots
Etching in HF
210mA/cm2, 150s
Lift-off
Ultrasonic fracture H2O, 24h
Si substrate (P++) Porous Si Free-standing film
Microparticles
Filtering
200nm pores
Nanoparticles
Activation
Luminescent Nanoparticles
Park, J.-H. et al. Nature Mater. 2009, 8, 331-336.
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Chem 6A Michael J. Sailor, UC San Diego
Extras
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