Chem 6A, Section D Oct 13, 2011 - Sailor Research Group...

Chem 6A Michael J. Sailor, UC San Diego

Chapter 7:Quantum Theory

Chem 6A, Section D Oct 13, 2011


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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2


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Problem: Fraunhofer Diffraction

Track spacing = d

Incident laser beam

beams in phase

CD surface

CD tracks

3

4


diffra

cted b

eam


Track spacing = d

beams in phase

CD surface

θ

θ

Additional path traveled



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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6


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


Discrete vs Continuous Spectra

aa

600550500450400Wavelength, nm

White LightSpectrum

Ne LineSpectrum

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0

0.2

0.4

0.6

0.8

1

0 500 1000 1500 2000 2500 3000 3500 4000

Inte

nsity

Frequency (Hz)

Frequency Spectra-Ne gas vs violinViolin G audible spectrum

0

0.2

0.4

0.6

0.8

1

3 1014 3.5 1014 4 1014 4.5 1014 5 1014 5.5 1014 6 1014

Inte

nsity

Frequency (Hz)

Ne gas visible spectrum


Chapter 7 (cont)Chem 6A, Section D Oct 18, 2011

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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


Quiz 3 score histogram

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F D C- C C+ B- B B+ A- A A+

num

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of s

tude

nts

Grades so far(after quizzes 1-3)


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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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:


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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UVA and UVB

UVA = 400 nm–315 nmUVB = 315 nm–280 nmUVC = 280 nm–100 nm


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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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


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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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


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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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


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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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


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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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.


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

€

1λ

=1.097 ×10−2 122−132

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =1.097 ×10

−2 14−19

⎛

⎝ ⎜

⎞

⎠ ⎟

λ = 656 nm

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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


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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


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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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


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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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


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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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



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

€

1λ

= RH122−1n22

⎛

⎝ ⎜

⎞

⎠ ⎟ = RH

14−1n22

⎛

⎝ ⎜

⎞

⎠ ⎟

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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


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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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


Solution:

E = -1310 Z 2

nfinal2 −Z 2

ninitial2

⎛

⎝ ⎜

⎞

⎠ ⎟

E = -1310 112

∞2 −112

32

⎛

⎝ ⎜

⎞

⎠ ⎟

E = -1310 −121

9⎛

⎝ ⎜ ⎞

⎠ =17, 612 kJ/mol

E = -1310(Z2/n2) kJ/mol

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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?


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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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.


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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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


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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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


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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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?


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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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


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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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


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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Quantum numbers


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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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


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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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


Quantum dots: Artificial atomshttp://www.invitrogen.com

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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


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


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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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


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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Extras

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Chem 6A, Section D Oct 13, 2011 - Sailor Research Group...

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