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Chem 4502 Quantum Mechanics & Spectroscopy (Jason Goodpaster)

Chap. 1: Dawn of Quantum Theory

Text: McQuarrie and Simon's "Physical Chemistry: A Molecular Approach", 1997. (All figures and tables are from our text unless otherwise noted.)

Lecture 2:

pp. 1-6 Blackbody radiation; Planck's constant, h

pp. 7-10 Photoelectric effect;

Einstein�s explanation; Ephoton = h ν

Lecture 3:

pp. 10-14 Hydrogen atom spectrum

Rydberg formula

pp. 15-18 de Broglie waves

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By 1900, many advances in physics • 1897 - JJ Thomson discovered

electron, measured its mass/charge ratio • Maxwell�s equations completed in

1873-4 partial differential equations that can solve all problems in classical electricity and magnetism

But there were a few experimental results that could not be explained...

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Blackbody Radiation

Figure: Engel�s �Quantum Chem�

Ideal Blackbody:

Absorbs and emits radiation without favoring particular frequencies

E.g., pinhole in sealed container

Radiation emitted is characteristic of the temperature of the body.

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Fig. 1.1 p. 3 Measured frequency distribution of emitted blackbody radiation

As T increases:

• maximum shifts to higher frequency (ν)

• total radiated energy (area under curve) increases

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Problem....

Dashed line: theoretical prediction (Rayleigh – Jeans Law)

�Ultraviolet Catastrophe�: energy density of radiation predicted to diverge as ν2

Predicts infinite total energy emitted!

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Blackbody radiation: Planck�s theory

Planck (unwillingly) introduced the concept of energy quantization into physics.

E = n h ν

ν frequency of light (sec-1) absorbed or emitted

by an electron oscillating in the blackbody

n positive integer (0, 1, 2, ...)

E energy of an oscillating electron; for a given ν,

the energies of the oscillators are discrete h a new constant; units must be energy x time Note: Planck didn�t quantize the light

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Blackbody radiation: Planck�s theory

Using statistical thermodynamics, he derived a formula for the energy density of the radiation.

The value of �h� was fit to the data.

Planck was able to fit the data, using the value:

h = 6.63 x 10-34 J·sec

This success alone did not convince people (including Planck) that energy is really quantized.

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Photoelectric Effect

Ejection of an electron (e-) from a metal by radiation.

Experiment is:

Shine light on a metal surface.

Measure the kinetic energies of the electrons ejected (if any).

See how their energies vary with the intensity and frequency of light.

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Early measurements of the photoelectric effect: Schematic diagram of a typical apparatus

light

from Karplus & Porter �Atoms and Molecules� p. 53

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Photoelectric Effect: Predictions of Classical Physics

Electron kinetic energy (eKE): • should increase with intensity of light (proportional to Amplitude2 of electric field) • should not depend on frequency of light As amplitude of oscillating electric field increases,

electrons on metal surface should oscillate more and escape with more energy.

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Fig. 1.4 p. 8 Experimental Observations: Photoelectric Effect (Sodium)

eKE :

• increases linearly with frequency of light

• does not depend on intensity of light

• no photoelectrons detected below a certain ν

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Photoelectric Effect: Einstein�s Theory

Particle model of light: light exists as small packages of energy, now called �photons�

Ephoton = hν

Ephoton energy in one photon ν frequency of the photon

This applies to light, not to matter (not to electrons, atoms, etc.)

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Photoelectric Effect: Einstein�s Theory

Conservation of energy: Ephoton = hν = Φ + ½ mev2

Ephoton energy in one photon absorbed by one electron in the metal Φ work function of the metal (minimum energy needed to free an electron) ½ mev2 kinetic energy of the photodetached electron

One electron in the metal absorbs one photon, in a sort of collision.

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� The next revolution in physics will occur when the properties of mind will be included in the equations of quantum theory. �

Eugene Wigner

1963 Physics Nobel Laureate

Quantum Quote of the Day

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Chap. 1: Dawn of Quantum Theory (continued)

Lecture 3:

pp. 15-18 de Broglie waves

pp. 10-14 Hydrogen atom spectrum

Rydberg formula

Lecture 4:

pp. 18-23 Bohr theory of the H atom

pp. 23-25 Uncertainty Principle

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Louis de Broglie

1892-1987

(Fig. 18.2 p. 38)

� After the first World War, I gave a great deal of thought to the theory of quanta. It was then that I had a sudden inspiration. Einstein�s wave-particle dualism for light was an absolutely general phenomenon extending to all physical nature.�

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de Broglie�s Hypothesis (1924)

Einstein showed that the energy of a relativistic particle moving freely in space is:

E = (m2c4 + p2c2)½ where m = mass, p = momentum

For a photon, m = 0 and E = hν, so

hν = ( 0 + p2c2)½ = p c

p = hν / c photon momentum (in the direction of propagation), another particle aspect of radiation

Also, ν = c / λ and 1 / λ = ν / c, where λ is the wavelength of light and ν is its frequency, so have:

p = h / λ λ = h / p for light

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de Broglie generalized this equation to matter as well:

λ = h / p

where p = classical momentum = m v (mass x velocity)

λ = de Broglie wavelength of the material particle

Matter waves ?!? �He [de Broglie] has lifted one corner of the great veil.�

Einstein

de Broglie�s Hypothesis

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de Broglie�s Hypothesis How to test this strange hypothesis?

But: to detect interference effects, must detect differences on the order of λ between the lengths of alternate �paths�.

Is this possible??

Recall 2-slit experiment with light:

Maxima on detector occur where r2 – r1 = nλ (n = 1, 2, 3, ...)

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de Broglie�s Hypothesis Can we observe a de Broglie wavelength?

Problem: What is the de Broglie wavelength of a golf ball (50 grams) traveling at 30 meters/second?

The size of a nucleus of an atom is about 10-15 meters. Conclude: golf ball displays only particle-like properties. (What if it�s at rest? For this, need uncertainty principle.)

λ = h / p h = 6.6 x 10-34 J∙s

λ = 4.4 x 10-34 meters = 4.4 x 10-22 picometers (pm)

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de Broglie�s Hypothesis

•  Problem: What is the de Broglie wavelength of an electron (me = 9.11 x 10-31 kg) accelerated through a potential of 54.0 Volts?

λ = h / p h = 6.63 x 10-34 J∙s e = 1.60 x 10-19 C

λ = 1.67 x 10-10 m = 1.67 Å = 167 pm This wavelength is on the scale of atomic diameters,

and it is in the x-ray region.

�Slits�: spacings between atoms in a crystal.

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Davisson-Germer experiment, 1927:

Scattering of an electron beam from a nickel crystal shows constructive and destructive interference (Atkins �Physical Chemistry� 6th Ed. p. 293)

Constructive interference occurs at angles at which the electron waves scattered by different Ni atoms have path lengths differing by nλ.

Destructive: nλ/2, n odd → �nodes�. 24

For single-plane diffraction: n λ = a sin θ a = distance between Ni atoms = 2.15 x 10-10 m (known from x-ray diffraction experiments) θ = angle at which intensity maximum is observed n = 1 for the strong 1st order peak For electrons accelerated to 54.0 V, D&G measured

the maximum scattering angle (n=1) to be 50.1o. What is the wavelength, λ, of the electrons ?

λ = a sin θ / n = 2.15 x 10-10 m (sin 50.1o) / 1

λ = 1.65 x 10-10 m

Same as we calculated from the de Broglie formula !

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Fig. 1.8 p. 17 Similarity between X-ray and Electron Diffraction Patterns (Aluminum Foil)

X-rays Electrons

George Thomson - used ≈ 10 keV electrons to penetrate thin metal films, recorded diffraction patterns

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These electron diffraction experiments confirmed the de Broglie wavelength:

λ = h / p

Nobel Prizes in Physics:

1929 de Broglie

1937 G. P. Thomson and C. J. Davisson

1906 JJ Thomson (George�s Dad) discovery of the electron as a subatomic particle (1987)

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The electron would land at a localized spot on the detector (no interference pattern).

What would we see if we passed

a single electron through Davisson and Germer�s or Thomson�s experiments ?

This is like the single photon 2-slit experiments –

wave-particle duality applies to both light and matter. 28

IT Characterization Facility 12 Shepherd Labs Transmission electron microscope (TEM) with an operating voltage range of 20 to 120 kV. Allows magnifications of thin samples (<500 nm) up to 700,000 times. Point-point resolution: 0.34 nm Line resolution: 0.2 nm

http://www.charfac.umn.edu/instruments/

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Chemical and Engineering News Sept. 2004

DUMBBELLS STEM image of a silicon crystal in the [112] orientation reveals pairs of atom columns in which the intrapair separation is 0.78 Å.

A milestone in electron microscopy--the first direct sub-angstrom imaging of a crystal lattice--has been reported by researchers at Oak Ridge National Laboratory and Nion, a company in Kirkland, Wash., that specializes in advanced electron-microscope optics [Science, 305, 1741 (2004)]. The researchers fitted a 300-kV scanning transmission electron microscope (STEM) at ORNL with a Nion aberration corrector...

SEE DOWN TO 0.6 Å Electron microscope achieves direct sub-angstrom imaging of a crystal

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Emission spectra of atoms (e.g., hydrogen):

another experiment that baffled classical physics.

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Fig. 1.7 p. 13 Hydrogen Atom Emission Spectrum

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Emission spectrum of H atom

The observed wavelengths emitted (�lines�) were found to fit the Rydberg formula:

ν = 1 / λ (cm) = 109,680 cm-1 ( 1 / n1

2 – 1 / n22)

where n1 and n2 are integers (n2 > n1)

Why would H atom care about integers?

∼

�wavenumbers�

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Rydberg formula for H atomν = 1 / λ (cm) = 109,680 cm-1 ( 1 / n1

2 – 1 / n22)

What are n1 and n2 for the shortest wavelength

emission line of the H atom? What is this wavelength, in nm? In what region of the electromagnetic spectrum is it?

∼

n1=1, n2 = ∞

ν = 109,680 cm-1

λ = 1 / ν = 1 / 109, 680 cm-1 = 91.2 x 10-7 cm

λ = 91.2 x 10-9 m = 91.2 nm UV region

∼

∼

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Note on use of wavenumbers (cm-1)

Spectroscopists often refer to cm-1 as an "energy" unit.

Really cm-1 is Energy / hc

What cm-1 of light corresponds to an energy of 1. eV?

For light, E = hc / λ . For E = 1. eV :

ν = 1 / λ = E / hc = 1.6022 x 10-19 J

(6.6261 x 10-34 Js)(2.9979 x 1010 cm/s)

ν = 8066 cm-1 corresponds to 1. eV of energy ∼

∼

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Chap. 1: Dawn of Quantum Theory (continued)

Lecture 4:

pp. 18-23 Bohr theory of the H atom pp. 23-25 Uncertainty Principle

Lecture 5:

Chap. 2 Classical Wave Equation:

pp. 39-49 Classical wave equation (1-dim.)

MathChapter A pp. 31-34 Complex numbers

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p. 190 Niels Bohr

1885-1962

Bohr designed his coat of arms with a yin-yang symbol. Inscription: CONTRARIA SUNT COMPLEMENTA (Opposites are Complements)

Energy levels, Principle of Complementarity 38

Bohr�s Model of the H Atom

• The electron can orbit around the nucleus only at discrete values of r, the orbital radius.

r • These �stationary orbits� must satisfy the condition:

mev r = n ℏ where n = 1, 2, 3,..

Here, ℏ (�h-bar�) is shorthand for h/2π.

( me v r ) is the angular momentum (L) of the electron (with me = e- mass, v = velocity, r = orbital radius).

Thus, Bohr assumed that the angular momentum is quantized.

quantum number

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Fig. 1.9 p. 19 de Broglie Waves in Bohr Orbits

Stable orbit - circumference must be: 2πr = nλ

Combine with de Broglie relation: λ = h / mev 2πr = n (h / mev) mevr = n (h/2π) = nℏ Same as

above

Stable Unstable: wave progressively disappears →

Can obtain Bohr�s mevr = nℏ by considering

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Bohr Model of the H Atom

Total energy (E) of the electron in the H atom: E = ½ mev2 + (-e2 / (4πεo r))

kinetic potential

For a stable orbit, must have a balance of forces: e2/(4πεo r2) = mev2/r

centrifugal (pseudo) outward force

Coulomb attraction between electron and proton

Multiply both sides by ½ r: e2/(8πεor) = ½ mev2

Substitute into equation for E above.

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Bohr Model of the H Atom

After substituting, the total energy of the electron is:

E = e2/(8πεor) + (-e2 / (4πεo r))

Simplifying: E = - e2 / (8πεo r)

kinetic potential

Negative energy → bound states

The electron is bound to the nucleus.

Recall: only discrete values of r are possible

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Bohr Model of the H Atom

E = -e2 / (8πεo r) Next: Obtain r in terms of fundamental constants.

We had: e2/(8πεor) = ½ mev2

Also: mev r = n h / 2π

Solve for v, then substitute here

Result: r = εoh2n2 / π me e2

Substituting values (p. 20 eqn. 1-18), obtain for n=1 the radius of the first (lowest energy) Bohr orbit:

r1 = 0.529 Å �ao, Bohr radius�

Now used as basic distance unit in �atomic units�.

r increases as n2

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Bohr Model of the H Atom

We had: E = -e2 / (8πεo r)

Substitute: r = εoh2n2 / π m e2 (n = 1, 2, ...)

Obtain: E = [ - m e4 / (8 εo2 h2) ] / n2

Quantized energy levels; energies go as 1/n2

where n (�quantum number�) is an integer (1, 2, 3, ...)

Recall: Rydberg Formula for H atomν = 1 / λ (cm) = 109,680 cm-1 ( 1 / n1

2 – 1 / n22)

empirical formula based on measurements

∼

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Bohr Model of the H Atom

Bohr prediction: E = [ - m e4 / (8 εo2 h2) ] / n2

Evaluate the constants in [brackets] (use m = 9.1044 x 10-31 kg, reduced mass of electron and proton)

E = - 2.18 x 10-18 J / n2 = - 13.6 eV / n2

I.e., 13.6/n2 eV is the energy (with respect to that of the

separated e- and nucleus) by which an electron in a state with quantum number n is bound to the nucleus.

Convert to wavenumbers (and use more sig figs):

E / hc = ν = - 109,680 cm-1 / n2

∼

Recall: Rydberg Formula ν = 1 / λ (cm) = 109,680 cm-1 ( 1 / n1

2 – 1 / n22)

SAME # !!

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Bohr Model of the H Atom Bohr prediction: E / hc = ν = -109,680 cm-1 / n2

To emit a photon, an H atom must make a transition

from a higher energy (less negative), higher n (�n2�) state (�E2�)

to a lower energy (more negative), lower n (�n1�) state (�E1�).

Then, the energy of the emitted photon is:

Ephoton = hν = ΔEsystem = E2 – E1 Bohr Frequency Condition

Or, in wavenumbers:

ν = 109,680 cm-1 (1 / n12 – 1 / n2

2)

in agreement with the Rydberg formula.

∼

∼

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Fig. 1.10 p. 21 Energy Level Diagram for the Hydrogen Atom

n=1, -109,680 cm-1

(-13.6 eV) r1 = 0.529 Å

n=∞, 0 cm-1

Ionization potential of H atom: 13.6 eV

n=2, -109,680/4 cm-1

r2 = 0.529*4=2.12 Å

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Bohr Model of the H Atom

Despite its success in predicting the H atom emission spectrum, the Bohr model had major problems:

☹ did not predict the effects of external magnetic fields on the H atom ☹ could not be successfully extended to atoms with more than 1 electron ☹ incompatible with uncertainty principle

In Chap. 6, we will revisit the H atom spectrum

using the Schrödinger equation. 48

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p. 114 Werner Heisenberg

1901-1976

Heisenberg�s Uncertainty Principle: Δx Δp ≥ h

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Heisenberg�s Uncertainty Principle (mid 1920�s) One form of the uncertainty principle:

Δx Δpx ≥ h (1-dimension)

Δx is the uncertainty in the particle�s location

Δpx ������������������������������������������� momentum

Δ px = m Δ vx

Therefore, the more narrowly the position is measured, the more uncertainty there is in the particle�s momentum (velocity), and vice versa.

If the position were known exactly, the momentum would be completely unknown.

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Heisenberg�s Uncertainty Principle

Δx Δp ≥ h

This can be understood in terms of fundamental limits on measurements.

It is rooted in the wave nature of matter.

�Complementary observables�

position and momentum

energy and time

some aspects of angular momentum (Ch. 6)

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Understanding the uncertainty principle in terms of superpositions of waves Atkins �Physical Chemistry� 5th Ed., p. 384

Δx Δ(1/λ) ≥ 1 (classical) smaller Δx requires larger range of λ�s

Add de Broglie:

1/ λ = p / h

gives

Δx Δp ≥ h

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Uncertainty Principle: Apply to electrons

What is the minimum uncertainty in the speed of an electron known to be located within 0.529 Å (Bohr radius for n=1) of the nucleus of a H atom?

[ h = 6.626x10-34 J·s; me = 9.11 x 10-31 kg ]

This is much larger than the e- speed in the Bohr model: mvr = nℏ so v = n ℏ / mr = 2 x 106 m/s

Bohr model isn�t consistent with the uncertainty principle.

Δx (mΔv) = h (minimum) so Δv = h / (m Δx)

Δv = 6.626x10-34 J·s / ((9.11x10-31 kg)(0.529 X 10-10 m))

Δv = 1.4 x 107 m/s

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Uncertainty Principle: Apply to Macroscopic Objects

What is the minimum uncertainty in the position of a golf ball (assume m=50. g exactly) known to be traveling at a speed of 30.00000±0.00001 m/s? [ h = 6.626x10-34 J·s ]

Δx (mΔv) = h (minimum) so Δx = h / m Δv

Δx = 6.626x10-34 J·s/ (0.050 kg)(0.00001 m)

Δx = 1 x 10-27 m This about 1012 times smaller than the size of a nucleus (∼10-15 m) !

The uncertainty principle applies to all matter, but its consequences are (usually) not observable for macroscopic objects.

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Uncertainty Principle: Questions to Ponder

Does the electron really have a well-defined position, speed and momentum at any given moment, but we can only measure it to within these uncertainties?

Or, are its position and momentum inherently �fuzzy� to these extents?

If you answered the former: what does it mean for a property to exist, if it cannot possibly be observed?

How does the uncertainty principle relate to the effects of the observer on the properties of that which is observed? 56

Uncertainty Principle: Questions to Ponder

What is the uncertainty in the position of a golf ball thrown up in the air, just at the point when it is changing direction and not moving?

Why does the transporter on Star Trek require Heisenberg compensators?

What would the world be like if h were a lot bigger?