Early Quantum Mechanics
Chapter 27
Three Major Discoveries
Light is both a Electron
wave and a
Orbitals are
particle
Quantized
Planck
Einstein
Compton
Bohr
The electron
(and all
matter) is both
a wave and a
particle
De Broglie
(Later
Schrodinger/Hei
senberg)
Wein’s Law
• Treat’s light solely as a wave
• Hot “blackbodies” radiate EM
• The hotter the object, the shorter the peak
wavelength
• Sun (~6000 K) emits in blue and UV
• A 3000 K object emits in IR
2.90 X 10-3 mK = lpeakT
Wein’s Law: Ex 1
Estimate the temperature of the sun. The Suns light
peak at about 500 nm (blue)
500 nm = 500 X 10-9 m
T = 2.90 X 10-3 mK
lpeak
T = 2.90 X 10-3 mK = 6000 K
500 X 10-9 m
Wein’s Law: Ex 2
Suppose a star has a surface temperature of about
3500 K. Estimate the wavelength of light
produced.
2.90 X 10-3 mK = lpeakT
l peak = 2.90 X 10-3 mK
T
l peak = 2.90 X 10-3 mK = 8.29 X 10-7 m = 830 nm
3500 K
Planck’s Quantum Hypothesis
• Energy of any atomic or molecular vibration is a
whole number
• Photon – the light particle
• Photons emitted come in “packets”
• E = hf
• h = 6.626 X 10-34 J s
(Planck’s constant)
Photons: Ex 1
Calculate the energy of a photon of wavelength 600
nm.
600 nm X 1 X 10-9m = 6 X10-7 m
1 nm
c = lf
f = c/l = (3X108 m/s)/(6 X10-7 m) = 5 X 1014 s-1
E = hf
E = (6.626 X 10-34 J s)(5 X 1014 s-1) = 3.3 X 10-19 J
Photons: Ex 1a
Convert your answer from the previous problem to
electron Volts (1 eV = 1.6 X 10-19 J)
Photons: Ex 2
Calculate the energy of a photon of wavelength 450
nm (blue light).
Photons: Ex 2a
Convert your answer from the previous problem to
electron Volts (1 eV = 1.6 X 10-19 J)
Photons: Ex 3
Estimate the number of photons emitted by a 100
Watt lightbulb per second. Assume each photon
has a wavelength of 500 nm.
500 nm X 1 X 10-9m = 5 X10-7 m
1 nm
c = lf
f = c/l = (3X108 m/s)/(5 X10-7 m) = 6 X 1014 s-1
100 Watts = 100 J/s (we are looking at 1 second)
E = nhf
n = E/hf
n = 100 J
(6.626 X 10-34 J s)(6 X 1014 s-1)
n = 2.5 X 1020
Photoelectric Effect (Einstein)
• When light shines on a
metal, electrons are
emitted
• Can detect a current from
the electrons
• Used in light meter,
scanners, digital cameras
(photodiodes rather than
tubes)
Three Key Points
1. Below a certain frequency, no electrons are
emitted
2. Greater intensity light produces more electrons
3. Greater Frequency light produces no more
electrons, but the come off with greater speed
Low Frequency
Not enough energy to
eject electron
High Frequency
Can eject electron
Energy of photon is
greater than W (ionization
energy
2. More intensity
– More photons
– More electrons ejected with same KE
3. Greater
Frequency
– No more
electrons ejected
– Electrons come
off with greater
speed (KE)
hf = KE + W
hf = energy of the photon
KE = Maximum KE of the emitted electron
W = Work function to eject electron
Metal
Na
Al
Cu
Zn
Ag
Pt
Pb
Fe
Work Function
(eV)
2.46
4.08
4.70
4.31
4.73
6.35
4.14
4.50
hf = KE + W: Ex 1
What is the maximum kinetic energy of an electron
emitted from a sodium atom whose work function
(Wo) is 2.28 eV when illuminated with 410 nm
light?
410 nm = 410 X 10-9 m or 4.10 X 10-7 m
2.28 eV X 1.60 X 10-19J
1 eV
=
3.65 X 10-19 J
c = lf
f = c/l
f = c/(4.10 X 10-7 m) = 7.32 X 1014 s-1
hf = KE + W
KE = hf – W
KE = (6.626 X 10-34 J s)(7.32 X 1014 s-1) - 3.65 X
10-19 J
KE = 1.20 X 10-19 J or 0.75 eV
hf = KE + W: Ex 2
What is the maximum kinetic energy of an electron
emitted from a sodium atom whose work function
(Wo) is 2.28 eV when illuminated with 550 nm
light?
ANS: 2.25 eV
hf = KE + W: Ex 3
What is the maximum wavelength of light (cutoff
wavelength) that will eject an electron from an
Aluminum sample? Aluminum’s work function
(Wo) is 4.08 eV?
4.08 eV X 1.60 X 10-19J
1 eV
=
6.53 X 10-19 J
hf = KE + W
hf = 0 + W
(looking for bare minimum)
c = lf
f = c/l
hf = W
hc = W
l
l = hc = (6.626 X 10-34 J s)(3.0 X 108 m/s)
W
6.53 X 10-19 J
l = 3.04 X 10-7 m = 304 nm (UV)
Photon/Matter Interactions
1. Electron excitation (photon disappears)
2. Ionization/photoelectric effect (photon
disappears)
3. Scattering by nucleus or electron
4. Pair production (photon disappears)
Electron Excitation
Ionization/Photoelectric Effect
•Photon is absorbed (disappears)
•Photon is absorbed (disappears)
•Electron jumps to an excited
state
•Electron is propelled out of the
atom
Scattering
•Photon collides with a nucleus
or electron
•Photon loses some energy
•Speed does not change, but the
wavelength increases
Pair Production
•Photon closely approaches a
nucleus
•Photon disappears
•An electron and positron are
created.
3. Scattering: Compton Effect
• Electrons and nuclie can scatter photons
• Scattered photon is at a lower frequency than
incident photon
• Some of the energy is transferred to the electron
or nucleus
l’ = l + h (1 – cos q)
moc
l’ = wavelength of
scattered photon
l = wavelength of
incident photon
mo = rest mass of particle
q = angle of incidence
Compton Effect: Ex 1
X-rays of wavelength 0.140 nm are scatterd from a
block of carbon. What will be the wavelength of
the X-rays scattered at 0o?
l’ = l + h (1 – cos q)
moc
l’ = 140 X10-9m + (6.626 X 10-34 J s)(1 – cos 0)
(9.11 X 10-31 kg)(3 X 108m/s)
l’ = 140 X10-9m + 0
l’ = 140 nm
Compton Effect: Ex 2
What will be the wavelength of the X-rays scattered
at 90o?
l’ = l + h (1 – cos q)
moc
l’ = 140 X10-9m + (6.626 X 10-34 J s)(1 – cos 90)
(9.11 X 10-31 kg)(3 X 108m/s)
l’ = 140 X10-9m + 2.4 X 10-12 m
l’ = 142 nm
Compton Effect: Ex 3
What will be the wavelength of the X-rays scattered
at 180o?
l’ = l + h (1 – cos q)
moc
l’ = 140 X10-9m + (6.626 X 10-34 J s)(1 – cos 180)
(9.11 X 10-31 kg)(3 X 108m/s)
l’ = 140 X10-9m + 4.8 X 10-12 m
l’ = 145 nm (straight back)
4. Pair Production
• Photon Disappears
• e- and e+ are produced
• They have opposite direction (law of
conservation of momentum)
• When e- and e+ collide they annihilate each other
 a new photon appears
Principle of Complimentarity
• Neils Bohr
• Any experiment can only observe light’s wave or
particle properties, not both
• Different “faces” that light shows
Wave
Particle
Prism
Blackbody Radiation
Photoelectric effect
The Discovery of the Electron
(Thomson)
• Cathode Ray Tube
• Charged particles produced (affected by magnetic
field)
• Concluded that atom must have positive and
negative parts
• Electron – negative part of the atom
• Only knew the e/m ratio
• Plum Pudding Model
Charge and Mass of the Electron
(Millikan)
• Oil drop experiment
• Determines charge on electron (uses electric field
to counteract gravity)
• Quantized
• e = 1.602 X 10-19 C
• m = 9.11 X 10-31 kg
The Nucleus (Rutherford)
• Gold Foil Experiment
• Discovers nucleus (disproves Plum Pudding
Model)
• Planetary Model
Wave Nature of Matter
• Everything has both wave and particle properties
• DeBroglie Wavelength
E2 = p2c2 + m2c4
E2 = p2c2
E = pc
E = hf
hf = pc
c = lf
(consider a photon)
(photon has no mass)
hf = plf
p = mv
(for a particle)
hf = mvlf
h = mvl
l= h
mv
Everything has a wavelength
Diffraction pattern of
electrons scattered off
aluminum foil
DeBroglie Wavelength: Ex 1
Calculate the wavelength of a baseball of mass 0.20
kg moving at 15 m/s
l= h
mv
l = (6.626 X 10-34 J s)
(0.20 kg)(15 m/s)
l = 2.2 X 10-34 m
DeBroglie Wavelength: Ex 2
Calculate the wavelength of an electron moving at
2.2 X 106 m/s
l= h
mv
l = (6.626 X 10-34 J s)
(9.11 X 10-31 kg)(2.2 X 106 m/s)
l = 3.3 X 10-10 m or 0.33 nm
DeBroglie Wavelength: Ex 3
Calculate the wavelength of an electron that has
been accelerated through a potential difference of
100 V
V = PE
q
V = 1 mv2
2 q
v2 = 2qV/m
(PE =KE)
v = (2qV/m)1/2
v=[(2)(1.602 X 10-19 C)(100V)/(9.11 X 10-31 kg)]1/2
v = 5.9 X 106 m/s
l= h
mv
l = (6.626 X 10-34 J s)
(9.11 X 10-31 kg)(5.9 X 106 m/s)
l = 3.3 X 10-10 m or 0.33 nm
Electron Microscope
• Electron’s
wavelength is
smaller than light
• Magnetic focusing
DNA
Line Spectra
• Discharge tube
– Low density gas (acts like isolated atoms)
– Run a high voltage through it
– Light is emitted
• Light is emitted only at certain (discrete)
wavelengths
• Gases absorb light at the same frequency that
they emit
Hydrogen
Helium
Solar absorption spectrum
Explaining the Lines
Lyman Series
1 = R1 l
12
1
n2
Balmer Series
1 = R1 l
22
1
n2
Paschen Series
1 = R1 1
l
32
n2
Bohr Model: Hydrogen
• Electrons orbit in ground state (without radiating
energy)
• Classically, electrons should radiate energy since
that are accelerating because they are changing
directions
• Jumps to excited state by absorbing a photon
• Returns to ground state by emitted a photon
Bohr Model
hf = Ee - Eg
Bohr’s Equation
• Worls only for H
and other 1electron atoms
(He+, Li2+, Be3+,
etc…)
• Energy of ionized
atom is set at 0
• Orbital energies
are below zero
En = -13.6 eV
n2
En
= Energy of an orbital
-13.6 eV = Orbital of hydrogen closest to nucleus
n
= Number of the orbital
(1 eV = 1.60 X 10-19 J)
Bohr’s Equation: Ex 1
Calculate the energy of the first three orbitals of
hydrogen
En = -13.6 eV
n2
E1 = -13.6 eV
12
E1 = -13.6 eV
E2 = -13.6 eV
22
E2 = -3.4 eV
E3 = -13.6 eV
32
E3 = -1.51 eV
Bohr’s Equation: Ex 2
What wavelength of light is emitted if a hydrogen
electron drops from the n=2 to the n=1 orbit?
E1 = -13.6 eV
E2 = -3.4 eV
DE = (-3.4eV - -13.6 eV) = 10.2 eV
DE = (10.2 eV)(1.60 X 10-19J/eV) = 1.63 X 10-18J
DE = hf
f = c/l
DE = hc/l
l = hc
DE
l = (6.626 X 10-34 J s)(3.0 X 108 m/s)
(1.63 X 10-18J)
l = 1.22 X 10-7 m = 122 nm (UV)
Bohr’s Equation: Ex 3
Calculate the wavelength of light emitted if a
hydrogen electron drops from the n=6 to the n=2
orbit?
(ANS: 410 nm (violet))
Bohr’s Equation: Ex 4
Calculate the wavelength of light that must be
absorbed to exite a hydrogen electron from the
n=1 to the n=3 orbit?
(ANS: 103 nm (UV))
Bohr Model: Other Atoms
En = (Z2)(-13.6 eV)
n2
Z = Atomic number of the element (H=1, He=2, etc)
Other Atoms: Ex 1
Calculate the ionization energy of He+. This is the
energy needed to move an electron from n=1 to
zero.
E1 = (22)(-13.6 eV)
12
E1 = -54.4 eV
Other Atoms: Ex 2
What wavelength of light would be required to
ionize He+
E1 = -54.4 eV  8.70 X 10-18 J
E = hf
E = hc/l
l = hc = (6.626 X 10-34 J s)(3.0 X 108 m/s)
E
(8.70 X 10-18 J)
l = 22.8 nm
Other Atoms: Ex 3
Calculate E1 for a Li2+ ion.
E1 = (32)(-13.6 eV)
12
E1 = -122.4 eV
Other Atoms: Ex 4
What wavelength of light would be emitted from a
n=3 to n=1 transition in He+ ?
(ANS: l = 34.2 nm)
Wave/Particle Duality
DeBroglie
• Each e- is actually a standing wave
• Only certain wavelengths produce resonance
Forbidden Zone
Circumference = 2pr
2prn = nl
l= h
mv
mvrn = nh
2p