Download From Last Time… Energy levels Emitting and absorbing light

Energy levels
From Last Time…
• Instead of drawing orbits, we can just indicate the energy
an electron would have if it were in that orbit.
• Hydrogen atom:
Zero energy
One electron orbiting around one proton (nucleus)
Electron can be in different “quantum states”
Quantum states labeled by integer n=1,2,3,4,…
In each different quantum state, electron has
n=4
n=3
n=2
• Different orbital radius
• Different energy
• Different wavelength
13.6
eV
32
E2 = "
13.6
eV
22
E1 = "
13.6
eV
12
!
– n=1 is lowest energy state,
energy depends on state as
13.6
" 2 eV
n
Fri. Mar. 24, 2006
!
E3 = "
Energy axis
–
–
–
–
n=1
Phy107 Lect. 24
1
Fri. Mar. 24, 2006
Phy107 Lect. 24
2
!
!
Emitting and absorbing light
Hydrogen atom
Zero energy
n=4
n=3
E3 = "
13.6
eV
32
n=2
E2 = "
13.6
eV
22
Photon
emitted
hf=E2-E1
n=1
!
!
E3 = "
13.6
eV
32
n=2
E2 = "
13.6
eV
22
Photon
absorbed
hf=E2-E1
E1 = "
13.6
eV
12
Photon is emitted when
!
electron drops
from one
quantum state to another
Fri. Mar. 24, 2006
n=4
n=3
n=1
!
!
E1 = "
An electron drops from an -1.5 eV energy level to one with
energy of -3.4 eV. What is the wavelength of the photon
emitted?
Zero energy
A. 650 nm
B. 400 nm
C. 250 nm
n=4
n=3
Photon
emitted
hf=E2-E1
13.6
eV
12
hf = hc/λ
= 1240 eV-nm/ λ
Absorbing a photon of correct
energy makes!electron jump to
higher quantum state.
Phy107 Lect. 24
3
n=2
E 3 = "1.5 eV
!
!
n=1
Fri. Mar. 24, 2006
E 2 = "3.4 eV
E1 = "13.6 eV
Phy107 Lect. 24
4
!
Energy conservation for Bohr atom
• Each orbit has a specific energy
En=-13.6/n2
• Photon emitted when electron
jumps from high energy to low
energy orbit.
Ei – Ef = h f
• Photon absorption induces
electron jump from
low to high energy orbit.
Ef – Ei = h f
• Agrees with experiment!
Fri. Mar. 24, 2006
Phy107 Lect. 24
Summary of Hydrogen atom
• Hydrogen atom:
–
–
–
–
One electron orbiting around one proton (nucleus)
Electron can be in different “quantum states”
Quantum states labeled by integer n=1,2,3,4,…
In each different quantum state, electron has
• Different orbital radius
• Different energy
• Different wavelength
– n=1 is lowest energy state,
energy depends on state as
13.6
" 2 eV
n
5
Fri. Mar. 24, 2006
Phy107 Lect. 24
6
!
1
Example: the Balmer series
Spectral Question
• All transitions terminate
at the n=2 level
• Each energy level has
energy En=-13.6 / n2 eV
• E.g. n=3 to n=2 transition
Compare the wavelength of a photon produced from
a transition from n=3 to n=1 with that of a photon
produced from a transition n=2 to n=1.
A. λ31 < λ 21
– Emitted photon has energy
C. λ31 > λ 21
## 13.6 & # 13.6 &&
E photon = %%" 2 ( " %" 2 (( = 1.89 eV
$$ 3 ' $ 2 ''
– Emitted wavelength
!
E photon
E31 > E21
hc
hc
1240 eV # nm
= hf = , " =
=
= 656 nm
"
E photon
1.89 eV
Fri. Mar. 24, 2006
n=3
n=2
B. λ31 = λ 21
Phy107 Lect. 24
7
so
λ31 < λ 21
Fri. Mar. 24, 2006
n=1
Phy107 Lect. 24
8
!
But why?
Resonance
• Most physical objects will vibrate at some set of
natural frequencies
• Why should only certain orbits be stable?
• Bohr had a complicated argument based on
“correspondence principle”
– That quantum mechanics must agree with classical
results when appropriate (high energies, large sizes)
• But incorporating wave nature of electron gives
a natural understanding of these
‘quantized orbits’
– Ringing bell
– Wine glass
– Musical instrument
• The electrons in an atom analogous to
sound waves in a musical instrument.
• In instrument, only certain pitches produced,
corresponding to particular vibration wavelengths.
• Since the electrons orbiting around the nucleus are
waves, only certain wavelengths are allowed.
Phy107 Lect. 24
9
Fri. Mar. 24, 2006
Phy107 Lect. 24
Resonances of a string
Resonance on a string
f=v/λ
λ/2
Fundamental,
wavelength 2L/1=2L,
frequency f
...
• Easier to think about in a normal wind instrument,
or vibrations of a string.
• Wind instrument with particular fingering plays a
particular pitch, particular wavelength.
• Guitar string vibrates at frequency, wavelength
determined by string length.
λ/2
wavelength 2L/2=L,
frequency 2f
λ/2
2nd harmonic,
λ=L/2
Phy107 Lect. 24
11
n=4
1st harmonic,
wavelength 2L/3,
frequency 3f
Fri. Mar. 24, 2006
10
Fri. Mar. 24, 2006
Phy107 Lect. 24
frequency
Fri. Mar. 24, 2006
n=3
n=2
n=1
Vibrational modes equally
spaced in frequency12
2
Not always equally spaced
String resonances
A string has a fundamental frequency of 440 Hz.
If I pluck it so that it vibrates at the first
harmonic (half the wavelength) what is the
frequency?
n=7
n=6
n=5
frequency
Wavelength has decreased by
factor of 2. Since f=v/λ,
frequency has gone up by
factor of two.
A. 440 Hz
B. 220 Hz
C. 880 Hz
n=4
n=3
n=2
Fri. Mar. 24, 2006
Phy107 Lect. 24
13
Fri. Mar. 24, 2006
•
Why not other wavelengths?
– In effect, the object “selects” the resonant
wavelengths by its physical properties.
Vibrational modes are accurately shown by means of
holographic interferometry, which displays a contour map of
the vibration. Several modes of vibration of a wine glass are
shown in the photo below. Points of maximum motion, which
occur around the rim, appear as "bull's eyes." The vibrational
amplitude changes by half a wavelength of light (316 nm) in
moving from one bright fringe to the next one. The
fundamental mode, designated as the (2,0) mode, has four
such regions, with the glass moving in alternate directions as it
vibrates.
• Electron is a wave.
• In the orbital picture, its propagation
direction is around the circumference
of the orbit.
• Wavelength = h / p
(p=momentum, and energy
determined by momentum)
• Reflection from ‘end’ interferes
destructively and ‘cancels out’ wave.
• Same happens in a wind instrument…
… and in an atom!
Phy107 Lect. 24
• How can we think about waves on a
circle?
15
Fri. Mar. 24, 2006
Waves on a circle
Wavelength
Vibrational modes
unequally spaced
14
Electron waves in an atom
• Waves not in the harmonic series are quickly
destroyed by interference
Fri. Mar. 24, 2006
Phy107 Lect. 24
Phy107 Lect. 24
16
Waves on a ring
• Here is my ‘ToneNut’
• Like a flute, but in the shape of a
doughnut.
• Produces particular pitch.
• Air inside must be vibrating at that
frequency
• Sound wave inside has wavelength
λ=v/f (red line).
Wavelength
• Condition on a ring
slightly different.
• Integer number of
wavelengths required
around circumference.
• Otherwise destructive
interference occurs when
wave travels around ring
and interferes with itself.
• What determines the
frequency/wavelength of the
sound?
Blow in here
Fri. Mar. 24, 2006
Phy107 Lect. 24
17
Fri. Mar. 24, 2006
Phy107 Lect. 24
18
3
Hydrogen atom music
Hydrogen atom music
• These are the five lowest energy
orbits for the one electron in the
hydrogen atom.
• Here the electron is in the
n=3 orbit.
• Three wavelengths fit along
the circumference of the
orbit.
• The hydrogen atom is playing
its third highest note.
• Highest note (shortest
wavelength) is n=1.
• Each orbit is labeled by the
quantum number n.
• The radius of each is n2ao.
• Hydrogen has one electron:
the electron must be in one of
these orbits.
• The smallest orbit has the lowest
energy. The energy is larger for
larger orbits.
Fri. Mar. 24, 2006
Phy107 Lect. 24
19
Fri. Mar. 24, 2006
Hydrogen atom music
• Here the electron is in the
n=5 orbit.
• Five wavelengths fit along the
circumference of the orbit.
• The hydrogen atom is playing
its next lowest note.
• The sequence goes on and on,
with longer and longer
wavelengths, lower and lower
notes.
Phy107 Lect. 24
21
Fri. Mar. 24, 2006
Hydrogen atom energies
• Wavelength gets longer in
higher n states, (electron
moving slower) so kinetic
energy goes down.
n=4
n=3
13.6
E 3 = " 2 eV
3
E2 = "
!
13.6
eV
22
Energy
n=2
!
• End result is
En = "
13.6
eV
n2
n=1
E1 = "
Phy107 Lect. 24
22
Hydrogen atom question
Zero energy
• But energy of Coulomb
interaction between electron
(-) and nucleus (+) goes up
faster with bigger n.
20
Hydrogen atom music
• Here the electron is in the
n=4 orbit.
• Four wavelengths fit along the
circumference of the orbit.
• The hydrogen atom is playing
its fourth highest note (lower
pitch than n=3 note).
Fri. Mar. 24, 2006
Phy107 Lect. 24
13.6
eV
12
Here is Peter Flanary’s
sculpture ‘Wave’ outside
Chamberlin Hall. What
quantum state of the
hydrogen atom could this
represent?
A. n=2
B. n=3
C. n=4
!
Fri. Mar. 24, 2006
!
Phy107 Lect. 24
23
Fri. Mar. 24, 2006
Phy107 Lect. 24
24
4
Another question
General aspects of Quantum Systems
Here is Donald Lipski’s sculpture
‘Nail’s Tail’ outside Camp Randall
Stadium.
What could it represent?
• System has set of quantum states, labeled by an integer
(n=1, n=2, n=3, etc)
• Each quantum state has a particular frequency and energy
associated with it.
A. A pile of footballs
• These are the only energies that the system can have:
the energy is quantized
B. “I hear its made of plastic. For 200
grand, I’d think we’d get granite”
• Analogy with classical system:
– System has set of vibrational modes, labeled by integer
fundamental (n=1), 1st harmonic (n=2), 2nd harmonic (n=3), etc
- Tim Stapleton (Stadium Barbers)
C. “I’m just glad it’s not my money”
– Each vibrational mode has a particular frequency and energy.
- Ken Kopp (New Orlean’s Take-Out)
– These are the only frequencies at which the system resonates.
Fri. Mar. 24, 2006
Phy107 Lect. 24
Fri. Mar. 24, 2006
25
Example: ‘Particle in a box’
Quantum Particle in a Box
– Wave reflects back and forth from the walls.
– Reflections cancel unless wavelength meets the
standing wave condition:
integer number of half-wavelengths fit in the tube.
L
• Classically, ball bounces back and forth in tube.
– No friction, so ball continues to bounce back and forth,
retaining its initial speed.
momentum
" = 2L
– Could label each state with a speed,
momentum=(mass)x(speed), or kinetic energy.
n=1
One halfwavelength
– This is a ‘classical state’ of the ball. A different classical state would
be ball bouncing back and forth with different speed.
!
"=L
27
Fri. Mar. 24, 2006
!
p=
n=2
Two halfwavelengths
– Any momentum, energy is possible.
Can increase momentum in arbitrarily small increments.
Phy107 Lect. 24
26
• In Quantum Mechanics, ball represented by wave
Particle confined to a fixed region of space
e.g. ball in a tube- ball moves only along length L
Fri. Mar. 24, 2006
Phy107 Lect. 24
!
Phy107 Lect. 24
28
Quantized energy levels
• Quantized momentum
h
h
p= =n
= npo
"
2L
• Energy = kinetic
A particle in a box has a mass m.
It’s energy is all energy of motion = p 2/2m.
We just saw that it’s momentum in state n is npo.
It’s energy levels
!
p 2 ( npo )
=
= n2Eo
2m
2m
• Or Quantized Energy
E=
Fri. Mar. 24, 2006
29
n=5
n=4
n=3
En = n2Eo
!
Energy
2
A. are equally spaced everywhere
B. get farther apart at higher energy
C. get closer together at higher energy.
Phy107 Lect. 24
momentum
h h
p = = = 2 po
" L
!
Particle in box question
Fri. Mar. 24, 2006
h h
=
# po
" 2L
n=2
n=1
Phy107 Lect. 24
30
!
5
The wavefunction of a particle
• We use a probabilistic interpretation
Particle in a box: Wavefunctions
Wavefunction
Probability = (Wavefunction)2
– The wavefunction Ψ(x) (psi) of a particle describes
the extended, wave-like properties.
– The square magnitude of the wavefunction |Ψ|2 gives
the probability of finding the particle at a particular
spatial location
• Similar to the interpretation used for light waves
– Square of the electric field gives
light intensity = number of photons / second.
Phy107 Lect. 24
31
Phy107 Lect. 24
Probability = (Wavefunction)2
0.5
Probability
0.0
• Now here is something unusual.
– In the middle of the box, probability of finding
the particle is ZERO!
– How can we understand this?
Fri. Mar. 24, 2006
Phy107 Lect. 24
33
Probability
Fri. Mar. 24, 2006
Probability
Tails
1 2 3 4
5 6
1/6
Phy107 Lect. 24
34
Particle in a box: Wavefunctions
Loaded die
Wavefunction
Probability
1/6
0.0
• “Continuous”
probability
distribution
Fri. Mar. 24, 2006
Heads
0.5
0.0
Discrete vs continuous
0.5
32
Understanding Probability
Next highest energy state
Wavefunction
Fri. Mar. 24, 2006
Probability
Fri. Mar. 24, 2006
• Ground state wavefunction and probability.
• Height of probability curve represents
likelihood of finding particle at that point.
1 2 3 4 5
Third
state
6
Next higher
state
0.5
1/6
0.0
Lowest energy
state
0 1 2 3 4 5 6
Phy107 Lect. 24
35
Fri. Mar. 24, 2006
Phy107 Lect. 24
36
6
Quantum Corral
Probability of finding electron
• Classically, equally likely to find particle anywhere
• QM - true on average for high n
D. Eigler (IBM)
Zeroes in the probability!
Purely quantum, interference effect
Fri. Mar. 24, 2006
Phy107 Lect. 24
37
• 48 Iron atoms assembled into a circular ring.
• The ripples inside the ring reflect the electron quantum states of
a circular ring (interference effects).
Fri. Mar. 24, 2006
Phy107 Lect. 24
38
7