Transcript Last Time…

From Last Time(s)…
Light shows both
particle and wave-like
properties
Atoms emit and
absorb photons
Einitial
Photon: E=hf
Photon
Efinal
Stable
orbit
Stable
orbit
Tues. Nov. 24, 2009
Phy208 Lect. 24
1
Exam 3 is
Thursday Dec. 3 (after Thanksgiving)
5:30-7 pm, Birge 145
Students w / scheduled academic
conflict please stay after class Tues.
Nov. 24 to arrange alternate time.
Covers: all material since exam 2.
Bring: Calculator
One (double-sided) 8 1/2 x 11 note sheet
Schedule:
Week14HW: assigned Thur. Nov. 19, due Fri. Dec. 4 (two weeks)
Exam 3 practice problems available at Mastering Physics
Last material for exam: Lecture of Tues. Nov. 24
Exam review: Tuesday, Dec. 1, in class
Tues. Nov. 24, 2009
Phy208 Lect. 24
2
Photon properties of light





Photon of frequency f has energy hf
 E
photon  hf  hc / 

h  6.626 1034 J  s  4.14 1015 eV  s

hc 1240eV  nm
Red light made of ONLY red photons

The intensity of the beam can be increased by
increasing the number of photons/second.

(#Photons/second)(Energy/photon) =
energy/second = power
Tues. Nov. 24, 2009
Phy208 Lect. 24
3
Emitting and absorbing light
Zero energy
n=4
n=3
13.6
E 3   2 eV
3
n=2
13.6
E 2   2 eV
2
Photon
emitted
hf=E2-E1
n=1


E3  
13.6
eV
32
n=2
E2  
13.6
eV
22
E1  
13.6
eV
12
Photon
absorbed
hf=E2-E1
E1  
13.6
eV
12
Photon is emitted when electron
drops fromone quantum
state to another
Thurs. Nov. 19, 2009
n=4
n=3
n=1


Absorbing a photon of correct

energy makes electron jump
to higher quantum state.
Phy208 Lect. 23
4
Matter waves


If light waves have particle-like properties,
maybe matter has wave properties?
de Broglie postulated that the
wavelength of matter
is related to momentum as
h

p

This is called
the de Broglie wavelength.
Tues. Nov. 24, 2009
Phy208 Lect. 24
Nobel prize, 1929
5
Why h / p ? Works for photons

Wave interpretation of light:



wavelength = (Speed of Light) / Frequency
=c/f
Particle interpretation of light (photons):

Energy = (Planck’s constant) x Frequency

E = hf, so f = E / h
c
c
h
Wavelength =  =


f E /h E /c
But photon momentum = p = E / c…
h

for a photon
p
Tues. Nov. 24, 2009
Phy208 Lect. 24
6


h
We argue that   applies to everything
p
Photons and footballs
both follow the same relation.


Everything has both
wave-like and particle-like properties
Tues. Nov. 24, 2009
Phy208 Lect. 24
7
Wavelengths of massive objects
h
deBroglie wavelength =  
p


p=mv
h

mv


Tues. Nov. 24, 2009
Phy208 Lect. 24
8
Matter Waves


deBroglie postulated that matter has wavelike
properties.
deBroglie wavelength   h / p
Example:
Wavelength of electron with 10 eV of energy:

Kinetic energy
p2
E KE 
 p  2mE KE
2m
h
hc
1240eV  nm



 0.39nm
2
6
2mE KE
2mc E KE
20.51110 eV 10eV 
Tues. Nov. 24, 2009
Phy208 Lect. 24
9
Wavelength of a football

Need m, v
to find 

Momentum:

Make the Right Call: The NFL's Own interpretations and
guidelines plus 100s of official rulings on game situations.
National FootBall League, Chicago. 1999:
"... short circumference, 21 to 21 1/4 inches;
weight, 14 to 15 ounces.”
(0.43 - 0.40 kg)
“Sometimes I don’t know how they catch that ball, because
Brett Aaron wings that thing 60, 70 mph,” Flanagan said.
(27 - 32 m/s)
Wells
mv  0.4 kg30 m /s  12 kg m /s
h 6.6 1034 J  s
35
26
 
 5.5 10 m  5.5 10 nm
p 12 kg m /s
Tues. Nov. 24, 2009
Phy208 Lect. 24
10
This is very small




1 nm = 10-9 m
Wavelength of red light = 700 nm
Spacing between atoms in solid ~ 0.25 nm
Wavelength of football = 10-26 nm
• What makes football wavelength so small?
h
h
 
p mv
Tues. Nov. 24, 2009
Large mass, large momentum
short wavelength
Phy208 Lect. 24
11
Suppose an electron is a wave…

Here is a wave:
h

p
x


…where is the electron?

Wave extends infinitely far in +x and -x direction
Tues. Nov. 24, 2009
Phy208 Lect. 24
12
Analogy with sound


Sound wave also has the same characteristics
But we can often locate sound waves


E.g. echoes bounce from walls. Can make a sound pulse
Example:




Hand clap: duration ~ 0.01 seconds
Speed of sound = 340 m/s
Spatial extent of sound pulse = 3.4 meters.
3.4 meter long hand clap travels past you at 340 m/s
Tues. Nov. 24, 2009
Phy208 Lect. 24
13
Beat frequency: spatial localization

What does a sound ‘particle’ look like?


Example:‘beat frequency’ between two notes
Two waves of almost same wavelength added.
Constructive
interference
Large
amplitude
Tues. Nov. 24, 2009
Destructive
interference
Small
amplitude
Phy208 Lect. 24
Constructive
interference
Large
amplitude
14
Making a particle out of waves
440 Hz +
439 Hz
440 Hz +
439 Hz +
438 Hz
440 Hz +
439 Hz +
438 Hz +
437 Hz +
436 Hz
Tues. Nov. 24, 2009
Phy208 Lect. 24
15
Adding many sound waves

Six sound waves with different wavelength added together
1=
4= /1.15
2= /1.05
5= /1.20
3= /1.10
6= /1.25
•Wave now resembles a particle, but what is the wavelength?
– Sound pulse is comprised of several wavelength
– The exact wavelength is indeterminate
8
4
0
-4
x
-8
-15
-10
-5
0
5
10
15
J
Tues. Nov. 24, 2009
Phy208 Lect. 24
16
Spatial extent of ‘wave packet’
8
4
0
-4
x
-8
-15
-10
-5
0
5
10
15
J


x = spatial spread of ‘wave packet’
Spatial extent decreases as the spread in
included wavelengths increases.
Tues. Nov. 24, 2009
Phy208 Lect. 24
17
Same occurs for a matter wave


Localized particle:
sum of waves with slightly different wavelengths.
 = h /p, each wave has different momentum.


There is some ‘uncertainty’ in the momentum
Still don’t know exact location of the particle!


Wave still is spread over x (‘uncertainty’ in position)
Can reduce x, but at the cost of increasing the spread in
wavelength (giving a spread in momentum).
Tues. Nov. 24, 2009
Phy208 Lect. 24
18
Heisenberg Uncertainty Principle

Using



x = position uncertainty
p = momentum uncertainty
Planck’s
constant
Heisenberg showed that the product
( x )  ( p ) is always greater than ( h / 4 )
Often write this as
x p ~
/2
h

is pronounced ‘h-bar’
2
where

Tues. Nov. 24, 2009
Phy208 Lect. 24
19
Uncertainty principle question
Suppose an electron is inside a box 1 nm in width.
There is some uncertainty in the momentum of
the electron. We then squeeze the box to make it
0.5 nm. What happens to the momentum
uncertainty?
A. Momentum becomes more uncertain
B. Momentum becomes less uncertain
C. Momentum uncertainty unchanged
Tues. Nov. 24, 2009
Phy208 Lect. 24
20
The wavefunction

Quantify this by giving a physical meaning to
the wave that describing the particle.

This wave is called the wavefunction.


Cannot be experimentally measured!
But the square of the wavefunction is a
physical quantity.

It’s value at some point in space
is the probability of finding the particle there!
Tues. Nov. 24, 2009
Phy208 Lect. 24
21
Electron waves in an atom


Electron is a wave.
Its ‘propagation direction’ is
around circumference of
orbit.

Wavelength = h / p

Waves on a circle?
Tues. Nov. 24, 2009
Phy208 Lect. 24
22
Waves on a circle
Wavelength





My ‘ToneNut’.
Produces particular pitch.
Sound wave inside has
wavelength =v/f (red line).
Integer number of
wavelengths required around
circumference
Otherwise destructive
interference

Blow in here
Tues. Nov. 24, 2009
wave travels around ring and
interferes with itself
Phy208 Lect. 24
23
Electron Standing Waves

Electron in circular orbit works same way

Integer number of deBroglie wavelengths
must fit on circumference of the orbit.

Circumference = (2)x(orbit radius) = 2r

h
h
So condition is 2r  n  n  n
p
mv

This says mvr  n
This is quantization
angular momentum (L=mvr)

Ln

Tues. Nov. 24, 2009
Phy208 Lect. 24
24
Electron standing-waves on an atom
Wave representing
electron


Electron wave extends around
circumference of orbit.
Only integer number of
wavelengths around orbit
allowed.
Tues. Nov. 24, 2009
Phy208 Lect. 24
Wave representing
electron
25
Hydrogen atom energies


Wavelength gets longer in higher n
states, (electron moving slower) so
kinetic energy goes down.
But energy of Coulomb interaction
between electron (-) and nucleus
(+) goes up faster with bigger n.
Zero energy
n=4
n=3
E3  
13.6
eV
32
n=2
E2  
13.6
eV
22
E1  
13.6
eV
12

Energy

End result is

13.6
E n   2 eV
n
n=1

Tues. Nov. 24, 2009
Phy208 Lect. 24
26
Hydrogen atom question
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
Tues. Nov. 24, 2009
Phy208 Lect. 24
27
Hydrogen atom music




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.
Tues. Nov. 24, 2009
Phy208 Lect. 24
28
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).
Tues. Nov. 24, 2009
Phy208 Lect. 24
29
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.
But Remember that these are higher and higher energies!
(Coulomb (electrostatic) potential energy dominates).
Tues. Nov. 24, 2009
Phy208 Lect. 24
30