from last time(s)…
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E initial. Photon. E final. Stable orbit. Stable orbit. From Last Time(s)…. Light shows both particle and wave-like properties. Atoms emit and absorb photons. Photon: E=hf. Exam 3 is Thursday Dec. 3 (after Thanksgiving). 5:30-7 pm, Birge 145. - PowerPoint PPT PresentationTRANSCRIPT
From Last Time(s)…
Tues. Nov. 24, 2009 Phy208 Lect. 24 1
Light shows both particle and wave-like properties
Photon: E=hf
Stable orbit
Stable orbit
Einitia
l
Efinal
Photon
Atoms emit and absorb photons
Tues. Nov. 24, 2009 Phy208 Lect. 24 2
Exam 3 is Thursday Dec. 3 (after Thanksgiving)
Students w / scheduled academic conflict please stay after class Tues. Nov. 24 to arrange alternate time.
5:30-7 pm, Birge 145
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 3
Photon properties of light
Photon of frequency f has energy hf
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
€
E photon = hf = hc /λ
€
h = 6.626 ×10−34 J ⋅s = 4.14 ×10−15eV ⋅s
€
hc =1240eV ⋅nm
Thurs. Nov. 19, 2009 Phy208 Lect. 23 4
Emitting and absorbing light
Photon is emitted when electron drops from one quantum state to another
Zero energy
n=1
n=2
n=3
n=4
€
E1 = −13.6
12 eV
€
E2 = −13.6
22 eV
€
E3 = −13.6
32 eV
n=1
n=2
n=3
n=4
€
E1 = −13.6
12 eV
€
E2 = −13.6
22 eV
€
E3 = −13.6
32 eV
Absorbing a photon of correct energy makes electron jump to higher quantum state.
Photon absorbed hf=E2-E1
Photon emittedhf=E2-E1
Tues. Nov. 24, 2009 Phy208 Lect. 24 5
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
This is called
the de Broglie wavelength.€
λ =h
p
Nobel prize, 1929
Tues. Nov. 24, 2009 Phy208 Lect. 24 6
Why h / p ? Works for photons Wave interpretation of light:
wavelength = (Speed of Light) / Frequency λ = c / f
Particle interpretation of light (photons): Energy = (Planck’s constant) x Frequency E = hf, so f = E / h
for a photon
€
λ =h
p
€
Wavelength=λ = c
f=c
E /h=h
E /c
But photon momentum = p = E / c…
Tues. Nov. 24, 2009 Phy208 Lect. 24 7
We argue that applies to everything
Photons and footballs both follow the same relation.
Everything has both wave-like and particle-like properties
€
λ =h
p
Tues. Nov. 24, 2009 Phy208 Lect. 24 8
Wavelengths of massive objects
deBroglie wavelength =
€
λ =h
p
p=mv
€
λ =h
mv
Tues. Nov. 24, 2009 Phy208 Lect. 24 9
Matter Waves
deBroglie postulated that matter has wavelike properties.
deBroglie wavelength
€
λ =h / p
Example: Wavelength of electron with 10 eV of energy:
Kinetic energy
€
EKE =p2
2m⇒ p = 2mEKE
λ =h
2mEKE=
hc
2mc 2EKE=
1240eV ⋅nm
2 0.511×106eV( ) 10eV( )= 0.39nm
Tues. Nov. 24, 2009 Phy208 Lect. 24 10
Wavelength of a football
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 wings that thing 60, 70 mph,” Flanagan said. (27 - 32 m/s)
Momentum:
€
mv = 0.4 kg( ) 30 m /s( ) =12 kg−m /s
€
λ =h
p=
6.6 ×10−34 J − s
12 kg−m /s= 5.5 ×10−35m = 5.5 ×10−26nm
Need m, v to find λ
AaronWells
Tues. Nov. 24, 2009 Phy208 Lect. 24 11
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?
€
λ =h
p=h
mvLarge mass, large momentumshort wavelength
Tues. Nov. 24, 2009 Phy208 Lect. 24 12
Suppose an electron is a wave…
Here is a wave:
…where is the electron? Wave extends infinitely far in +x and -x direction
λ
x
€
λ =h
p
Tues. Nov. 24, 2009 Phy208 Lect. 24 13
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 14
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 interferenceLarge
amplitude
Constructive interference
Large amplitude
Destructive interference
Small amplitude
Tues. Nov. 24, 2009 Phy208 Lect. 24 15
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 16
Adding many sound waves
Six sound waves with different wavelength added togetherλ1=λ λ2= λ/1.05 λ3= λ/1.10 λ4= λ/1.15 λ5= λ/1.20 λ6= λ/1.25
-8
-4
0
4
8
-15 -10 -5 0 5 10 15J
x
•Wave now resembles a particle, but what is the wavelength?– Sound pulse is comprised of several wavelength– The exact wavelength is indeterminate
Tues. Nov. 24, 2009 Phy208 Lect. 24 17
Spatial extent of ‘wave packet’
x = spatial spread of ‘wave packet’ Spatial extent decreases as the spread in
included wavelengths increases.
-8
-4
0
4
8
-15 -10 -5 0 5 10 15J
x
Tues. Nov. 24, 2009 Phy208 Lect. 24 18
Same occurs for a matter wave
Localized particle:sum of waves with slightly different wavelengths.
λ = 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 19
Heisenberg Uncertainty Principle
Using x = position uncertainty p = momentum uncertainty
Heisenberg showed that the product
( x ) ( p ) is always greater than ( h / 4 )
Often write this as
where is pronounced ‘h-bar’
Planck’sconstant
€
x( ) Δp( ) ~ h /2
€
h=h
2π
Tues. Nov. 24, 2009 Phy208 Lect. 24 20
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 21
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 22
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 23
Waves on a circle
My ‘ToneNut’. Produces particular pitch. Sound wave inside has
wavelength λ=v/f (red line). Integer number of
wavelengths required around circumference
Otherwise destructive interference wave travels around ring and
interferes with itselfBlow in here
Wavelength
Tues. Nov. 24, 2009 Phy208 Lect. 24 24
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) = 2r
So condition is
This says
This is quantization angular momentum (L=mvr)
€
2πr = nλ = nh
p= n
h
mv
€
mvr = nh
€
L = nh
Tues. Nov. 24, 2009 Phy208 Lect. 24 25
Wave representingelectron
Electron standing-waves on an atom
Wave representingelectron
Electron wave extends around circumference of orbit.
Only integer number of wavelengths around orbit allowed.
Tues. Nov. 24, 2009 Phy208 Lect. 24 26
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.
End result is
Zero energy
n=1
n=2
n=3
n=4
€
E1 = −13.6
12 eV
€
E2 = −13.6
22 eV
€
E3 = −13.6
32 eV
En
erg
y
€
En = −13.6
n2 eV
Tues. Nov. 24, 2009 Phy208 Lect. 24 27
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=2B. n=3C. n=4
Tues. Nov. 24, 2009 Phy208 Lect. 24 28
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 29
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 30
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).