Physics 102: Lecture 23, Slide 1
De Broglie Waves, Uncertainty, and Atoms
Physics 102: Lecture 23
Physics 102: Lecture 23, Slide 2
Three Early Indications of Problems with Classical Physics
• Blackbody radiation• Photoelectric effect• Wave-particle duality
Lecture 22:Quantum Mechanics
• Compton scattering• DeBroglie• Heisenberg Uncertainty Principle
Today
Physics 102: Lecture 23, Slide 3
Experiment: Outgoing photon has longer wavelength λ′
Recoil electron carries some momentum and KE
Incoming photon has momentum p, and wavelength λ
This experiment really shows photon momentum!
Electron at rest
Compton Scattering
Pincoming photon + 0 = Poutgoing photon + Pelectron
λhchfE ==
λhp =
Photon energy Photon momentum
⇒ E = pc
Physics 102: Lecture 23, Slide 4
Compton Scattering
• Incident photon loses momentum, since it transfers momentum to the electron
• Lower momentum means longer wavelength• This is proof that a photon has momentum
λhp =
Physics 102: Lecture 23, Slide 5
Is Light a Wave or a Particle?• Wave
– Electric and Magnetic fields act like waves– Superposition, Interference, and Diffraction
• Particle– Photons– Collision with electrons in photo-electric effect– Compton scattering from electrons
BOTH Particle AND Wave
Physics 102: Lecture 23, Slide 6
ACT: Photon CollisionsPhotons with equal energy and momentum hit both
sides of the plate. The photon from the left sticks to the plate, the photon from the right bounces off the plate. What is the direction of the net impulse on the plate?
1) Left 2) Right 3) Zero
Physics 102: Lecture 23, Slide 7
Incident photons
Radiometer
Preflight 23.1Photon A strikes a black surface and is absorbed. Photon B strikes a shiny surface and is reflected back. Which photon imparts more momentum to the surface?
Photon A Photon B
Black side (absorbs)
Shiny side (reflects)
Physics 102: Lecture 23, Slide 8
Photons bouncing off shiny side and sticking to black side. Shiny side gets more momentum so it should rotate with the black side leading
Ideal Radiometer
Physics 102: Lecture 23, Slide 9
Our RadiometerBlack side is hotter: gas molecules bounce off it with more momentum than on shiny side-this is a bigger effect than the photon momentum
Physics 102: Lecture 23, Slide 10
Electrons are Particles and Waves!• Depending on the experiment electron can behave
like– wave (interference) – particle (localized mass and charge)
• Recall Young’s double slit experiment: – If we measure which slit the electron went through,
then there is no interference pattern!!
Physics 102: Lecture 23, Slide 11
λhp =
So far only photons have wavelength, but De Broglie postulated that it holds for any object with momentum- an electron, a nucleus, an atom, a baseball,…...
Explains why we can see interference and diffraction for material particles like electrons!!
De Broglie Waves
ph
=λ
Physics 102: Lecture 23, Slide 12
Which baseball has the longest De Broglie wavelength?
(1) A fastball (100 mph)
(2) A knuckleball (60 mph)
(3) Neither - only curveballs have a wavelength
Preflight 23.3
Physics 102: Lecture 23, Slide 13
ACT: De Broglie Wavelength
A stone is dropped from the top of a building.
1. It decreases
2. It stays the same
3. It increases
What happens to the de Broglie wavelength of the stone as it falls?
Physics 102: Lecture 23, Slide 14
Some Numerology
• 1 eV = energy gained by a charge +e when accelerated through a potential difference of 1 Volt– e = 1.6 x 10-19 C so 1 eV = 1.6 x 10-19 J
• h = 6.626 x 10-34 J·sec• c = 3 x 108 m/s
– hc = 1.988 x 10-25 J·m = 1240 eV·nm• mass of electron m = 9.1 x 10-34 kg
– mc2 = 8.2 x 10-13 J = 511,000 eV = 511 keV
Standard units (m, kg, s) are not convenient for talking about photons & electrons
Physics 102: Lecture 23, Slide 15
• Photon with 1 eV energy:
Comparison:Wavelength of Photon vs. Electron
λhcE =
Ehc
=⇒ λ nm 1240eV1
nmeV1240==
You have a photon and an electron, both with 1 eV of energy. Find the de Broglie wavelength of each.
• Electron with 1 eV kinetic energy:
KE =
12
mv 2 and p = mv, so KE = p2
2mK.E.)(2mp =Solve for
KE)(2mh
=λKE)(2 2mc
hc=
eV) 1)(eV 000,511(2nmeV1240
= nm23.1=
Big difference!
Eq
uat
ion
s a
re d
iffe
ren
t -
be
ca
refu
l!
Physics 102: Lecture 23, Slide 16
X-ray diffraction e– diffraction
From College Physics, Vol. Two
Identical pattern emerges if de Broglie wavelength of e– equals the X-ray wavelength!
X-ray vs. electron diffraction
Demo
Physics 102: Lecture 23, Slide 17
Preflights 23.4, 23.5
Photon A has twice as much momentum as Photon B. Compare their energies.
• EA = EB
• EA = 2 EB
• EA = 4 EB
Electron A has twice as much momentum as Electron B. Compare their energies.
• EA = EB
• EA = 2 EB
• EA = 4 EB
Physics 102: Lecture 23, Slide 18
ACT: De Broglie
Compare the wavelength of a bowling ball with the wavelength of a golf ball, if each has 10 Joules of kinetic energy.
(1) λbowling > λgolf
(2) λbowling = λgolf
(3) λbowling < λgolf
Physics 102: Lecture 23, Slide 19
Rough idea: if we know momentum very precisely, we lose knowledge of location, and vice versa.
Heisenberg Uncertainty Principle
π2hypy ≥ΔΔ
Recall: Quantum Mechanics tells us nothing is certain, everything is probability
Uncertainty in position (along y)
Uncertainty in momentum (along y)
Physics 102: Lecture 23, Slide 20
Number of electrons arriving at screen
screen
w
x
y
θΔpy = p sinθ
θ
= Δy
θλθ
sinsin pypy =ΔΔ pλ= h=
w = λ/sinθ
electron beam
Electron diffractionElectron beam traveling through slit will diffract
Recall single-slit diffraction 1st minimum:sinθ = λ/w
Using de Broglie λ
Single slit diffraction pattern
Physics 102: Lecture 23, Slide 21
Electron entered slit with momentum along x direction and no momentum in the y direction. When it is diffracted it acquires a pywhich can be as big as h/w.
The “Uncertainty in py” is Δpy ≈ h/w.An electron passed through the slit somewhere along the y direction. The “Uncertainty in y” is Δy ≈ w.
electron beam
screen
Number of electrons arriving at screen
w
x
y py ⋅w = h
py
∴ Δpy ⋅ Δy ≈h
Physics 102: Lecture 23, Slide 22
electron beam
screen
Number of electrons arriving at screen
w
x
y
py
∴ Δpy ⋅ Δy ≈h
If we make the slit narrower (decrease w =Δy) the diffraction peak gets broader (Δpy increases).
“If we know location very precisely, we lose knowledge of momentum, and vice versa.”
Physics 102: Lecture 23, Slide 23
to be precise... ΔpyΔy ≥ h
2π
Of course if we try to locate the position of the particle along the x axis to Δx we will not know its x component of momentum better than Δpx, where
ΔpxΔx ≥ h
2πand the same for z.
Preflight 23.7According to the H.U.P., if we know the x-position of a particle, we can not know its:
(1) y-position (2) x-momentum
(3) y-momentum (4) Energy
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