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De Broglie Waves, Uncertainty, and Atoms - Lecture Slides | PHY 1161, Exams of Physics

Material Type: Exam; Class: Principles of Physics II; Subject: Physics; University: Eastern Illinois University; Term: Unknown 1989;

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Physics 1161: Lecture 23, Slide 1
De Broglie Waves, Uncertainty, and Atoms
Today’s Lecture will cover textbook sections 30-5 – 30-7
Physics 1161: Lecture 23
Final
Exam
Physics 1161: Lecture 23, Slide 2
Outgoing photon
has momentum p
′′
and wavelength λ
λλ
λ
′′
Recoil electron
carries some
momentum and KE
Incoming photon has
momentum, p, and
wavelength λ
λλ
λ
This experiment really shows photon momentum!
Electron at
rest
Compton Scattering
P
incoming photon
+ 0 = P
outgoing photon
+ P
electron
λ
hc
hfE ==
λ
h
p=
Energy of a photon
Physics 1161: Lecture 23, Slide 3
ACT: Photon Collisions
Photons 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
pf3
pf4
pf5
pf8

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Physics 1161: Lecture 23, Slide 1

De Broglie Waves, Uncertainty, and Atoms

  • Today’s Lecture will cover textbook sections 30-5 – 30-

Physics 1161: Lecture 23 Final

Exam

Physics 1161: Lecture 23, Slide 2

Outgoing photon has momentum p′′′′ and wavelength λλλλ′′′′

Recoil electron carries some momentum and KE

Incoming photon has momentum, p, and wavelength λλλλ

This experiment really shows photon momentum!

Electron at rest

Compton Scattering

Pincoming photon + 0 = Poutgoing photon + Pelectron

hc

E = hf =

h

p =

Energy of a photon

Physics 1161: Lecture 23, Slide 3

ACT: Photon Collisions

Photons 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 1161: Lecture 23, Slide 4

Black side (absorbs)

Shiny side (reflects)

Incident photons

Radiometer

Preflight 23.

Photon 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

Physics 1161: Lecture 23, Slide 5

λ

h p =

So far only for 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

p

h λ =

Physics 1161: Lecture 23, Slide 6

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.

Physics 1161: Lecture 23, Slide 10

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 1161: Lecture 23, Slide 11

Rough idea: if we know momentum very precisely, we lose knowledge of location, and vice versa.

If we know the momentum p, then we know the wavelength λ, and that means we’re not sure where along the wave the particle is actually located!

λλ λλ

y

Heisenberg Uncertainty Principle

h

∆py ∆ y≥

Physics 1161: Lecture 23, Slide 12

Number of electrons arriving at screen sin θ =

λ w

⇒ w =

λ sin θ

screen

w

x

y

pθθ θθ

p θ ∆py = p sinθθθθ θθθ

Heisenberg Test

∆y = w = λ/sinθ

h

∆py ∆ y≥

electron beam

Physics 1161: Lecture 23, Slide 13

Electron entered slit with momentum along x direction and no momentum in the y direction. When it is diffracted it acquires a py which can be as big as h/w.

The “Uncertainty in py” is ∆∆∆∆py≈≈≈≈ h/w. An electron passed through the slit somewhere along the y direction. The “Uncertainty in y” is ∆∆∆∆y ≈≈≈≈ w.

Electron diffraction

electron beam

screen

Number of electrons arriving at screen

w

x

y (^) p (^) y ⋅ w = h

py

∴ ∆py ⋅ ∆y ≈ h

Physics 1161: Lecture 23, Slide 14

electron beam

screen

Number of electrons arriving at screen

w

x

y

py

∴ ∆py ⋅ ∆y ≈ h

If we make the slit narrower (decrease w=∆∆∆∆y) the diffraction peak gets broader (∆∆∆∆py increases).

“If we know location very precisely, we lose knowledge of momentum, and vice versa.”

Remember earlier we saw that a particle whose momentum (and therefore wavelength) is known precisely is very uncertain in position.

Physics 1161: Lecture 23, Slide 15

to be precise... ∆p^ y∆y^ ≥^

h 2 π

Of course if we try to locate the position of the particle along the x axis to ∆∆∆∆x we will not know its x component of momentum better than ∆px, where

∆p (^) x∆x ≥

h 2 π and the same for z.

Preflight 23.

According 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

Physics 1161: Lecture 23, Slide 19

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 1161: Lecture 23, Slide 20

Our Radiometer

Black 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 1161: Lecture 23, Slide 21

Recap

  • Photons carry momentum p=h/λ
  • Everything has wavelength λ=h/p
  • Uncertainty Principle ∆p∆x > h/(2π)
  • Atom
    • Positive nucleus 10-15^ m
    • Electrons “orbit” 10-10^ m
    • Classical E+M doesn’t give stable orbit
    • Need Quantum Mechanics!

Physics 1161: Lecture 23, Slide 22

See you Wednesday!

  • Read Textbook Sections 30-5 – 30-