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Chapter 28: Quantum Physics

For Wed recitation:• Online Qs• Practice Problems:

# 3, 6, 13, 21, 25

Lab: 2.16 (Atomic Spectra) • Do Pre-Lab & turn in• Next week optional 2.03

Final Exam: Tue Dec 11 3:30-5:30 pm @220 MSC • 200 pts: Chs.25,27,28,(26)• 200 pts: OQ-like on 12,16-24

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§28.1 Wave-Particle DualityLight is both wave-like (interference & diffraction) and particle-like (photoelectric effect).

Double slit experiment: allow only 1 photon at a time, but:

• still makes interference pattern!

• can’t determine which slit it will pass thru

• can’t determine where it will hit screen

• can calculate probability:

• higher probability higher intensity

• IE2, so E2 probability of striking at a given location; E represents the wave function.

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If a wave (light) can behave like a particle, can a particle act like a wave?

Double slit experiment w/ electrons:

• interference pattern! Wave-like!

Allow only 1 e– at a time:

• still makes interference pattern

• can still calculate probabilities

Add detector to see which slit used:

• one slit or other, not both

• interference pattern goes away!

• wave function “collapses” to particle!!

§28.2 Matter Waves

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Diffraction (waves incident on a crystal sample)

Electrons: X-rays:

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Δl = mλ = 2d sinθ

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Like photons, “matter waves” have a wavelength:

“de Broglie wavelength”

Momentum:

Electron beam defined by accelerating potential, gives them Kinetic Energy:

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λ =h

p

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p = mvNote: need a relativistic correction if v~c (Ch.26)

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p = 2mK

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Example (PP 28.8): What are the de Broglie wavelengths of electrons with kinetic energy of (a) 1.0 eV and (b) 1.0 keV?

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§28.3 Electron Microscope

Resolution (see fine detail):

• visible light microscope limited by diffraction to Δ~1/2 λ (~200 nm).

• much smaller (0.2-10 nm) using a beam of electrons (smaller λ).

€

asinΔθ ≥1.22λ

Fig. 28.06

Transmission Electr. Micr.

Scanning Electr. Micr.

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Example: We want to image a biological sample at a resolution of 15 nm using an electron microscope.

(a)What is the kinetic energy of a beam of electrons with a de Broglie wavelength of 15.0 nm?

(b) Through what potential difference should the electrons be accelerated to have this wavelength?

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§28.4 Heisenberg’s Uncertainty Principle

Sets limits on how precise measurements of a particle’s position (x) and momentum (px) can be:

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1ΔΔ xpx

where2

h

.2

1ΔΔ tE

The energy-time uncertainty principle:

wave packet

Uncertainty in position

& momentum

Superposition

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Δx

€

Δx

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Example: We send an electron through a very narrow slit of width 2.010-8 m. What is the uncertainty in the electron’s y-component momentum?

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Example: An electron is confined to a “quantum wire” of length 150 nm.

(a)What is the minimum uncertainty in the electron’s component of momentum along the wire?

(b)In its velocity?

13§28.

5 W

ave

Fun

ctio

ns f

or a

Con

fined

Par

ticle

Conclude: A confined particle has quantized energy levels

Analogy: standing wave on a string:

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λn =2L

n

Same for electron in a quantum wire (particle in a 1D box), so

& particle’s KE is

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pn =h

λ n=n

2L

⎛

⎝ ⎜

⎞

⎠ ⎟h

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En = n2 h2

8mL2

⎛

⎝ ⎜

⎞

⎠ ⎟= n2E1

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Electron cloud represents the electron probability density for an H atom (the electron is confined to its orbit):

.2

1ΔΔ tE

Energy states and durations are “blurred”

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Example: We want to image a biological sample at a resolution of 15 nm using an electron microscope.

(a)What is the kinetic energy of a beam of electrons with a de Broglie wavelength of 15.0 nm?

(b) Through what potential difference should the electrons be accelerated to have this wavelength?

-

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λ =h

p=

h

2mK Square both sides, solve for K:

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K =h2

2mλ2=

(6.626 ×10−34 Js)2

2(9.11×10−31kg)(15 ×10−9m)2=1.07x10-21 J = 0.0067 eV (low E!)

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K final =U initial = −qΔV = eΔV(b)

so

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ΔV =K

e=

0.0067eV

e= 0.0067 V (low Voltage, easy desktop machine!)

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Example: We send an electron through a very narrow slit of width 2.010-8 m. What is the uncertainty in the electron’s y-component momentum?

Key idea: electron goes through slit; maybe through center, or ±a/2 above/below it,so use Δy = a/2! Then H.E.P. says so

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ΔyΔpy ≥h

2

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Δpy ≥h

2Δy=

h

2 a2( )

=h

2πa=

6.626 ×10−34 Js( )

2π (2.0 ×10−8m)= 5.3×10−27 kgm

s

Notice: This uncertainty in the electron’s vertical momentum means it can veer off its straight-line course; many veered electrons diffraction pattern!!

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Example: An electron is confined to a “quantum wire” of length 150 nm.

(a)What is the minimum uncertainty in the electron’s component of momentum along the wire?

(b)In its velocity?

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Δpx ≥h

2Δx=

h

2 l2( )

=h

2πl=

6.626 ×10−34 Js( )

2π (150 ×10−9m)= 7 ×10−28 kgm

s

Key idea: electron w/in wire; maybe at center, or ±l/2 from center, so use Δx = l/2! Then use H.E.P.

(b) Solve for the velocity:

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Δv =Δp

m=

6.626 ×10−34 Js

9.11 ×10−31kg= 770 m

s ≈1 kms