General Physics (PHY 2140) - Physics & Astronomy - Wayne ...apetrov/PHY2140/Lecture29.pdf ·...

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1 1 11/11/2003 11/11/2003 General Physics (PHY 2140) Lecture 28 Lecture 28 Modern Physics Quantum Physics Compton scattering Chapter 27 http://www.physics.wayne.edu/~apetrov/PHY2140/

Transcript of General Physics (PHY 2140) - Physics & Astronomy - Wayne ...apetrov/PHY2140/Lecture29.pdf ·...

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General Physics (PHY 2140)

Lecture 28Lecture 28

Modern Physics

Quantum PhysicsCompton scattering

Chapter 27

http://www.physics.wayne.edu/~apetrov/PHY2140/

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If you want to know If you want to know your progressyour progress so far, please so far, please send me send me an email requestan email request atat

[email protected]@physics.wayne.edu

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Lightning ReviewLightning Review

Last lecture:

1.1. Quantum physicsQuantum physicsPhotoeffect

, 1,2,3,...nE nhf n= =

Photoeffect KE hf= −Φ

Review Problem: A xenon arc lamp is covered with an interference filter that only transmits light of 400- nm wavelength. When the transmitted light strikes a metal surface, a stream of electrons emerges from the metal. If the intensity ofthe light striking the surface is doubled,

1. more electrons are emitted in a given time interval.2. the electrons that are emitted are more energetic.3. both of the above.4. neither of the above.

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Review problem: photoeffectReview problem: photoeffect

Molybdenum has a work function of 4.20 eV. Calculate the stoppinMolybdenum has a work function of 4.20 eV. Calculate the stopping g potential if the incident light has a wavelength of 180 nm.potential if the incident light has a wavelength of 180 nm.

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Molybdenum has a work function of 4.20 eV. Calculate the stoppinMolybdenum has a work function of 4.20 eV. Calculate the stopping potential if the g potential if the incident light has a wavelength of 180 nm.incident light has a wavelength of 180 nm.

Recall that KEKEmaxmax==hfhf -- ΦΦ. This can be used to obtain kinetic energy, which will be equal to electric potential energy

Given:

Φ = 4.40 eVλ = 180 nm

Find:

Vs = ?

( )max sKE e V= ∆

shf hcVe e e eλ

Φ Φ∆ = − = −Thus,

Or numerically,

( )( )( )( )( )( )

34 8

9 19

19

19

6.63 10 3.00 10

180 10 1.60 10

4.20 1.60 102.71

1.60 10

s

J s m sV

m C

eV J eVV

C

− −

× ⋅ ×∆ =

× ×

×− =

×

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27.4 X27.4 X--RaysRays

Electromagnetic radiation with short wavelengthsElectromagnetic radiation with short wavelengthsWavelengths less than for ultravioletWavelengths less than for ultravioletWavelengths are typically about 0.1 nmWavelengths are typically about 0.1 nmXX--rays have the ability to penetrate most materials with relative rays have the ability to penetrate most materials with relative easeease

Discovered and named by Roentgen in 1895Discovered and named by Roentgen in 1895

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Production of XProduction of X--raysrays

XX--rays are produced when highrays are produced when high--speed electrons are suddenly speed electrons are suddenly slowed downslowed down

Can be caused by the electron Can be caused by the electron striking a metal targetstriking a metal target

A current in the filament causes A current in the filament causes electrons to be electrons to be emittedemittedThese freed electrons are These freed electrons are accelerated toward a dense accelerated toward a dense metal targetmetal targetThe target is held at a higher The target is held at a higher potential than the filamentpotential than the filament

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Production of XProduction of X--raysrays

An electron passes near a An electron passes near a target nucleustarget nucleus

The electron is deflected from The electron is deflected from its path by its attraction to the its path by its attraction to the nucleusnucleus

This produces an accelerationThis produces an acceleration

It will emit electromagnetic It will emit electromagnetic radiation radiation when it is acceleratedwhen it is accelerated

The maximum x-ray energy, and minimum wavelength results when the electron loses all its energy in a single collision, such that

e∆V = hfmax = hc/λmin or therefore min

hce V

λ =∆

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Diffraction of XDiffraction of X--rays by Crystalsrays by Crystals

For diffraction to occur, the spacing between the lines For diffraction to occur, the spacing between the lines must be approximately equal to the wavelength of the must be approximately equal to the wavelength of the radiation to be measuredradiation to be measuredFor XFor X--rays, rays, the regular array of atoms in a crystal can act the regular array of atoms in a crystal can act as a threeas a three--dimensional grating for diffracting Xdimensional grating for diffracting X--raysrays

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Schematic for XSchematic for X--ray Diffractionray Diffraction

A continuous beam of XA continuous beam of X--rays rays is incident on the crystalis incident on the crystal

The diffracted radiation is very The diffracted radiation is very intense in certain directionsintense in certain directions

These directions correspond These directions correspond to constructive interference to constructive interference from waves reflected from the from waves reflected from the layers of the crystallayers of the crystal

The diffraction pattern is The diffraction pattern is detected by photographic filmdetected by photographic film

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Photo of XPhoto of X--ray Diffraction Patternray Diffraction Pattern

The array of spots is called a The array of spots is called a LaueLaue patternpatternThe crystal structure is The crystal structure is determined by analyzing the determined by analyzing the positions and intensities of the positions and intensities of the various spotsvarious spotsThis is for NaClThis is for NaCl

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Bragg’s LawBragg’s Law

The beam reflected from the lower The beam reflected from the lower surface travels farther than the surface travels farther than the one reflected from the upper one reflected from the upper surfacesurface

If the path difference equals some If the path difference equals some integral multiple of the wavelength, integral multiple of the wavelength, constructive interference occursconstructive interference occurs

Bragg’s LawBragg’s Law gives the conditions gives the conditions for for constructive interferenceconstructive interference

2 d sin θ = m λ, m = 1, 2, 3…2 d sin θ = m λ, m = 1, 2, 3…

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Problem: XProblem: X--ray diffractionray diffraction

XX--rays of wavelength 0.140 nm are reflected from a certain crystalrays of wavelength 0.140 nm are reflected from a certain crystal, , and the firstand the first--order maximum occurs at an angle of 14.4°. What order maximum occurs at an angle of 14.4°. What value does this give for the value does this give for the interplanarinterplanar spacing of this crystal?spacing of this crystal?

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27.4 The Compton Effect27.4 The Compton Effect

Compton directed a beam of xCompton directed a beam of x--rays toward a block of graphiterays toward a block of graphiteHe found that the scattered xHe found that the scattered x--rays had a slightly longer wavelength rays had a slightly longer wavelength that the incident xthat the incident x--raysrays

This means they also had less energyThis means they also had less energyThe amount of energy reduction depended on the angle at which thThe amount of energy reduction depended on the angle at which the e xx--rays were scatteredrays were scatteredThe change in wavelength is called the The change in wavelength is called the Compton shiftCompton shift

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Compton ScatteringCompton Scattering

Compton assumed the Compton assumed the photons acted like other photons acted like other particles in collisionsparticles in collisions

Energy and momentum Energy and momentum were conservedwere conserved

The shift in wavelength isThe shift in wavelength is )cos1(cm

he

o θ−=λ−λ=λ∆

Compton wavelength

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Compton ScatteringCompton Scattering

The quantity h/mThe quantity h/meec is called the c is called the Compton wavelengthCompton wavelengthCompton wavelength = 0.00243 nmCompton wavelength = 0.00243 nmVery small compared to visible lightVery small compared to visible light

The Compton shift depends on the The Compton shift depends on the scattering anglescattering angle and and not not on on the the wavelengthwavelengthExperiments confirm the results of Compton scattering and Experiments confirm the results of Compton scattering and strongly support the photon conceptstrongly support the photon concept

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Problem: Compton scatteringProblem: Compton scattering

A beam of 0.68A beam of 0.68--nm photons undergoes Compton scattering from free nm photons undergoes Compton scattering from free electrons. What are the energy and momentum of the photons that electrons. What are the energy and momentum of the photons that emerge at a 45° angle with respect to the incident beam? emerge at a 45° angle with respect to the incident beam?

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QUICK QUIZ 1

An x-ray photon is scattered by an electron. The frequency of the scattered photon relative to that of the incident photon (a) increases, (b) decreases, or (c) remains the same.

(b). Some energy is transferred to the electron in the scattering process. Therefore, the scattered photon must have less energy (and hence, lower frequency) than the incident photon.

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QUICK QUIZ 2

A photon of energy E0 strikes a free electron, with the scattered photon of energy E moving in the direction opposite that of the incident photon. In this Compton effect interaction, the resulting kinetic energy of the electron is (a) E0 , (b) E , (c) E0 − E , (d) E0 + E , (e) none of the above.

(c). Conservation of energy requires the kinetic energy given tothe electron be equal to the difference between the energy of the incident photon and that of the scattered photon.

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27.8 Photons and Electromagnetic Waves27.8 Photons and Electromagnetic Waves

Light has a dual nature. It exhibits both wave and particle Light has a dual nature. It exhibits both wave and particle characteristicscharacteristics

Applies to all electromagnetic radiationApplies to all electromagnetic radiation

The photoelectric effect and Compton scattering offer evidence fThe photoelectric effect and Compton scattering offer evidence for or the particle nature of lightthe particle nature of light

When light and matter interact, light behaves as if it were compWhen light and matter interact, light behaves as if it were composed of osed of particlesparticles

Interference and diffraction offer evidence of the wave nature oInterference and diffraction offer evidence of the wave nature of lightf light

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28.9 Wave Properties of Particles28.9 Wave Properties of Particles

In 1924, Louis de Broglie postulated that In 1924, Louis de Broglie postulated that because because photons have wave and particle characteristics, perhaps photons have wave and particle characteristics, perhaps all forms of matter have both propertiesall forms of matter have both properties

Furthermore, the frequency and wavelength of matter Furthermore, the frequency and wavelength of matter waves can be determinedwaves can be determinedThe The de de BroglieBroglie wavelengthwavelength of a particle isof a particle is

The frequency of matter waves ismvh

=λThe frequency of matter waves is

hEƒ =

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The DavissonThe Davisson--Germer ExperimentGermer Experiment

They scattered lowThey scattered low--energy electrons from a nickel targetenergy electrons from a nickel target

They followed this with extensive diffraction measurements from They followed this with extensive diffraction measurements from various materialsvarious materials

The wavelength of the electrons calculated from the diffraction The wavelength of the electrons calculated from the diffraction data data agreed with the expected de Broglie wavelengthagreed with the expected de Broglie wavelength

This confirmed the wave nature of electronsThis confirmed the wave nature of electrons

Other experimenters have confirmed the wave nature of other Other experimenters have confirmed the wave nature of other particlesparticles

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QUICK QUIZ 3

A non-relativistic electron and a non-relativistic proton are moving and have the same de Broglie wavelength. Which of the following are also the same for the two particles: (a) speed, (b) kinetic energy, (c) momentum, (d) frequency?

(c). Two particles with the same de Broglie wavelength will have the same momentum p = mv. If the electron and proton have the same momentum, they cannot have the same speed because of the difference in their masses. For the same reason, remembering that KE = p2/2m, they cannot have the same kinetic energy. Because the kinetic energy is the only type of energy an isolated particle can have, and we have argued that the particles have different energies, Equation 27.15 tells us that the particles do not have the same frequency.

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If you want to know If you want to know your progressyour progress so far, please so far, please send me send me an email requestan email request atat

[email protected]@physics.wayne.edu