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MyChapter 27

Lecture

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Chapter 27: Early Quantum Physics and the Photon

•Blackbody Radiation

•The Photoelectric Effect

•Compton Scattering

•Early Models of the Atom

•The Bohr Model for the Hydrogen Atom

•Pair Production/Annihilation

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§27.1 Quantization

A quantity is quantized if its possible values are limited to a discrete set.

An example from classical physics is the allowed frequencies of standing waves on a stretched string. Only integer multiples of the fundamental frequency produce standing waves.

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§27.2 Blackbody Radiation

A blackbody emits a continuous spectrum of radiation. The spectrum is determined only by the temperature of the blackbody.

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To correctly explain the shape of the blackbody spectrum Planck proposed that the energy absorbed or emitted by oscillating charges came in discrete bundles called quanta. The energy of the quanta are

hfE =0where h = 6.62610-34 J s is called Planck’s constant.

The quantum of EM radiation is the photon.

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§27.3 The Photoelectric Effect

Under certain circumstances EM radiation incident on a metal will eject electrons from the metal. This is the photoelectric effect.

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Experiments show:

1. Brighter light causes more electrons to be ejected, but not with more kinetic energy.

2. The maximum KE of ejected electrons depends on the frequency of the incident light.

3. The frequency of the incident light must exceed a certain threshold, otherwise no electrons are ejected.

4. Electrons are ejected with no observed time delay regardless of the intensity of the incident light.

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The wave theory of light says EM waves carry energy. The energy is absorbed by electrons in the metal target which are then ejected when they accumulate enough energy to escape. However the wave theory is unable to completely explain the photoelectric effect. Einstein proposed a particle theory of light.

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Wave theory predicts a more intense beam of light, having more energy, should cause more electrons to be emitted and they should have more kinetic energy.

Particle theory predicts a more intense beam of light to have more photons so more electrons should be emitted, but since the energy of a photon does not change with beam intensity, the kinetic energy of the ejected electrons should not change.

The particle theory is consistent with observation 1.

Observation 1

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Wave theory cannot explain the frequency dependence of the maximum kinetic energy.

Particle theory predicts the maximum kinetic energy of the ejected electrons to show a dependence on the frequency of the incident light. Each electron in the metal absorbs a whole photon: some of the energy is used to eject the electron and the rest goes into the KE of the electron.

Observation 2

The particle theory is consistent with observation 2.

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The maximum KE of an ejected electron is

φ−=hfKEmax

where is called the work function and is the energy needed to break the bond between the electron and the metal.

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Particle theory predicts a threshold frequency is needed. Only the incident photons with f > fthreshold will have enough energy to free the electron from the metal.

Wave theory can offer no explanation.

Observation 3

The particle theory is consistent with observation 3.

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The electron is ejected from the metal when the energy supplied by the photon exactly equals the work function. This defines the threshold frequency.

hf

hf

φφ

=

=−

threshold

threshold 0

Here it is often convenient to use h = 4.13610-15 eV s.

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Wave theory predicts that if the intensity of the light is low, then it will take some time before an electron absorbs enough energy to be ejected from the metal.

Particle theory predicts a low intensity light beam will just have a low number of photons, but as long as f > fthreshold an electron that absorbs a whole photon will be ejected; no time delay should be observed.

Observation 4

The particle theory is consistent with observation 4.

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The particle theory of light is needed to explain the photoelectric effect (and Compton scattering and pair production). A wave theory of light is needed to explain interference patterns. Both are correct (wave-particle duality).

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Example (text problem 27.1): A 200 W infrared laser emits photons with a wavelength of 2.010-6 m while a 200 W ultraviolet laser emits photons with a wavelength of7.010-8 m.

λhc

hfE == The UV photon has the greater energy; its wavelength is smallest.

(b) What is the energy of a single infrared photon and the energy of a single ultraviolet photon?

eV 0.62J 109.9

eV 18J 108.2

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IRIR

18

UVUV

=×==

=×==

−

−

λ

λhc

E

hcE

(a) Which has greater energy, a single infrared photon or a single ultraviolet photon?

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Example (text problem 27.4): The photoelectric threshold frequency of silver is 1.041015 Hz. What is the minimum energy required to remove an electron from silver?

( )( )eV 4.30J 1089.6

Hz 1004.1Js 10626.6

0

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1534

threshold

thresholdmax

=×=

××=

=

=−=

−

−

hf

hfKE

φ

φ

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Example (text problem 27.11): Two different monochromatic light sources, one yellow (580 nm) and one violet (425 nm), are used in a photoelectric effect experiment. The metal surface has a photoelectric threshold frequency of6.201014 Hz.

The frequency of each source is

Hz. 1006.7

Hz 1017.5

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violetviolet

14

yellowyellow

×==

×==

λ

λ

cf

cf

Only the violet light is above the threshold frequency.

(a) Are both sources able to eject photoelectrons from the metal? Explain.

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(b) How much energy is required to eject an electron from the metal?

Example continued:

( )( )eV 2.56J 1011.4

Hz 1020.6Js 10626.6

0

19

1434

threshold

thresholdmax

=×=

××=

=

=−=

−

−

hf

hfKE

φ

φ

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§27.5 Compton Scattering

x

y

Before Collision After Collision

Photon (E0, p0)

Free electron at rest

Photon (E1, p1)

Free electron (K, p)

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Conserve momentum and energy during the collision:

φφ

sinsin0 :

coscos :

1

10

ppy

pppx

fi

−=

+=

= pp

)( 1010

21

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ppcEEK

cmKEcmE

EE

ee

fi

−=−=∴++=+

=

E=pc for a photon

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( )

( ).cos1

cos101

θλλ

θλλ

−=Δ

−=−

c

ecm

h

The Compton wavelength

λ is the Compton shift.

Manipulating the previous expressions gives

pm 426.2 ==cmh

ecλ

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Example (text problem 27.27): A photon is incident on an electron at rest. The scattered photon has a wavelength of 2.81 pm and moves at an angle of 29.5 with respect to the direction of the incident photon.

( )( )( )

pm. 315.0

5.29cos1pm 43.2

cos1

=

°−=

−=Δ θλλ cThe Compton shift is

pm. 2.50 pm 0.314pm 81.210

=−=−= λλλ

The incident wavelength is

(a) What is the wavelength of the incident photon?

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Example continued:

(b) What is the final kinetic energy of the electron?

keV 55 J 1077.8

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)(

15

1010

10

=×=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

−=

−

λλλλhc

hhc

ppcK

The final kinetic energy of the electron is equal to the change in the photon’s energy.

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§27.6 Spectroscopy and Early Models of the Atom

A hot, solid object will emit a continuous spectrum. A hot gas will show an emission or line spectrum (dark background with bright lines). Each element has its own unique set of spectral lines.

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An absorption spectrum (bright background with dark lines) is seen if a hot source is viewed though a gas.

Examples of emission spectra:

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Before the structure of the atom was known, an empirical result was derived for the wavelengths of the spectral lines of hydrogen in the visible portion of the spectrum (the Balmer series).

⎟⎠

⎞⎜⎝

⎛ −=2

1

4

11

nR

λ

Where R = 1.097107, m-1 is the Rydberg constant and n 3.

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The Thomson model of the atom had a volume of positive charge with the negatively charged electrons embedded within the volume.

Scattering experiments by Rutherford led to the conclusion that an atom had a very small nucleus of positive charge (10-5 times the size of the atom containing nearly all of the mass) that was surrounded by the electrons.

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It was thought that the electrons in their orbits should radiate (they are accelerated) causing the electron’s orbit to decay, implying that atoms are not stable. This is obviously false. Any model of the atom must also explain the line spectra of the elements.

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§27.7 The Bohr Model of the Hydrogen Atom

The Bohr model assumes:

The electron is allowed to be in only one of a discrete set of states called stationary states. The electron orbits have quantized radii, energy, and angular momentum.

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Newtonian physics applies to an electron in a stationary state.

The electron can transition between one stationary state and another provided it can absorb/emit a photon of energy equal to the energy difference between the states. E = hf.

The stationary states have quantized angular momentum in the amount

;2

hnh

nLn ==π

n is an integer.

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The allowed radii are

02

2

22

ankem

nr

en ==

h

where a0 = 52.9 pm is the Bohr radius.

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The energy levels are given by 21

22

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2 n

E

n

ekmE en =−=

h

where E1 = 13.6 eV is the energy of the ground state, the lowest possible energy of the electron. When n > 1 the electron is in an excited state. The quantity n is an integer and is the principal quantum number.

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Energy level diagram for hydrogen

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The energy of a photon emitted (absorbed) by an electron during a transition is

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−==

221

221

111

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if

iffi

nnhc

E

nnEEE

hcE

λ

λ

Rhc

E=×=− −171 m 10097.1where is the Rydberg constant.

When nf=2, the above result reduces to the Balmer formula.

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The Bohr model correctly predicted the wavelengths of the spectral lines of hydrogen in the visible. There are several problems with the Bohr model.

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Bohr’s model, while successful at predicting the spectrum of hydrogen, fails at predicting the spectra of most other elements. Only hydrogenic atoms (atoms that only have one electron; Li2+ for example) can have their spectra computed using the Bohr model.

.02

2

22

Z

an

kZem

nr

en ==

h

2H1

2

22

422

2 n

EZ

n

eZkmE en =−=

h

The allowed radii are

The energy levels are

where Z is the atomic number of the atom and E1H = 13.6 eV .

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Example (text problem 27.34): Find the Bohr radius of doubly ionized lithium (Li2+).

Z

anrn

02

=What is r1?

pm 6.173

1

3

10

02

1 === aa

r

The inner most energy level is closer to the nucleus than in an H atom.

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Example (text problem 27.35): Find the energy in eV required to remove the remaining electron from a doubly ionized lithium (Li2+) atom.

The electron is in the ground state (n = 1), so

eV. 12291

3H12

H12

1 −=== EE

E

To remove the electron will require the electron be given 122 eV of energy.

2H1

2

n

EZEn =

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Example (text problem 27.43): A hydrogen atom has an electron in the n = 5 level.

One photon; the electron may transition from the n = 5 level to the n = 1 level.

(b) What is the maximum number that might be emitted?

Four photons; the electron may cascade from n = 5 to 4 to 3 to 2 to n = 1.

(a) If the electron returns to the ground state, what is the minimum number of photons that can be emitted?

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§27.8 Pair Annihilation and Pair Production

A photon can interact with an atomic nucleus and change itself into an electron-positron pair (or some particle-antiparticle pair.) A positron is an antielectron. The nucleus is needed to ensure that momentum is conserved.

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The energy of the photon must be at least 2mec2. If E > 2mec2 , then the additional energy goes into the kinetic energy of the electron-positron pair. This is pair production.

The inverse process is photons. 2→+ +− ee

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The process of pair production created protons, neutrons, and electrons in the earliest moments after the Big Bang.

To have enough energy, the photons must be “hot” enough. Electrons need T~1010 K and for protons/neutrons T~1013 K. The early Universe must have been much hotter than it is today.

Pair production creates equal amounts of matter and antimatter. Where in the Universe is all of the antimatter?

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Example (text problem 27.57): A muon and an antimuon, each with mass 207 times greater than an electron, were at rest when they annihilated and produced two photons of equal energy. What is the wavelength of each of the photons?

For an electron-positron pair MeV. 022.12 2e =cm

For a muon-antimuon pair

( ) MeV. 2122 207

22

e

2

== cm

cmμ

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Example continued:

The created photons each have 106 MeV of energy. Their wavelengths are

m. 1017.1nm 1017.1

eV 10106

nm eV 1240

145

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−− ×=×=×

==Ehcλ

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Summary

•Quantization

•The Photoelectric Effect

•Compton Scattering

•Pair Production/Annihilation

•Spectroscopy

•Bohr Model for Hydrogen

}These processes are explained by light behaving like a particle, not as a wave.