Moderna Bno

7/29/2019 Moderna Bno

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Lecture 15

Experimental Evidence ofQuantum Theory IV


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Assignments

Reading: Ch. 3.7, 3.8, 3.9.

Homework 4: due on Friday (02/25).

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Photoelectric Effect

Methods of electron emission:

Thermionic emission: Application of heat allows electrons to gainenough energy to escape.

Secondary emission: The electron gains enough energy bytransfer from another high-speed particle that strikes the material

from outside.

Field emission: A strong external electric field pulls the electron outof the material.

Photoelectric effect: Incident light (electromagnetic radiation)shining on the material transfers energy to the electrons, allowingthem to escape. The ejected electrons are called photoelectrons.

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Work Function

Work Function: the minimum extra kinetic energy that allows electronsto escape from the material. It is the minimum binding energy of theelectron to the material.

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Experimental Setup

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Classical Theory of the Photoelectric Effect

Initial observations by Hertz in 1887

The kinetic energy of the

photoelectrons should increase withthe light intensity and not depend onthe light frequency.

Classical theory also predicted that theelectrons absorb energy from the beamat a fixed rate. So, for extremely lowlight intensities, a long time wouldelapse before any one electron could

obtain sufficient energy to escape. Heinrich Hertz(1857-1894)

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Experimental Results

Experiment: The maximum kinetic energies of thephotoelectrons are independent of the light intensity.Classical Theory: The maximum kinetic energyincreases with intensity.

Emax

k = eV

0

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Experiment: The maximum kinetic energy of thephotoelectrons depends only on the frequency of the light.Classical Theory: The maximum kinetic energy isindependent of frequency.

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Emax

k = eV

0

I


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Experiment: The smaller thework functionof the emitter

material, the smaller is thethreshold frequency of the lightthat can eject photoelectrons.

Classical Theory: Nothreshold frequency.


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The number of photoelectronsis proportional to the intensity

of light.

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Experiment: The photoelectrons are emitted instantlyfollowing illumination, independent of the light intensity.Classical Theory: Slow production of photoelectrons forweak intensity, may take hours/days.


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Quiz

10

As the wavelength (intensity) of the light incident on a metalsurface is increased, the maximum kinetic energy of

photoelectrons emitted form the surface

A. Increases (Increases)B. Decreases (Decreases)C. Increases (Does not Change)D. Decreases (Does not Change)E. Does not Change (Does not Change)

Answer: D


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Einsteins Theory

Einstein suggested that the electromagnetic radiationfield is quantized into particles called photons. Each

photon has the energy quantum:

where fis the frequency of the light and h is Plancksconstant. (Nobel Prize in 1921)

The photon travels at the speed of light in a vacuum,

and its wavelength is given by

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E = hf

f = c


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Conservation of energy yields:

Energy before (phonon) = Energy after (electron)

where is the work function of the metal.

The retarding potentials (V0) measured in the photoelectric effectare the opposing potentials needed to stop the most energetic

electrons:

Einsteins Theory

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hf= + KE = +1

2mvmax

2

eV0=1

2mv

max

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Electron

kine

ticenergy

Quantum Interpretation

The kinetic energy of the electron doesnot depend on the light intensity, butonly on the light frequency and the

work function of the material.

Einstein in 1905 predicted that thestopping potential was linearly

proportional to the light frequency, witha slope h, the same constant found

by Planck.

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1

2

mvmax

2= eV

0= hf

= hf0

eV0= h( f f

0)


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Quiz

14

No photoelectrons are emitted from tungsten unless thewavelength of light is less than 270 nm (1 nm = 10-9m). Nowif we need photoelectrons with maximum kinetic energy of

2.0 eV, what frequency of light should be used to illuminatetungsten? (h = 6.63 10-34 J s = 4.14 10-15eVs)

A. 0.48 1015 HzB. 0.96 1015 HzC.

1.11

10

15

HzD. 1.59 1015 HzE. 2.07 1015 Hz

Answer: D

hf = + Ek = hf0 + Ek

f = f0+

Ek

h

=c

0+

Ek

h= 1.1110

15Hz + 0.4810

15Hz

= 1.591015

Hz


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Photons also have momentum

p=E

c=

hf

c=

h

Use our expression for the relativisticenergy to find the momentum of a photon,

which has no mass:

Solar SailsCosmos 1

Stellar interiorsresists gravity

Momentum oflight causesradiation

pressure

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p= h/


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Example: Radiation Pressure on Earth Surface

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The power received by earth surface from the sun is 1400 W/m2. Estimate the radiation pressure.

Assume all photons have a wavelength of green light:=

510 nm

Energy of a photon:E = hf=

hc

=

1240 eV nm

510 nm= 2.43 eV = 3.891019 J

Number of photons per second per m2:N =1400 J/ 3.891019 J = 3.61021

momentum of a photon: p= h/ = 6.631034 J s / 510109 m

= 1.31027 N s

Finally, the radiation pressure: P = Np= 3.61021 1.31027 N s/(sm2)

= 4.7106 N/m2 = 4.7106 Pa

Compared with the atomsphere pressure on earth surface: 1.01105 Pa


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Compton Effect

When a photon enters matter,

it can interact with an electron.

E=

hc / =

hfp= h/

This yields the change in wavelength of the scattered photon, knownas the Compton effect:

Photons have energy andmomentum:

The laws ofconservation of energy and momentum apply, as in anyelastic collision between two particles.

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Compton Effect: the Derivation

E

e

2= (mc2)2 + p

e

2c2

hf+mc2

= hf'+ Ee

h

=h

'cos+ p

ecos

h'sin= pesin

pe

2= (p

esin)2 + (p

ecos)2 = (

h

'sin)2 + (

h

h

'cos)2 = (

h

')2 + (

h

)2 2(

h

')(h

)cos

= c / f

[h( ff')+mc

2

]2

= (mc2

)2

+ (hf)2

+ (hf')2

2(hf)(hf')cos

mc2( f f') = hff'(1 cos)

h

mc2(1 cos) =

f f'

ff'=

1

c(' )

=

' =

h

mc(1 cos) 18


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Example

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Derive the electron recoil kineticenergyEk and angle Ek = hf hf'= (1 f'/ f)hf

= (1 / ')hf= (1

+ )hf

=

+ hf=

/

1+ / hf

h

=h

'cos+ p

ecos

h

'sin= p

esin

cot=

h

h

'cos

h

'sin

='

1

sin cot=

+

1

sin cot

= [1+h

mc(1 cos)]

1

sin cot= 1+

hf

mc2

1 cossin

= (1+hf

mc2)tan

2


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Exercise

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A photon of wavelength 2.0 nm Compton-scatters from anelectron at an angle of 90. What is the modified wavelengththe the percentage change, ? /

=h

mc(1 cos) =

hc

mc2(1 cos)

= 1240 eV nm0.511 MeV

(1 cos90) = 2.43103 nm

'= + = 2.00243 nm

=

2.43103

nm

2 nm= 0.122%


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Pair Production and Annihilation

In 1932, C. D. Anderson observed apositively charged electron (e+) incosmic radiation. This particle, called apositron, had been predicted to existseveral years earlier by P. A. M. Dirac(Nobel Prize in 1936).

A photons energy can be convertedentirely into an electron and a positronin a process called pair production:

Paul Dirac(1902 - 1984)

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But it cannot occur in vacuum.


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Pair Production in Empty Space

Conservation of energy for pairproduction in empty space is:

hf = E++ E

This yields a lowerlimit on the photon energy: hf> p

c+ p+c

hf pc+ p

+cThis yields an upperlimit on the photon energy:

hf= p

ccos(

)+ p+

ccos(+

)Momentum conservation yields:

A contradiction! Hence the conversion of energy and momentum

for pair production in empty space is impossible!

hfE+

E

The total energy for a particle is:

So:

E

2= p

2c2 +m2c4

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Pair Production in Matter

In the presence of matter, thenucleus absorbs some energy

and momentum.

The photon energy required forpair production in the presence

of matter is:

hf = E++ E

+ K.E.(nucleus)

hf > 2mec2=1.022MeV

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Pair Annihilation

A positron passing through matterwill likely annihilate with anelectron. The electron and positroncan form an atom-like configurationfirst, called positronium.

Pair annihilation in empty spaceproduces two photons to conservemomentum. Annihilation near a

nucleus can result in a singlephoton.

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Pair Annihilation

Conservation of energy:

Conservation of momentum:

f1= f

2= f

hf1

c

hf2

c= 0

2m

ec2 hf

1+ hf

2

hf =mec2= 0.511MeV

So the two photons will have thesame frequency:

The two photons from positroniumannihilation will move in oppositedirections with an energy:

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