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					         The curious case of the nature of light

Summary.

Light exhibits both diffraction and interference and

therefore exhibits wave-like properties. In this lecture

topic, we will see that light can also exhibit particle

properties. These “ light ” particles” are called photons.

Light can behave like a wave or like a particle.



Background reading: Tipler sections 34.1, 34.2


Is light a wave (for a definition of a wave, see

http://www.kettering.edu/~drussell/Demos/waves-

intro/waves-intro.html ) or a particle? These are very

different entities.



   Particles are discrete, with their energy concentrated

     into a finite space. They exist at a specific location.
     Interactions between particles are governed by laws,

     such as the conservation of energy and momentum.



   The energy of waves cannot be considered to exist

     in a single place. A wave can propagate until it

     exists in all locations. The energy carried by a wave

     depends on its intensity. One of the best-known

     interactions between waves is the phenomenon of

     interference.

By the end of the 19th century, it had been firmly

established that light was a wave. Then came the

photoelectric effect.



When light falls on a metal surface, electrons are emitted

(photoelectrons).
                   monochromatic
                   light



                                                 evacuated tube



       A

                      anode                cathode




                         variable voltage V




The experiment is shown above. Light falls on the

electrode called the cathode and electric current flows in

the external circuit.

It is found that

   Current is directly proportional to light intensity

   Current appears within nanoseconds of light hitting

     the cathode

   Photoelectrons are emitted only if the frequency of

     the light is greater than some threshold frequency f0

   If V > 0 (anode positive with respect to cathode),

     current does not change as V increases. If V < 0,
     current decreases until V = -Vstop, the stopping

     potential

   Vstop is independent of light intensity



http://phet.colorado.edu/simulations/sims.php?sim=Phot

oelectric_Effect

Under classical wave theory, an atom will absorb energy

from an incident wave and will continue to absorb

energy until it has enough energy to be emitted from the

solid. Classical wave theory expected that the energy of

the light source should be determined by its intensity.

Hence, the energy required to eject a photoelectron

should be supplied by light of high intensity, no matter

how low the frequency of the radiation. Thus, there

should be no threshold frequency, below which no

electrons are emitted. Moreover, the kinetic energy of the

electrons should increase with intensity, not with light
frequency. These predictions are not observed, so the

results are counter to physical intuition.

Einstein saw the incident light as a stream of discrete

bundles of energy or photons. It is these photons which

are absorbed or emitted when a body absorbs from or

emits energy to the incident light. A photon is one

emitted or absorbed quantum of electromagnetic energy.

The energy of a photon is related to the frequency of the

light by

                           E = hf

h is known as Planck’s constant and has a value of 6.63 x

10-34 Js. These photons also carry momentum, which,

from relativity, can be shown to be

                           E = cp

Using this concept, we can explain the features of the

photoelectric effect thus. We assume that a metal

contains large numbers of free electrons. To remove an
electron from the surface of a metal takes a minimum

energy, known as the work function Electrons that are

in the bulk of the metal take more energy.

     Material               (eV)
       Na                   2.28
       Al                   4.08
       Co                   3.90
       Cu                   4.70
       Zn                   4.31
      Ag                    4.73
       Pt                   6.35
       Pb                   4.14



An electron will be emitted whenever a photon is

absorbed of energy E = hf ≥ . If hf > , an electron will

be emitted and the difference in energy between the

photon energy and the energy needed to free the electron

from the metal will appear as kinetic energy. The

maximum kinetic energy of the electrons is

                      Kmax = hf – 
This kinetic energy can be determined from the stopping

potential since Kmax = eVstop.

The cut-off frequency occurs when the photon energy

just equals the work function.

                              hf0 = 

Worked example

The work function of tungsten metal is 4.52 eV. What is

the cut-off wavelength for tungsten? (b)What is the

maximum kinetic energy of photoelectrons when light of

wavelength 200 nm falls on a tungsten surface? (c) What

is the stopping potential in this case?

Solution

(a) hf0 = hc/= 

giving 0 = 274 nm

(b) Kmax = hf - 

 6.2 eV - 4.52 eV

           = 2.68 eV
(c) The stopping potential is

 Vstop = Kmax/e = 2.68 volts
                   The Compton Effect

The Compton effect is a dramatic illustration of the

particle-like nature of light. A beam of X-rays or -rays

falls on a target (e.g. graphite) and the intensity of the

radiation scattered at various angles is measured as a

function of wavelength.

(http://physics.berea.edu/~king/Teaching/ModPhys/QM/

Compton/compton.html)




The results of such an experiment is shown below
These data show that, although the incident beam

consists essentially of a single wavelength, the scattered

beam has intensity peaks at two wavelengths. There is a

shifted component at longer wavelength. The Compton

shift '– varies with the angle at which the

scattered beam is observed.
The shifted component at ’ cannot be explained

classically. It can be interpreted by considering the

incident beam as a collection of photons colliding with

free electrons in the target as in a collision between

billiard balls. In such a collision, the photon looses some

energy to the (recoiling) electron and so shifts to longer

wavelength. Note that the photon is scattered and not

absorbed. (The electrons are not really free but are the

loosely bound outer or valence electrons in atoms. Since

the energy of the X-rays is very much greater than the

electron binding energy, the electrons can be considered

as free. The assumption is further justified by noting that

the wavelength of the scattered beam ’ is independent

of the target material.)
Consider the collision between a photon of initial

momentum p0, energy E0 with a stationary electron of

rest mass m0. The scattered photon has a momentum p1,

energy E1 whilst the electron recoils with momentum p

and kinetic energy K.

Conserving momentum in the x- and y-directions gives

                   p0 = p1cos + pcos

                        p1sin = psin

Squaring and adding gives

                p02 + p12 –2p0p1cos = p2

Conservation of energy gives

                         E0 – E 1 = K
For a photon, E = cp. Therefore,

                       K = c(p0 – p1)

For the electron, the total energy E is related to

momentum p by the relativistic formula

              E2 = c2p2 + m02c4 = (K + m0c2)2

which gives

        p2 = K2/c2 + 2Km0 = p02 + p12 –2p0p1cos

from which we get

                  1 1     1
                            1  cos θ 
                  p1 p 0 m 0c

Since E = hf = cp for a photon, then

                 = '–  = c(1 – cos)

c is known as the Compton wavelength and equals

0.00243nm.

The unshifted peak is the result of scattering from the

whole atom. Since the mass of the atom is many
thousands of times more massive than an electron, the

resulting Compton shift from this scattering is negligible.

Worked Example

X-rays of wavelength 0.2400 nm are Compton scattered

and the scattered beam is observed at an angle of 600

relative to the incident beam. Find (a) the wavelength of

the scattered X-rays (b) the energy of the scattered X-ray

photons (c) the kinetic energy of the scattered electrons

(d) the direction of travel of the scattered electrons



Solution

  (a) ’ –  =c(1 – cos)

  Thus, ’ =  + c(1 – cos)

            = 0.24 nm + 0.00243 nm( 1 – cos 60)

             = 0.2412 nm

  (b) The energy of the scattered photons is
       hc
E1 =     '
           = 5141 eV
       λ

(c) The kinetic energy of the electron is

 K = E0 – E1 = 5167 eV – 5141 eV = 26 eV

(d) From conservation of momentum,

                 p0 = p1cos + pcos

                   p1sin = psin

                               p1sinθ
                   tan  =
                             p p1cosθ

Now E = cp for the photons and the momentum of the

recoiling electrons can b e calculated form their kinetic

energy.

Calculations give  = 59.70.
                       What is light?

Light behaves with wave-like properties (diffraction,

interference) and with particle-like properties

(photoelectric effect). Light is not either particles or

waves, but somehow both particles and waves. This is

known as wave-particle duality. It shows only one or

other aspect, depending on the kind of experiment we are

doing. Neither the wave nor the particle picture is correct

all the time; both are needed for a complete description

of physical phenomena, and the two are complimentary

with one another. Modern quantum theory emphasises

the primacy of measurement and not attributing

properties to objects beyond what can be measured.

Hence the concept of wave-particle duality arose: it is

not necessary, or useful, to say that light is a particle - or

a wave - just that in certain circumstances it behaves like

a wave, and in others like a particle.
                    General Problems

1. When a metal is illuminated with light of wavelength

 < 388 nm, photoelectrons are observed. Determine the

work function of the metal.

(Ans. 3.2 eV)

2. A stopping potential of 2.0 V is measured when light

of 290 nm illuminates a metal.

(i) Determine the work function of the metal

(ii) Determine the stopping potential if the intensity of

the light is doubled

(Ans. (i) 2.28 eV (ii) 2.0 V)

				
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