Download The Compton Effect

Graphical Analysis and Applications
of Photoelectric Effect
EMR 2
Review of Photoelectric Effect
Graphical Analysis
Photoelectric Current
 Number of electrons that move from a cathode to an anode
in some time interval
 Current equation: I = q/t (charge is dependent upon the
number of electrons being emitted)
Graphs
1. Energy vs frequency
2.
Current vs frequency
Slope
=h
Ek
1019 J
fo
f (1014 Hz )
W
f  fo
Graphs
3. Current vs Intensity
4. Current vs Stopping Voltage
Wave Particle Duality Theory
 Theory indicates that particles can behave as waves
and waves can behave as particles.
 to describe the photoelectric effect, light or
radiation which are described by a wave model can
be considered a particle with a discrete amount of
energy
Applications of Photoelectric Effect
 practical applications include the photocell, photoconductive
devices and solar cells.
 A photocell is usually a vacuum tube with two electrodes. One
is a photosensitive cathode which emits electrons when exposed
to light and the other is an anode which is maintained at a
positive voltage with respect to the cathode. Thus when light
shines on the cathode, electrons are attracted to the anode.
 An electron current flows in the tube from cathode to anode. The
current can be used to operate a relay, which might turn a motor
on to open a door or ring a bell in an alarm system.
 The system can be made to be responsive to light, as above, or
sensitive to the removal of light as when a beam of light incident
on the cathode is interrupted, causing the current to stop.
Photocell
Photocells in Medicine
 Absorption of light by a
bacteria cell
causes a drop in the
number of photons
absorbed by the photocell
and a drop in the current
Photocells in Garage Door Openers
 Light to photocell
is interrupted, and
the corresponding
drop in
photocurrent
signals the motor
to
reverse.
Photocells in Movie Film
 Optical sound track is
like a bar-code, but
much more
detailed. The track
modulates the
intensity of the light at
a frequency which is
the same as the sound
which was used to
produced the
track.
The Compton Effect
EMR 2
Compton Effect
 Einstein predicted that photons should also possess
momentum which is a particle like property
p  mv
E  hf
E
hc

EM waves
vc
E
v
2
c
E
p 2c
c
E
p
E  pc
c
c f
p
hf  pc
p
hf
c
E  mc 2
m
hc

 pc
p
h

E
c2
Two equations that describe
the momentum of photons
making no reference to mass
Examples
 Determine the momentum of a
a. Photon of wavelength 250 nm
p
h

6.63  1034 Js
p
2.50  107 m
p  2.65  1027 kgm / s
b. Electron moving at 4.00 x 105 m/s
p  mv
p  9.11  1031 kg  4.00  105 m / s
p  3.64  1025 kgm / s
Example:
Compton (1922) – indicated experimentally
that photons have momentum
 directed x-rays at graphite atoms and detected the
scattered rays
Scattered
x-ray
Incident
beam
Graphite
or
carbon
atom
 the energy and momentum gained by the electron within
the atom equals the energy and momentum lost by the
photon
 the results obey the laws of conservation of energy and
momentum
 Compton Effect – the scattering of an x-ray by an electron
resulting in a reduced frequency of the x-ray (increase in
wavelength)
p
 Momentum of an x-ray (EMR)
h

h  p
• Compton showed how the change in wavelength ( ) of the
scattered photonis related to the angle at which the x-ray is
scattered

e


Scattered x-ray
Equation
   f  i
Equation :
h
 
(1  cos  )
mc
Speed of
light
Mass of
scattering
electrons
 Energy and
momentum are
conserved, so the
collision is elastic
Eg) Determine the maximum change in wavelength of a
0.050 nm x-ray scattered by an electron
(maximum scattering is at 180⁰ where the photon bounces
back)
 ( occurs when 1 – cosθ is a maximum value at θ = 180⁰)
h
(1  cos  )
mc
6.63  1034 Js
 
(1  cos180)
31
8
(9.11  10 kg )(3.0  10 m / s)
 
  2.4259  1012 m(1  1)
  2.4259  1012 m(2)
  4.85  1012 m
Eg) An x-ray photon with a wavelength of 0.0150
nm scatters at 60.0⁰ after contacting an electron.
Determine the wavelength of the scattered
photon.
h
 
(1  cos  )
mc
6.63  10 34 Js
 
(1  cos 60)
31
8
(9.11  10 kg )(3.0  10 m / s)
  2.4259  1012 m(0.500)
  1.21  10 12 m
   f  i
1.21  1012 m   f  1.50  10 11 m
 f  1.62  1011 m
Wave Like Behavior of Matter
EMR 2
Wave-like Behaviour of Matter
 De-Broglie (1923) said wave-particle
duality is a basic property of quantum
mechanics
 Suggested particles have wave properties
De Broglie wave equation
p
p
h

h

wave or photon momentum
p  mv
mv 
h

h
 
mv
 De Broglie wavelength is more significant for small masses traveling at
high speeds rather than large masses traveling at low speeds
Eg) Determine the De Broglie wavelength for
an alpha particle traveling at 0.015c.
p
h

p  mv
mv 
h

h
 
mv
6.63  1034 Js

27
8
6.65  10 kg (0.015)(3.00  10 m / s)
  2.2  10
14
m
Eg) An electron is accelerated by a potential
difference of 220V. Determine the De Broglie
wavelength for the electron.
E
q
Ek  Vq
1 2
mv
2
V
Ek 
Ek  1.6  1019 C (220V )
1
3.52  10 J  (9.11  1031 kg )(v 2 )
2
v  8.79  106 m / s
17
Ek  3.52  1017 J
h
p
h

mv

6.63  1034 Js

(9.11  1031 kg )(8.79  106 m / s )
  8.28  1011 m
 http://www.kcvs.ca/site/projects/physics.html
 - De Broglie wanted to show experimentally that electrons
diffract.
 - when an electron beam is directed onto a crystal lattice
structure with atomic spacings of 10-10 m the electron beam
diffracted similar to x-rays or other EM waves
 This indicates that electrons have a wavelength and travel ina
standing wave pattern
Diffraction Pattern
Review Questions:
 Light with a frequency of 5.50 x 1014 Hz is incident on a
photoelectric surface, the same photoelectric tube has a
stopping voltage of 0.540V.
What is the threshold frequency of the photoelectric surface?
Ek max  hf  W
Ek max  qVstop
Ek max  1.60  10
19
W  hf  Ek max
  0.540 W   6.63  10 5.50 10   8.60 10
Ek max  8.60  1020 J
34
W  hf 0  2.78  1019 J
2.78  1019 J
fo 
h
f 0  4.20  1014 Hz
14
20
2 D Scattering
 An incident photon with a wavelength of 4.50 x 10-11 m
collides with an electron that is at rest. The photon is
scattered at an angle of 62.0° and has a final momentum and
energy.
λ = 4.50 x 10-11 m
62.0°
  4.63 1011 m
 What is the momentum and the velocity of the scattered
electron?
 Show that energy is conserved.