UNIT 24 : QUANTIZATION OF LIGHT
(3 hours)
24.1 Planck’s Quantum
Theory
24.2 The Photoelectric
Effect
1
SUBTOPIC :
24 .1 Planck’s Quantum Theory
½ hour
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Distinguish between Planck’s quantum
theory and classical theory of energy.
b) Use Einstein’s formulae for
hc
E

hf

a photon energy,
.

2
Einstein memorial in Washington, D.C
3
Fig 39-CO, p.1244
Albert Einstein
German-American
Physicist
(1879–1955)
4
p.1251
Max Planck
German Physicist
(1858–1947)
5
p.1288
24.1 Planck’s Quantum Theory
• The foundation of the Planck’s quantum theory is a
theory of black body radiation.
• Black body is defined as an ideal system or object
that absorbs and emits all the em radiations that is
incident on it.
• The electromagnetic radiation emitted
by the black body is called black body
radiation.
• In an ideal black body, incident light is
completely absorbed.
• Light that enters the cavity through the
small hole is reflected multiple times
black body
from the interior walls until it is
6
completely absorbed.
• The spectrum of electromagnetic radiation emitted
by the black body (experimental result) is shown in
figure 1.
Experimental
result
Rayleigh Jeans theory
Wien’s theory
Classical
physics
7
Figure 1 : Black Body Spectrum
• Rayleigh-Jeans and Wien’s theories (classical
physics) failed to explain the shape of the black
body spectrum or the spectrum of light emitted by
hot objects.
• Classical physics predicts a black body radiation
curve that rises without limit as the f increases.
•
The classical ideas are :



Energy of the e.m. radiation is related to the
amplitude and does not depend on its frequency
or wavelength.
An object at any temperature emits radiation
(energy)
Energy of the e.m. radiation is continuously.
8
• In 1900, Max Planck proposed his theory that is
fit with the experimental curve in figure 1 at all
wavelengths known as Planck’s quantum
theory.
• The assumptions made by Planck in his theory
are :

The e.m. radiation emitted or absorb by the
black body not in a continuous stream of
waves but in discrete little bundles
(separate) packets of energy or quanta,
known as photon.
 This means the energy of e.m. radiation is
quantised. Not all values of energy are
possible
9
 The energy size of the radiation depends
on its frequency.
•
The energy is proportional to the frequency
(E  f ) of the radiation. The greater the
frequency , the greater the quantum of energy
(f ↑, E↑).
• Each discrete energy value represents a
different quantum state for the molecule.
•When molecule is in n = 1 quantum state, its
energy is hf. When molecule is in n = 2 quantum
state, its energy is 2hf and so on.
10
Comparison between Planck’ quantum theory and classical theory of energy.
Planck’s Quantum
Theory
Energy of the e.m radiation is
quantised. (discrete)
Classical theory
Energy of the e.m radiation is
continously.
Photon
Energy of e.m radiation
depends on its frequency or
wavelength
E  hf
Energy of e.m radiation does
not depend on its frequency
or wavelength (depends on
Intensity) I  A2
Eclassical k BT k B  Boltzman' s constant
T  temperature
11
• According to this assumptions, the quantum E of
the energy for radiation of frequency f is given
by
E  hf
where
c  f
h : Planck constant  6.63 1034 J s
hc
E

Planck’s quantum
theory
12
Photons
• In 1905, Albert Einstein proposed that light comes in
bundle of energy (light is transmitted as tiny
particles), called photons.
• Photon is defined as a particle with zero mass
consisting of a quantum of electromagnetic
radiation where its energy is concentrated.
Quantum means “fixed amount”
13
• In equation form, photon energy (energy of photon)
is
c
E  hf  h

• Unit of photon energy is J or eV.
• The electronvolt (eV) is a unit of energy that can
be defined as the kinetic energy gained by an
electron in being accelerated by a potential
difference (voltage) of 1 volt.
• Unit conversion : 1 eV  1.60  1019 J
• Photons travel at the speed of light in a vacuum.
• Photons are required to explain the photoelectric
effect and other phenomena that require light to
14
have particle property.
Higher frequency photons have more energy. A
photon of blue light is more energetic than a
photon of red light.
f blue  f red ,
Eblue  Ered
A photon travels at the speed of light (c) in vacuum
Example 24.1
Calculate the energy of a photon of blue light,
  450 nm .
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s
1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
17
Example 24.2
(a) What is the energy of a photon of red light of
wavelength 650 nm ?
(b) What is the wavelength of a photon of energy
2.40 eV ?
Solution
18
(b) Converting energy in eV to Joule (J)
19
Exercise
A photon have an energy of 3.2 eV. Calculate
the frequency, vacuum wavelength and energy
in joule of the photon.
(7.72 x 1014 Hz ,389 nm, 5.12 x10-19 J)
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s
1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
20
SUBTOPIC :
24 .2 The Photoelectric Effect
2 ½ hours
LEARNING OUTCOMES :
At the end of this lesson, the students should
be able to :
a) Explain the phenomenon of photoelectric effect.
b) Define and determine threshold frequency, work
function and stopping potential.
c) Describe and sketch diagram of the photoelectric
effect experimental set-up.
d) Explain the failure of wave theory to justify the
photoelectric effect.
21
SUBTOPIC :
24 .2 The Photoelectric Effect
2 ½ hours
LEARNING OUTCOMES :
At the end of this lesson, the students should be able to :
e) Explain by using graph and equations the
observations of photoelectric effect experiment in
terms of the dependence of :
i ) kinetic energy of photoelectron on the frequency
of light;
½ mvmax2 = eVs = hf – hfo
ii ) photoelectric current on intensity of incident light;
iii) work function and threshold frequency on the
types of metal surface; Wo =hfo
f) Use Einstein’s photoelectric effect equation,
Kmax = eVs = hf – Wo
22
24 .2 The photoelectric effect
• The photoelectric effect is the emission of electrons
from the metal surface when electromagnetic
radiation of enough frequency falls/strikes/
incidents /shines on it.
• A photoelectron is an electron ejected due to
photoelectric effect (an electron emitted from
the surface of the metal when light strikes its surface).
em radiation
photoelectron
(light)
- - - - -- - - - - Metal surface
23
Free electrons
9.2 The photoelectric effect
• The photoelectric effect can be measured using a
device like that pictured in figure below.
e.m. radiation (incoming light)
Cathode (emitter
or target metal)
Anode(collector)
photoelectron
-
-
vacuum
glass
A
V
A
power supply
rheostat
The photoelectric effect’s experiment
24
9.2 The photoelectric effect
• A negative electrode (cathode or target metal or
emitter) and a positive electrode (anode or
collector) are placed inside an evacuated glass
tube.
• The monochromatic light (UV- incoming light) of
known frequency is incident on the target metal.
• The incoming light ejects photoelectrons from a
target metal.
• The photoelectrons are then attracted to the
collector.
• The result is a photoelectric current flows in
the circuit that can be measured with an ammeter.
25
9.1 The photoelectric effect
• When the positive voltage (potential difference)
is increased, more photoelectrons reach the
collector , hence the photoelectric current also
increases.
• As positive voltage becomes sufficiently large, the
photoelectric current reaches a maximum
constant value Im, called saturation current.
Saturation current is defined as the maximum
constant value of photocurrent in which when
all the photoelectrons have reached the
anode.
26
9.2 The photoelectric effect
• If the positive voltage is gradually decreased, the
photoelectric current I also decreases slowly.
Even at zero voltage there are still some
photoelectrons with sufficient energy reach the
collector and the photoelectric current flows is Io .
Photoelect ric current , I
Im
Graph of
photoelectric current
I0
against voltage for
photoeclectric
effect’s experiment
 Vs
0
B
(After)
Voltage,V
A (Before reversing the terminal)
27
9.2 The photoelectric effect
• When the voltage is made negative by reversing
the power supply terminal as shown in figure
below, the photoelectric current decreases since
most photoelectrons are repelled by the collector
which is now negative electric potential.
Cathode (emitter
or target metal)
e.m. radiation (incoming light)
Anode(collector)
photoelectron
-
-
vacuum
glass
A
V
power supply
B
rheostat
Reversing power
supply terminal
(to determine the
stopping potential)
28
• If this reverse voltage is small enough, the fastest
electrons will still reach the collector and there will
be the photoelectric current in the circuit.
• If the reverse voltage is increased, a point is
reached where the photoelectric current reaches
zero – no photoelectrons have sufficient kinetic
energy to reach the collector.
• This reverse voltage is called the stopping
potential , Vs.
Vs is defined as the minimum reverse potential (voltage)
needed for electrons from reaching the collector.
• By using conservation of energy :
(loss of KE of photoelectron = gain in PE) ;
K.Emax = eVs
1 2
eVs  mv
2
29
Einstein’s theory of Photoelectric Effect
 According to Einstein’s theory, an electron is
ejected/emitted from the target metal by a
collision with a single photon.
 In this process, all the photon energy is
transferred to the electron on the surface of metal
target.
 Since electrons are held in the metal by attractive
forces, some minimum energy,Wo (work function,
which is on the order of a few electron volts for
most metal) is required just enough to get an
electron out through the surface.
30
Einstein’s theory of Photoelectric Effect
 If the frequency f of the incoming light is so low
that is hf < Wo , then the photon will not have
enough energy to eject any electron at all.
 If hf > Wo , then electron will be ejected and
energy will be conserved (the excess energy
appears as kinetic energy of the ejected electron).
 This is summed up by Einstein’s photoelectric
equation , E  W  K .E
o
max
1
2
hf  W0  mvmax
2
hf  W0  eVs
but
1 2
eVs  mv
2
31
Einstein’s theory of Photoelectric Effect
E  Wo  K .Emax
1
2
hf  W0  mvmax
2
E  hf  h
c
Einstein’s
photoelectric
equation
= photon energy

f = frequency of em radiation /incoming light
K .Emax
1 2
 mv max= maximum kinetic energy of
2
ejected electron.
vmax = maximum speed of the photoelectron
32
Einstein’s theory of Photoelectric Effect
hf > Wo
vmax
hf
-
Metal
Electron is
emitted
hf
- v=0
W0
-
1
2
hf  W0  mvmax
2
Metal
hf  Wo
-
W0
Electron is
ejected.
hf
hf < Wo
Metal
-
W0
No electron is ejected.
33
Terms in photoelectric effect
Work Function, W0
is the minimum energy required (needed) to
eject an electron from the surface of target
metal
W0  hf0
where
h
= Planck’s Constant
f0
= Threshold frequency
• W0 depends on the type metal used
Metal
Potassium
Natrium
Zink
Copper
Platinum
Work Function, W0 (eV)
2.24
2.46
3.57
4.16
6.30
Terms in photoelectric effect
Threshold Frequency, f 0
Is the minimum frequency of e.m required to eject
an electrons from surface of the target metal
If the frequency of the incident radiation is less than
the threshold frequency (f < f0 ) then electrons would
not be removed from the metal surface.
Depends on the type of metal used
c threshold wavelength.  o
fo 
o maximum wavelength of e.m. radiation required to
eject an electron from the surface of the target
metal.
Terms in photoelectric effect
Maximum Kinetic Energy, Kmax
K max
1
2
 mv max  eV s  hf  hf 0
2
where;
1eV = 1.6 x 10-19 J
Kmax of the photoelectrons depends on the
frequency of the incident radiation
Kmax of the photoelectrons increases when
the frequency of the incident radiation
increases
Example 24 .3
The work function for a silver surface is Wo = 4.74 eV.
Calculate the
a) minimum frequency that light must have to eject
electrons from the surface.
b) maximum wavelength that light must have to eject
electrons from the surface.
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s
1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
38
Example 24.4
What is the maximum kinetic energy of electrons
ejected from calcium by 420 nm violet light, given
the work function for calcium metal is 2.71 eV?
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s
1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
39
Example 24.5
Sodium has a work function of 2.30 eV. Calculate
a. its threshold frequency,
b. the maximum speed of the photoelectrons
produced when the sodium is illuminated by
light of wavelength 500 nm,
c. the stopping potential with light of this
wavelength.
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s
1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
Solution 24.5
40
Solution 24.5
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s
1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
b.
c.
41
Example 24.6
In an experiment of photoelectric effect, no current
flows through the circuit when the voltage across
the anode and cathode is -1.70 V. Calculate
a. the work function, and
b. the threshold wavelength of the metal (cathode)
if it is illuminated by ultraviolet radiation of
frequency 1.70 x 1015 Hz.
(Given : c = 3.00 x 108 m s-1,
h = 6.63 x 10-34 J s , 1 eV=1.60 x 10-19 J,
me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
42
Solution 24.6
a) W0  8.55 1019 J
b) 0  2.33107 m
43
Example 24.7
The energy of a photon from an electromagnetic
wave is 2.25 eV
a. Calculate its wavelength.
b. If this electromagnetic wave shines on a metal,
photoelectrons are emitted with a maximum
kinetic energy of 1.10 eV. Calculate the work
function of this metal in joules.
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s ,
1 eV=1.60 x 10-19 J, mass of electron m = 9.11 x
10-31 kg, e = 1.60 x 10-19 C)
44
Solution 24.7
Ans. : 553 nm, 1.84 x 10-19 J
45
Graphs in Photoelectric Effect
 Generally, Einstein’s photoelectric equation;
E  Wo  K .Emax
K .Emax  E  Wo
K .Emax  hf  Wo
K.Emax
 m x c
y
f ↑ K.Emax ↑
f
0
Wo
fo
46
K .Emax  hf  Wo
Graphs in Photoelectric Effect
eVs  hf  Wo
Wo
h
Vs  f 
e
e
y  mx  c
Stopping voltage ,Vs
0
W0

e
fo
f ↑ Vs ↑
frequency , f
47
Graphs in Photoelectric Effect
Variation of stopping voltage Vs with frequency f of
the radiation for different metals but the intensity
is fixed.
Stopping voltage ,Vs
W01
W02
K .Emax  hf  Wo
eVs  hf  Wo
Wo
h
Vs  f 
e
e
y  mx  c
0
f 01
W0  f0
f 02
W02 > W01
frequency , f
f02 > f01
(1) Kmax of photoelectron varies linearly with light frequency.
(2) Different threshold frequency for different metal
48
Graphs in Photoelectric Effect
Variation of photoelectric current I with voltage V for
the radiation of different intensities but its
frequency and metal are fixed.
Photoelect ric current , I
Intensity 2x
Intensity 1x
Vs
0
Voltage,V
When intensity is increased the maximum current attained is
higher showing that more e- are emitted.
Vs remains the same shows that the Kmax of photoelectron
independent of intensity of light
49
Notes:
Classical physics
Light intensity , I 
energy
time  area
Quantum physics
number of photons
Light intensity , I 
time  area
Light intensity  number of photons
Light intensity ↑ ,
number of photons ↑ ,
number of electrons ↑ ,
current ↑
.
(If light intensity ↑, photoelectric current ↑).
50
Graphs in Photoelectric Effect
Variation of photoelectric current I with voltage V
for the radiation of different frequencies but its
intensity and metal are fixed.
Photoelect ric current , I
Im
f2 > f1
f2
Vs2 > Vs1
f1
Voltage,V
 Vs 2 Vs10
K .Emax  hf  Wo
eVs  hf  Wo
Wo
h
Vs  f 
e
e
f ↑ Vs ↑
51
Graphs in Photoelectric Effect
Variation of photoelectric current I with voltage V
for the different metals but the intensity and
frequency of the radiation are fixed.
Photoelect ric current , I
Im
W01
W02
 Vs1  Vs 20
K .Emax  hf  Wo
eVs  hf  Wo
Wo
h
Vs  f 
e
e
Wo ,Vs 
Voltage,V
W02 > W01
Vs1 > Vs2
52
Example 24.8
K.Emax (x 10-19 J)
f(x 1014 )Hz
0
4 .8
Use the graph above to find the value of
i) work function and
ii) the threshold wavelength.
53
K .Emax  E  Wo
Solution 24.8
K.Emax (x 10-19 J)
K .Emax  hf  Wo
y
 m x c
f(x 1014 )Hz
0
4 .8
54
OBSERVATIONS
of the photoelectric effects experiment
1. Electrons are emitted immediately
2. Stopping potential does not depend on the
intensity of light.
3. Threshold frequency of light is different for
different target metal.
4. Number of electrons emitted of the
photoelectron current depend on the intensity of
light.
55
EXPLAIN the failure of classical theory to justify the
photoelectric effect.
1. MAXIMUM KINETIC ENERGY OF PHOTOELECTRON
Clasiccal prediction Experimental
Result
Modern Theory
The higher the
intensity, the
greater the energy
imparted to the
metal surface for
emission of
photoelectrons.
•The higher the
intensity of light
the greater the
kinetic energy
maximum of
photoelectrons.
Based on Einstein’s
photoelectric equation:
Very low intensity
but high frequency
radiation could
emit
photoelectrons.
The maximum
kinetic energy of
photoelectrons is
independent of
light intensity.
K max  hf  W0
The maximum kinetic
energy of photoelectron
depends only on the light
frequency .
The maximum kinetic
energy of photoelectrons
DOES NOT depend on
light intensity.
56
Dependence of and on the Types of Metal
Surface
2. EMISSION OF PHOTOELECTRON (energy)
Clasiccal
prediction
Emission of
photoelectro
ns occur for
all
frequencies
of light.
Energy of
light is
independent
of frequency.
Experimental
Result
Emission of
photoelectrons
occur only when
frequency of
the light
exceeds the
certain
frequency
which value is
characteristic
of the material
being
illuminated.
Modern Theory
When the light frequency is
greater than threshold
frequency, a higher rate of
photons striking the metal
surface results in a higher
rate of photoelectrons
emitted. If it is less than
threshold frequency no
photoelectrons are emitted.
Hence the emission of
photoelectrons depend on
the light frequency.
58
3. EMISSION OF PHOTOELECTRON ( time )
Clasiccal prediction
Experimental
Result
Modern Theory
Light energy is spread
over the wavefront, the
amount of energy
incident on any one
electron is small. An
electron must gather
sufficient energy before
emission, hence there is
time interval between
absorption of light
energy and emission.
Time interval increases
if the light intensity is
low.
Photoelectrons
are emitted
from the
surface of the
metal almost
instantaneously
after the
surface is
illuminated, even
at very low light
intensities.
The transfer of
photon’s energy to an
electron is
instantaneous as its
energy is absorbed in
its entirely, much
like a particle to
particle collision. The
emission of
photoelectron is
immediate and no
time interval
between absorption
of light energy and
emission.
59
4. ENERGY OF LIGHT
Clasiccal
prediction
Experiment Modern Theory
al Result
Energy of light
depends only
on amplitude
( or intensity)
and not on
frequency.
Energy of
light
depends
on
frequency
According to Planck’s
quantum theory which is
E=hf
Energy of light depends
on its frequency.
60
Dependence of photoelectric current on
intensity of incident light
• Experimental observations deviate from
classical predictions based on Maxwell’s
e.m. theory. Hence the classical physics
cannot explain the phenomenon of
photoelectric effect.
• The modern theory based on Einstein’s
photon theory of light can explain the
phenomenon of photoelectric effect.
• It is because Einstein postulated that light is
quantized and light is emitted, transmitted
and reabsorbed as photons.
62
SUMMARY : Comparison between classical physics
and quantum physics about photoelectric effect experiment
Feature
Classical physics
Quantum physics
Threshold
frequency
An incident light of
any frequency can
eject electrons (does
not has threshold
frequency), as long as
the beam has
sufficient intensity.
To eject an electron, the
incident light must have a
frequency greater than a
certain minimum value,
(threshold frequency) , no
matter how intense the
light.
Maximum kinetic
energy
of photoelectrons
Depends on the light
Depends only on the light
frequency .
Emission of
photoelectrons
There should be some
delays to emit
electrons from a
metal surface.
Energy of light
Depends on the light
intensity.
intensity.
Electrons are emitted
spontaneously.
Depends only on the light
63
frequency .
Exercise
(Given c = 3.00 x 108 m s-1, h = 6.63 x 10-34 J s
1 eV=1.60 x 10-19 J, me = 9.11 x 10-31 kg, e = 1.60 x 10-19 C)
1. Find the energy of the photons in a beam whose
wavelength is 500 nm. ( 3.98 x 10 -19 J)

2. Determine the vacuum wavelength corresponding to a -ray
energy of 1019 eV. (1.24 x10-25 m)
3. A sodium surface is illuminated with light of wavelength
300 nm. The work function for sodium metal is 2.46 eV.
Calculate
a) the kinetic energy of the ejected photoelectrons
b) the cutoff wavelength for sodium
c) maximum speed of the photoelectrons.
-1)
(1.68 eV, 505 nm, 7.68 x 105 ms64
4. Radiation of wavelength 600 nm is incidents upon the
surface of a metal. Photoelectrons are emitted from the
surface with maximum speed 4.0 x 105 ms-1. Determine
the threshold wavelength of the radiation. (7.7 x 10-7 m)
5. Determine the maximum kinetic energy, in eV, of
photoelectrons emitted from a surface which has a work
function of 4.65 eV when electromagnetic radiation of
wavelength 200 nm is incident on the surface. (1.57 eV)
6. When light of wavelength 540 nm is incident on the
cathode of photocell, the stopping potential obtained is
0.063 V. When light of wavelength 440 nm is used, the h
stopping potential becomes 0.865 V. Determine the ratio .
e
( 6.35 x 10-15 J s C-1)
65
7. In an experiment on the photoelectric effect, the following
data were collected.
Wavelength of e.m.
radiation, (nm)
350
450
Stopping
potential, Vs (V)
1.70
0.900
a. Calculate the maximum velocity of the photoelectrons
when the wavelength of the incident radiation is 350 nm.
b. Determine the value of the Planck constant from the
above data.
(7.73 x 105 m s-1, 6.72 x 10-34 J s)
66
8. In a photoelectric effect experiment it is observed that
no current flows unless the wavelength is less than 570
nm. Calculate
a. the work function of this material in electronvolts.
b. the stopping voltage required if light of wavelength
400 nm is used.
(2.18 eV, 0.92 V)
9. In a photoelectric experiments, a graph of the light frequency f
is plotted against the maximum kinetic energy Kmax of the
photoelectron as shown in figure below.
f  1014 Hz
 2.0
Kmax (eV )
Based on the graph, for the light frequency of 6.00 x 1014 Hz,
calculate
a. the threshold frequency.
b. the maximum kinetic energy of the photoelectron.
c. the maximum velocity of the photoelectron.
f 0  4.83 1014 Hz , K max  7.78 1020 J , v  4.13  10 5 m s -1
68
10. A photocell with cathode and anode made of the same metal
connected in a circuit as shown in the figure below.
Monochromatic light of wavelength 365 nm shines on the
cathode and the photocurrent I is measured for various values
of voltage V across the cathode and anode. The result is
shown in the graph.
I (nA)
365 nm
5
G
V
1
0
V (V )
a. Calculate the maximum kinetic energy of the photoelectron.
b. Deduce the work function of the cathode.
c. If the experiment is repeated with monochromatic light of
wavelength 313 nm, determine the new intercept with the
V-axis for the new graph.
69
-19
-19
(1.60 x 10 J, 3.85 x 10 J, -1.57 V)