This document provides an overview of Planck's quantum theory and the photoelectric effect. It begins by outlining the key learning outcomes for understanding Planck's quantum theory, which distinguished energy of electromagnetic radiation as quantized rather than continuous. It then describes the photoelectric effect and defines important concepts like work function and stopping potential. Finally, it presents Einstein's explanation of the photoelectric effect using photon energy and provides examples demonstrating how to use the photoelectric equations.

1. 1
Chapter 29 Waves and Particles:
29.1 Planck’s Quantum Theory
29.2 The Photoelectric Effect

2. 2
29 .1 Planck’s Quantum Theory
SUBTOPIC :
LEARNING OUTCOMES :
a) Distinguish between Planck’s quantum
theory and classical theory of energy.
b) Use Einstein’s formulae for
a photon energy, .
At the end of this lesson, the students should
be able to :
hc
E hf

3. 3
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.
black body
• In an ideal black body, incident light is
completely absorbed.
• Light that enters the cavity through the
small hole is reflected multiple times
from the interior walls until it is
completely absorbed.

4. 4
Experimental
result
Rayleigh -
Jeans theory
Wien’s theory
Classical
physics
• The spectrum of electromagnetic radiation emitted
by the black body (experimental result) is shown in
figure 1.
Figure 1 : Black Body Spectrum

5. 5
• 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 does not
depend on its frequency or wavelength.
 Energy of the e.m. radiation is
continuously.

6. 6
• 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 by the black body
is a discrete (separate) packets of energy
known as quanta. This means the energy of
e.m. radiation is quantised.
 The energy size of the radiation depends
on its frequency.

7. 7
Comparison between Planck’ quantum theory and classical theory of energy.
Planck’s Quantum
Theory
Classical theory
Energy of the e.m radiation is
quantised. (discrete)
Energy of the e.m radiation is
continously.
Energy of e.m radiation
depends on its frequency or
wavelength
Energy of e.m radiation does
not depend on its frequency
or wavelength (depends on
Intensity)
Photon
2
AIhfE
TkE Bclassical
etemperatur
constantsBoltzman'
T
kB

8. 8
• According to this assumptions, the quantum E of
the energy for radiation of frequency f is given
by
hfE
where constantPlanck:h J s10636 34
.
Planck’s quantum
theory
fc
hc
E

9. 9
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”

10. 10
• 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
have particle property.
• In equation form, photon energy (energy of photon)
is c
hhfE
• 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 : J10601eV1 19
.

11. 11
Example 24.1
Calculate the energy of a photon of blue light,
.nm450
(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)

12. 12
Example 24.2
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)

13. 13
29 .2 The Photoelectric Effect
SUBTOPIC :
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.

14. 14
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 mvmax
2 = 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
The Photoelectric Effect

15. 15
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
(light)
photoelectron
- - - - - - - - - -
Metal surface
Free electrons
-

16. 16
• The photoelectric effect can be measured using a
device like that pictured in figure below.
9.2 The photoelectric effect
Anode(collector)
Cathode (emitter
or target metal)
photoelectron
glass
-
-
-
rheostat
power supply
e.m. radiation (incoming light)
vacuum A
V
The photoelectric effect’s experiment
A

17. 17
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.

18. 18
• 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.
9.1 The photoelectric effect
Saturation current is defined as the maximum
constant value of photocurrent in which when
all the photoelectrons have reached the
anode.

19. 19
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.
Anode(collector)
Cathode (emitter
or target metal)
photoelectron
glass
-
-
-
rheostat
power supply
e.m. radiation (incoming light)
vacuum A
V
B
Reversing power
supply terminal
(to determine the
stopping potential)

20. 20
• 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 2
s mv
2
1
eV

21. 21
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.
Einstein’s theory of Photoelectric Effect

22. 22
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 ,
Einstein’s theory of Photoelectric Effect
2
max0
2
1
mvWhf
max.EKWE o
2
s mv
2
1
eV
seVWhf 0
but

23. 23
= photon energy
2
maxmax
2
1
. mvEK
Einstein’s theory of Photoelectric Effect
= maximum kinetic energy of
ejected electron.
f = frequency of em radiation /incoming light
vmax = maximum speed of the photoelectron
Einstein’s
photoelectric
equation
c
hhfE
2
max0
2
1
mvWhf
max.EKWE o

24. 24
fo = threshold frequency.
= minimum frequency of e.m. radiation
required to eject an electron from the
surface of the metal.
o
Wo = the work function of a metal.
= the minimum energy required (needed) to
eject an electron from the surface of
target metal.
Einstein’s theory of Photoelectric Effect
= threshold wavelength.
= maximum wavelength of e.m. radiation
required to eject an electron from the
surface of the target metal.
o
o
c
f
max.EKWE o
2
max0
2
1
mvWhf
o
oo
hc
hfW

25. 25
-hf
vmax
-Metal
W0
-
hf
v=0
-Metal
W0
hf
-Metal
W0
oWhf2
max0
2
1
mvWhf
hf < Wo
hf > Wo
Electron is
emitted
Electron is
ejected.
No electron is ejected.
Einstein’s theory of Photoelectric Effect

26. 26
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.
nmb)
Hz1.14x10
a)
15
263o
o
oo
f
hfW
(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)

27. 27
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?
K.Emax = E – Wo
(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)

28. 28
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.
00 hfWa.
(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

29. 29
Solution 24.5
b. 2
0
2
1
mvW
hc
2
2
1
mveVs
c.
(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)

30. 30
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)
Example 24.6

31. 31
Solution 24.6
JWa 19
0 1055.8)
mb 7
0 1033.2)

32. 32
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)

33. 33
Solution 24.7
Ans. : 553 nm, 1.84 x 10-19 J

34. 34
Graphs in Photoelectric Effect
Generally, Einstein’s photoelectric equation;
max.EKWE o
cmxy
WhfEK
WEEK
o
o
max
max
.
.
f ↑ K.Emax ↑
K.Emax
f
of
oW
0

35. 35
Graphs in Photoelectric Effect
f,frequency
sV,voltageStopping
e
W0
0
cmxy
e
W
f
e
h
V
WhfeV
WhfEK
o
s
os
omax.
f ↑ Vs ↑
of

36. 36
Graphs in Photoelectric Effect
Variation of stopping voltage Vs with frequency f of
the radiation for different metals but the intensity
is fixed.
f,frequency
sV,voltageStopping
0
01f
W01
02f
W02
cmxy
e
W
f
e
h
V
WhfeV
WhfEK
o
s
os
omax.
00 fW
W02 > W01
f02 > f01

37. 37
Graphs in Photoelectric Effect
Intensity 2x
Intensity 1x
I,currentricPhotoelect
V,Voltage
0
Variation of photoelectric current I with voltage V for
the radiation of different intensities but its
frequency and metal are fixed.
Vs

38. 38
Notes:
Classical physics
Light intensity ,
areatime
energy
I
Quantum physics
Light intensity ,
areatime
photonsofnumber
I
photonsofnumberintensityLight
Light intensity ↑ ,
number of photons ↑ ,
number of electrons ↑ ,
current ↑ .
(If light intensity ↑, photoelectric current ↑).

39. 39
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.
f2
mI
1sV
I,currentricPhotoelect
V,Voltage0
f1
f2 > f1
2sV
e
W
f
e
h
V
WhfeV
WhfEK
o
s
os
omax.
f ↑ Vs ↑
Vs2 > Vs1

40. 40
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.
W01
1sV
mI
I,currentricPhotoelect
V,Voltage02sV
W02
W02 > W01
e
W
f
e
h
V
WhfeV
WhfEK
o
s
os
omax.
so VW ,
Vs1 > Vs2

41. 41
Example 24.8
K.Emax (x 10-19 J)
f(x 1014 )Hz
8.4
0
Use the graph above to find the value of
i) work function and
ii) the threshold wavelength.

42. 42
Solution 24.8
K.Emax (x 10-19 J)
f(x 1014 )Hz
8.4
0
cmxy
WhfEK
WEEK
o
o
max
max
.
.
m10x6.25
h,wavelengtThreshold
10x3.18
x104.8
whengraphtheFrom
7-
19-
14
o
o
ooo
f
c
J
HzfhfW
EK ,0. max

43. 43
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.

44. 44
EXPLAIN the failure of classical theory to justify the
photoelectric effect.
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.
Very low intensity
but high
frequency
radiation could
emit
photoelectrons.
The maximum
kinetic energy of
photoelectrons is
independent of
light intensity.
Based on Einstein’s
photoelectric equation:
The maximum kinetic
energy of photoelectron
depends only on the
light frequency .
The maximum kinetic
energy of
photoelectrons DOES
NOT depend on light
intensity.
0WhfKmax
1. MAXIMUM KINETIC ENERGY OF PHOTOELECTRON

45. 45
Clasiccal
prediction
Experimental
Result
Modern Theory
Emission of
photoelectro
ns occur for
all
frequencies
of light.
Energy of
light is
independent
of
frequency.
Emission of
photoelectrons
occur only
when
frequency of
the light
exceeds the
certain
frequency
which value is
characteristic
of the material
being
illuminated.
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.
2. EMISSION OF PHOTOELECTRON ( energy )

46. 46
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.
3. EMISSION OF PHOTOELECTRON ( time )

47. 47
Clasiccal
prediction
Experiment
al Result
Modern Theory
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.
4. ENERGY OF LIGHT

48. 48
• 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.

49. 49
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
intensity.
Depends only on the light
frequency .
Emission of
photoelectrons
There should be some
delays to emit
electrons from a metal
surface.
Electrons are emitted
spontaneously.
Energy of light Depends on the light
intensity.
Depends only on the light
frequency .
SUMMARY : Comparison between classical physics
and quantum physics about photoelectric effect experiment

50. 50
Exercise
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.
(1.68 eV, 505 nm, 7.68 x 105 ms-1)
(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)

51. 51
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
stopping potential becomes 0.865 V. Determine the ratio
( 6.35 x 10-15 J s C-1)
.
h
e

52. 52
7. In an experiment on the photoelectric effect, the following
data were collected.
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.
Wavelength of e.m.
radiation, (nm)
Stopping
potential, Vs (V)
350 1.70
450 0.900
(7.73 x 105 m s-1, 6.72 x 10-34 J s)

53. 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)

54. 54
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.
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.
Hz10f 14
02.
)(max eVK
,1083.4 14
0 Hzf ,1078.7 20
max JK 1-5
sm10134v .

55. 55
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.
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.
365 nm
V
G
5
1
)(nAI
)(VV0
(1.60 x 10-19 J, 3.85 x 10-19 J, -1.57 V)