chapter 9: quantization of light - ysl...
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Chapter 9
1
Chapter 9: Quantization of Light
9.1 Planck’s Quantum Theory
L.O 9.1.1 Distinguish between Planck’s quantum theory and classical theory of
energy
• The foundation of the Planck’s quantum theory is a theory of black body radiation.
• Black body is defined as an ideal system that absorbs the entire radiation incident on it.
The electromagnetic radiation emitted by the black body is called black body radiation.
• From the black body experiment, the distribution of energy in black body depends only
on the temperature. If the temperature increases thus the energy of the black body
increases and vice versa.
• The spectrum of electromagnetic radiation emitted by the black body (experimental result)
is shown in figure
• The Rayleigh-Jeans and Wien’s theories failed to fit the experimental curve because this
two theories based on classical ideas which are:
i. Energy of the EM radiation does not depend on its frequency or wavelength.
ii. Energy of the EM radiation is continuously.
• In 1900, Max Planck proposed his theory that is fit with the experimental curve in figure
above at all wavelengths known as Planck’s quantum theory.
• The assumptions made by Planck in his theory are:
i. The EM. 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 quantized, not all values of
energy are possible
ii. The energy size of the radiation depends on its frequency.
Failed to explain
the shape
Chapter 9
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• According to Planck’s assumptions, the quantum E of the energy for radiation of
frequency f is given by
hfE
• Since the speed of electromagnetic wave in a vacuum is fc , then equation can also be
written as
hcE
L.O 9.1.2 Use Einstein’s equation for a photon energy
• 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”)
• In equation form, photon energy (energy of photon) is hfE .
• Unit of photon energy is Joule (J) or electron-volt (eV).
• The electron-volt (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.
J 1060.1eV 1 19
• Photons travel at the speed of light in a vacuum. They are required to explain the
photoelectric effect and other phenomena that require light to have particle property.
Example
Question Solution
A photon of the green light has a wavelength
of 740 nm. Calculate
a. the photon’s frequency
b. the photon’s energy in joule and
electron-volt
(Given the speed of light in the vacuum,
c = 3.00108 m s1 and Planck’s constant,
h = 6.631034 J s)
Planck’s quantum
theory
Chapter 9
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9.2 The Photoelectric Effect
L.O 9.2.1 Explain the phenomenon of photoelectric effect
The photoelectric effect is defined as the emission of electron from the surface of a metal
when the EM radiation (light) of higher frequency strikes its surface.
Figure below shows the emission of the electron from the surface of the metal after shining by
the light.
Photoelectron is defined as an electron emitted from the surface of the metal when EM
radiation (light) strikes its surface.
L.O 9.2.2 Describe and sketch diagram of the photoelectric effect experiment set-up
• When a monochromatic light of known frequency (or wavelength) shines on the cathode,
photoelectrons are emitted.
• These photoelectrons are attracted to the anode and give rise to a photoelectric current or
photocurrent I which is detected by the galvanometer.
• When the positive voltage (potential difference) is increased, more photoelectrons reach
the anode, hence the photoelectric current also increase.
• As positive voltage becomes sufficiently large, the photoelectric current reaches a
maximum constant value Im, called saturation current (the maximum constant value of
photocurrent when all the photoelectrons have reached the anode).
• If the positive voltage is gradually decreased, the photoelectric current I also decrease
slowly. Even at zero voltage there are still some photoelectrons with sufficient energy
reach the anode and the photoelectric current flows is I0.
Chapter 9
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Reversing power supply terminal
• When the voltage is made negative by reversing the power supply terminal as shown in
figure below, the photoelectric current decreases even further to very low values since
most photoelectrons are repelled by anode which is now negative.
• As the potential of the anode becomes more negative, less photoelectrons reach the anode
thus the photoelectric current drops until its value equals zero (no photoelectrons have
sufficient kinetic energy to reach the collector).
• The electric potential at this moment is called stopping potential (voltage) Vs (the
minimum value of reverse potential (voltage) when there are no photoelectrons
reaching the anode).
• By using conservation of energy :
(loss of KE of photoelectron = gain in PE)
UK max
SeVmv 2
max2
1
• The variation of photoelectric current I as a function of the voltage V can be shown
through the graph in figure below.
At stopping voltage
Chapter 9
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L.O 9.2.3 Define threshold frequency, work function and stopping potential
L.O 9.2.6 Use Einstein’s photoelectron equation
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 (E = hf ) 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.
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).
If hf = Wo, then electron will be ejected but the kinetic energy is equal to zero.
This is summed up by Einstein’s photoelectric equation:
oWKE max
oWmvhf max2
1 or os WeVhf
Einstein’s
photoelectric equation
Chapter 9
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• Work function W0 of a metal
• Is defined as the minimum energy of e.m. radiation required to emit an electron
from the surface of the metal.
• It depends on the metal used.
• Equation :
oo hfEW min
• fo is known as threshold frequency
• Is defined as the minimum frequency of EM radiation required to emit an electron
from the surface of the 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.
• Since fc , then
o
o
cf
• o is known as threshold wavelength
• Is defined as the maximum wavelength of EM radiation required to emit an electron
from the surface of the metal.
• If the wavelength of the incident radiation is greater than the threshold wavelength
( > o) then electrons would not be removed from the metal surface.
Example
Question Solution
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 = 3.00×108 m s-1, h = 6.63×10-34 J s,
1 eV=1.60×10-19 J, me = 9.11×10-31 kg, e =
1.60×10-19 C)
Chapter 9
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Question Solution
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 = 3.00×108 m s-1, h = 6.63×10-34 J s,
1 eV=1.60×10-19 J, me = 9.11×10-31 kg, e =
1.60×10-19 C)
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
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
electron-volts
b. the stopping voltage required if light of
wavelength 400 nm is used
Exercise
Question
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.
Answer: 553 nm, 1.84×10-19 J
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
b. the threshold wavelength of the metal (cathode)
If it is illuminated by ultraviolet radiation of frequency 1.70 x 1015 Hz.
Answer: 8.55×10-19 J, 2.33×10-7 m
Chapter 9
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L.O 9.2.4 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
ii. photoelectric current on intensity of incident light
iii. work function and threshold frequency on the types of metal surface
Generally, Einstein’s photoelectric equation:
oWKE max
oWEK max
cmxy
WfhK o
max
• Variation of stopping voltage Vs with frequency f of the radiation for different metals but
the intensity is fixed:
cmxy
e
Wf
e
hV o
s
seVK max
max , Kf
Vs
f f0
‒Wo
Gradient =
sVf ,
Vs
f
f01 f02 f03
W01 W02 W03
Kmax
f f0 ‒W0
Gradient = h
Since oo hfW , oo fW
12 oo WW
12 oo ff
Different threshold frequency
for different metal
Chapter 9
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• Variation of photoelectric current I with voltage V for the radiation of different intensities
but its frequency and metal are fixed.
Extra Knowledge
Classical physics Quantum physics
AreaTime
EnergyintensityLight
AreaTime
photon ofNumber intensityLight
Since photon ofnumber intensity light
Light intensity ↑, number of photons ↑, number of electrons ↑, current ↑.
→ If light intensity ↑, photoelectric current ↑
• Variation of photoelectric current I with voltage V for the radiation of different
frequencies but its intensity and metal are fixed.
• Variation of photoelectric current I with voltage V for the different metals but the
intensity and frequency of the radiation are fixed.
When intensity is increased, the
maximum current attained is higher showing that more electrons are emitted.
Vs remains the same shows that the Kmax
of photoelectron independent of intensity
of light
photonsofnumber intensity Light
, e
Wf
e
hV o
s fVs
12 ss VV
12 ff
, e
Wf
e
hV o
s o WVs
12 ss VV
0102 WW
Chapter 9
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Example
Question Solution
Use the graph above to find the value of
a. work function and
b. the threshold wavelength
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
Explain why
a. the graphs are parallel
b. the visible light cause photoemission
from caesium but not from zinc
Vs
f 4.8
W02 W03
Chapter 9
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Exercise
Question
In an experiment on the photoelectric effect, the following data were collected.
Wavelength of EM
radiation, (nm)
Stopping potential, VS
(V)
350 1.70
450 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.
Answer: 7.73 x 105 m s-1, 6.72 x 10-34 J s
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.
Answer: 1.60×10-19 J, 3.85×10-19 J, ‒1.57 V
Chapter 9
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L.O 9.2.5 Explain failure of classical theory to justify photoelectric effect
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 depends on the intensity of light.
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/ independent of
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.
• Experimental observations deviate from classical predictions based on Maxwell’s EM
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.