PHYSICS
CHAPTER 9
CHAPTER 9: Quantization of light (4 Hours)
PHYSICS CHAPTER 9 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
9.1 9. 1 Pl Plan anck ck’s ’s qua quant ntum um the theor ory y ( 1 hour) At the end of this cchapter, hapter, students should be able to: this chapter, Explain briefly Planck’s quantum theory and classical theory of energy. Write and use Einstein’s formulae for photon energy,
E = hf =
hc λ
PHYSICS CHAPTER 9 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
9.1 9. 1 Pl Plan anck ck’s ’s qua quant ntum um the theor ory y ( 1 hour) At the end of this cchapter, hapter, students should be able to: this chapter, Explain briefly Planck’s quantum theory and classical theory of energy. Write and use Einstein’s formulae for photon energy,
E = hf =
hc λ
PHYSICS
CHAPTER 9
9.1 Planck’s quantum theory 9.1. 9. 1.1 1 Cl Clas assi sica call theo theory ry of of blac black k body body rad radia iatio tion n
Black body is defined as an ideal system that absorbs all the radiation incident on it. it . The electromagnetic (EM) radiation emitted by the black body is called black body radiation. radiation .
From the black body experiment, the distribution of energy in black body, E depends only on the temperature, T .
E = k BT
(9.1)
where k B : Boltzmann's constant
T : temperatur e in kelvin
If the temperature increases thus the energy of the black body increases and vice versa.
PHYSICS
CHAPTER 9
The spectrum of EM radiation emitted by the black body (experimental result) is shown in Figure 9.1.
Experimental result Rayleigh -Jeans theory Wien’s theory
Classical physics
Figure 9.1 From the curve, Wien’s theory was accurate at short wavelengths but deviated at longer wavelengths whereas the reverse was true for the Rayleigh-Jeans theory.
PHYSICS
CHAPTER 9
The Rayleigh-Jeans and Wien’s theories failed to fit the experimental curve because this two theories based on classical ideas which are
Energy of the EM radiation is not depend on its frequency or wavelength. wavelength Energy of the EM radiation is continuously. continuously
9.1.2 Planck’s quantum theory
In 1900, Max Planck proposed his theory that is fit with the experimental curve in Figure 9.1 at all wavelengths known as Planck’s quantum theory.
The assumptions made by Planck in his theory are :
The EM radiation emitted by the black body is in discrete (separate) packets of energy. energy Each packet is called a quantum of energy. energy This means the energy of EM radiation is quantised. quantised The energy size of the radiation depends on its frequency. frequency
PHYSICS
CHAPTER 9
According to this assumptions, the quantum of the energy E for radiation of frequency f is given by
E = hf
(9.2)
−34 h : Planck' s constant 6 . 63 10 Js = × where
Since the speed of EM radiation in a vacuum is
c = λ f then eq. (9.2) can be written as
E =
hc λ
(9.3)
From eq. (9.3), the quantum of the energy E for radiation is inversely proportional to its wavelength. wavelength
PHYSICS
CHAPTER 9
It is convenient to express many quantum energies in electronvolts.
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. volt Unit conversion:
1 eV
=
1.60 ×10
19
−
J
In 1905, Albert Einstein extended Planck’s idea by proposing that electromagnetic radiation is also quantised. It consists of particle like packets (bundles) of energy called photons of electromagnetic radiation. Note:
For EM radiation of n packets, the energy E n is given by
E n
=
nhf
(9.4)
where n : real number = 1,2,3,... 7
PHYSICS
CHAPTER 9
9.1.3 Photon
Photon is defined as a particle with zero mass consisting of a quantum of electromagnetic radiation where its energy is concentrated. concentrated A photon may also be regarded as a unit of energy equal to hf . 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. property Table 9.1 shows the differences between the photon and electromagnetic wave.
PHYSICS
CHAPTER 9 EM Wave
Energy of the EM wave depends on the intensity of the wave. Intensity of the wave I is proportional to the squared of its amplitude A2 where
I ∝ A
Photon
Energy of a photon is proportional to the frequency of the EM wave where
E ∝ f
2
Its energy is continuously and spread out through the medium as shown in Figure 9.2a.
Its energy is discrete as shown in Figure 9.2b.
Photon Figure 9.2b
Figure 9.2a Table 9.1
PHYSICS
CHAPTER 9
Example 1 : 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 =3.00× 108 m s−1 and Planck’s constant, h =6.63× 10−34 J s) −9 Solution : λ 740 10 m
=
×
a. The frequency of the photon is given by
c = λ f
3.00 × 108 = ( 740 × 10 −9 ) f f = 4.05 × 1014 Hz
b. By applying the Planck’s quantum theory, thus the photon’s energy in joule is
E = hf
E = ( 6.63 × 10 −34 )( 4.05 × 1014 ) E = 2.69 × 10 −19 J
and its energy in electron-volt is
E =
2.69 × 10 −19
1.60 × 10
−19
E = 1.66 eV
PHYSICS
CHAPTER 9
Example 2 : For a gamma radiation of wavelength 4.62 × 10−12 m propagates in the air, calculate the energy of a photon for gamma radiation in electron-volt. (Given the speed of light in the vacuum, c =3.00× 108 m s−1 and Planck’s constant, h =6.63× 10−34 J s) −12 Solution : λ 4.62 10 m
=
×
By applying the Planck’s quantum theory, thus the energy of a photon in electron-volt is
E =
hc λ
( 6.63 × 10 −34 )( 3.00 × 108 ) E = 4.62 × 10 −12 E = 4.31 × 10 −14 J
=
4.31 × 10 −14 1.60 × 10 −19
E = 2.69 × 10 5 eV
PHYSICS CHAPTER 9 Learning Outcome: s c i s y h p / y m .. u d e . ki r t a m . h p m .k w w w
9.2 The photoelectric effect (3 hours) At the end of this this chapter, chapter, students should be able to: Explain the phenomenon of photoelectric effect.
Define threshold frequency, work function and stopping potential. Describe and sketch diagram of the photoelectric effect experimental set-up. Explain by using graph and equations the observations of photoelectric effect experiment in terms of the dependence of : kinetic energy of photoelectron on the frequency of light;
1
2
mvmax
2
=
eV s
=
hf − hf 0
PHYSICS CHAPTER 9 Learning Outcome: s c i s y h p / y m .. u d e . ki r t a m . h p m .k w w w
9.2 The photoelectric effect (3 hours) At the end of this this chapter, chapter, students should be able to: photoelectric current on intensity of incident light;
work function and threshold frequency on the types of metal surface.
W 0
=
hf 0
Explain the failure of wave theory to justify the photoelectric effect.
PHYSICS
CHAPTER 9
9.2 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. surface Figure 9.3 shows the emission of the electron from the surface of the metal after shining by the light.
EM radiation
- - - - -
-
photoelectron
- - - - -
Metal Free electrons
Figure 9.3 Photoelectron is defined as an electron emitted from the surface of the metal when the EM radiation (light) strikes its
PHYSICS
CHAPTER 9
9.2.1 Photoelectric experiment
The photoelectric effect can be studied through the experiment made by Franck Hertz in 1887.
Figure 9.4a shows a schematic diagram of an experimental arrangement for studying the photoelectric effect. EM radiation (light) cathode anode
photoelectron
-
vacuum
glass
G
V power supply
Figure 9.4a
rheostat
PHYSICS
CHAPTER 9
The set-up apparatus as follows:
Two conducting electrodes, the anode (positive electric potential) and the cathode (negative electric potential) are encased in an evacuated tube (vacuum).
The monochromatic light of known frequency and intensity is incident on the cathode.
Explanation of the experiment When a monochromatic light of suitable frequency (or wavelength) shines on the cathode, photoelectrons are emitted.
These photoelectrons are attracted to the anode and give rise to the photoelectric current or photocurrent I which is measured by the galvanometer.
When the positive voltage (potential difference) across the cathode and anode is increased, more photoelectrons reach the anode , thus the photoelectric current increases.
PHYSICS
CHAPTER 9
As positive voltage becomes sufficiently large, the photoelectric current reaches a maximum constant value I m, called saturation current. current Saturation current is defined as the maximum constant value of photocurrent when all the photoelectrons have reached the anode. anode 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 anode and the photoelectric current flows is I 0. Finally, when the voltage is made negative by reversing the power supply terminal as shown in Figure 9.4b, the photoelectric current decreases even further to very low values since most photoelectrons are repelled by anode which is now negative electric potential.
PHYSICS
CHAPTER 9 EM radiation (light) cathode
anode
photoelectron
-
vacuum
glass
G
V power supply
rheostat
Figure 9.4b: reversing power supply terminal As the potential of the anode becomes more negative, less photoelectrons reach the anode thus the photoelectric current drops until its value equals zero which the electric potential at this moment is called stopping potential (voltage) V s.
Stopping potential is defined as the minimum value of negative voltage when there are no photoelectrons reaching the anode
PHYSICS
CHAPTER 9
The potential energy U due to this retarding voltage V s now equals the maximum kinetic energy K max of the photoelectron.
U = K max
eV s
=
1
mv max
2
(9.5)
2 where m : mass of the electron The variation of photoelectric current I as a function of the voltage V can be shown through the graph in Figure 9.4c. Photoelectric current, I
I m I 0
Stimulation 9.1 Figure 9.4c
−
V s After
0
Voltage,V Before reversing the terminal
PHYSICS
CHAPTER 9
9.2.2 Einstein’s theory of photoelectric effect
A photon is a ‘packet’ of electromagnetic radiation with particle-like characteristic and carries the energy E given by
E = hf and this energy is not spread out through the medium. medium Work function W 0 of a metal
Is defined as the minimum energy of EM radiation required to emit an electron from the surface of the metal. metal It depends on the metal used. used Its formulae is W 0 = E min and E min = hf 0
W 0
=
hf 0
(9.6)
where f 0 is called threshold frequency and is defined as the minimum frequency of EM radiation required to emit an electron from the surface of the metal. metal
PHYSICS
CHAPTER 9
Since c=λ f then the eq. (9.6) can be written as
W 0
=
hc
(9.7)
λ 0
where λ 0 is called threshold wavelength and is defined as the maximum wavelength of EM radiation required to emit an electron from the surface of the metal. metal Table 9.2 shows the work functions of several elements. Element
Work function (eV)
Aluminum
4.3
Sodium
2.3
Copper
4.7
Gold
5.1
Silver
4.3 Table 9.2
PHYSICS
CHAPTER 9
Einstein’s photoelectric equation In the photoelectric effect, Einstein summarizes that some of the energy E imparted by a photon is actually used to release an electron from the surface of a metal (i.e. to overcome the binding force) and that the rest appears as the maximum kinetic energy of the emitted electron ( photoelectron). photoelectron It is given by 1 E = K max + W 0 where E = hf and K max = mv max 2
hf =
1 2
2
mv max
2
+
W 0
(9.8)
where eq. (9.8) is known as Einstein’s photoelectric equation.
Since K max=eV s then the eq. (9.8) can be written as
hf = eV s
+
W 0
where V s : stopping voltage
(9.9)
e : magnitude for charge of electron
PHYSICS
CHAPTER 9 Note:
1st case: hf > W 0 OR f > f 0
hf Figure 9.5a
Metal
-
W 0
K max Electron is emitted with maximum kinetic energy. energy
2nd case: hf = W 0 OR f = f 0
hf Figure 9.5b
-
vmax
3rd case:
Metal
K max = 0 - v =0 W 0 Electron is emitted but maximum kinetic energy is zero. zero -
hf < W 0 OR f < f 0
hf
No electron is emitted. Figure 9.5c
Metal
-
W 0
PHYSICS
CHAPTER 9
Example 3 : Cadmium has a work function of 4.22 eV. Calculate a. its threshold frequency, b. the maximum speed of the photoelectrons when the cadmium is shined by UV radiation of wavelength 275 nm, c. the stopping potential. (Given c =3.00× 108 m s−1, h =6.63× 10−34 J s, me=9.11× 10−31 kg and
e=1.60× 10− C) −19 W 4 . 22 1 . 60 10 = × Solution : 0 19
(
) = 6.75 × 10 −19 J
a. By using the equation of the work function, thus
= hf 0 6.75 × 10 −19 = ( 6.63 × 10 −34 ) f 0 f 0 = 1.02 × 1015 Hz W 0
PHYSICS
CHAPTER 9
(
Solution : W 0 4.22 1.60 10 −19 −9 b. Given λ 275 10 m
=
=
×
×
) = 6.75 × 10 −19 J
By applying the Einstein’s photoelectric equation, thus
E = K max
hc λ
=
1 2
+ W 0
mvmax
2
+ W 0
( 6.63 × 10 −34 )(3.00 × 108 ) = 1 (9.11 × 10 −31 )v 275 × 10
−9
vmax
2 = 3.26 × 10 5 m s −1
c. The stopping potential is given by
K max
=
1
2
mvmax
2
eV s
=
1 2 1
2 max
+ 6.75 × 10 −19
mvmax
2
(1.60 × 10 )V s = (9.11 × 10 )(3.26 × 10 ) −19
2 V = 0 303 V
− 31
5 2
PHYSICS
CHAPTER 9
Example 4 : A beam of white light containing frequencies between 4.00 × 1014 Hz and 7.90 × 1014 Hz is incident on a sodium surface, which has a work function of 2.28 eV. a. Calculate the threshold frequency of the sodium surface. b. What is the range of frequencies in this beam of light for which electrons are ejected from the sodium surface? c. Determine the highest maximum kinetic energy of the photoelectrons that are ejected from this surface. (Given c =3.00× 108 m s−1, h =6.63× 10−34 J s, me=9.11× 10−31 kg and
e=1.60× 10− C) 19
PHYSICS
CHAPTER 9
Solution : W 0
= 2.28(1.60 × 10 −19 ) = 3.65 × 10 −19 J
a. The threshold frequency is
= hf 0 3.65 × 10 −19 = ( 6.63 × 10 −34 ) f 0 f 0 = 5.51 × 1014 Hz W 0
b. The range of the frequencies that eject electrons is 5.51
1014 Hz and 7.90
c. For the highest K max, take
1014 Hz
f = 7.90 × 1014 Hz
By applying the Einstein’s photoelectric equation, thus
E = K max + W 0 1 2 hf = mvmax + W 0 2
(6.63 × 10 −34 )(7.90 × 1014 ) = K max + 3.65 × 10 −19 K
= 1 59 × 10 −19 J
PHYSICS
CHAPTER 9
Exercise 9.1 : Given c =3.00× 10 m s− , h =6.63× 10− J s, me=9.11× 10− kg and 8
1
34
31
e=1.60× 10− C 19
1.
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, electrons are emitted with a maximum kinetic energy of 1.10 eV. Calculate the work function of this metal in joules. ANS. : 553 nm; 1.84 10 19 J 2. In a photoelectric effect experiment it is observed that no current flows when the wavelength of EM radiation is greater 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. (Physics for scientists & engineers, 3 rd edition, Giancoli, Q15, p.974)
PHYSICS
CHAPTER 9
Exercise 9.1 : 3.
In an experiment on the photoelectric effect, the following data were collected. Wavelength of EM radiation, λ (nm)
Stopping potential, V (V) s
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. ANS. : 7.73 105 m s 1; 6.72 10 34 J s
PHYSICS
CHAPTER 9
9.2.3 Graph of photoelectric experiment Variation of photoelectric current I with voltage V
for the radiation of different intensities but its frequency is fixed. fixed I
−
V s
2 I m
Intensity 2x
I m
Intensity 1x
0
V
Figure 9.6a Reason: From the experiment, the photoelectric current is directly proportional to the intensity of the radiation as shown in Figure 9.6b.
PHYSICS
CHAPTER 9 I
2 I m I m 0
1×
2×
Light intensity
Figure 9.6b for the radiation of different frequencies but its intensity is I fixed. fixed
I m
f 2 > f 1
f 2 f 1
− V s2 − V s1 0
V
PHYSICS
CHAPTER 9
Reason: From the Einstein’s photoelectric equation,
Stopping voltage,V s
hf = eV s
+ W 0
V s2
W 0 h V s = f − e e
y
m x
=
+
c
V s1
0 W 0
−
f 0 f 1 f 2
frequency, f If V s=0 ,
e Figure 9.7b
hf = e(0) + W 0 W 0 = hf f 0
PHYSICS
CHAPTER 9
For the different metals of cathode but the intensity and frequency of the radiation are fixed. fixed
I
I
m
W 02 > W 01 W 01 Figure 9.8a
W 02
V
− V s1 − V s2 0 Reason: From the Einstein’s photoelectric equation,
V s hf e V s1
hf = eV s
+ W 0
y
V s2
0
1 hf V s = − W 0 + e e
W 01 W 02
hf E =
m
=
W 0
x
c
+
Energy of a photon in EM radiation
PHYSICS
CHAPTER 9
Variation of stopping voltage V s with frequency f of the radiation
for different metals of cathode but the intensity is fixed. fixed
V s
W 01 W 02
W 03
W 03 >W 02 > W 01 Figure 9.9
0
f 01 f 02 f 03
Reason: Since W 0=hf 0 then
hf = eV s
+ W 0
f
W 0
∝
h W 0 V s = f − e e
y
m x
=
+
c
Threshold (cut-off) frequency hf = e(0) + W 0 If V s=0 ,
f 0
W 0
= hf
f 0
PHYSICS
CHAPTER 9
9.2.4 Failure of wave theory of light
Table 9.3 shows the classical predictions (wave theory), photoelectric experimental observation and modern theory explanation about photoelectric experiment.
Classical predictions
Emission of photoelectrons occur for all frequencies of light. Energy of light is independent of frequency.
Experimental observation
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
PHYSICS
CHAPTER 9
Classical predictions
Experimental observation
Modern theory
The higher the intensity, the greater the energy imparted to the metal surface for emission of photoelectrons. When the intensity is low, the energy of the radiation is too small for emission of electrons.
Very low intensity but high frequency radiation could emit photoelectrons. The maximum kinetic energy of photoelectrons is independent of light intensity.
The intensity of light is the number of photons radiated per unit time on a unit surface area. area Based on the Einstein’s photoelectric equation:
K max
= hf − W 0
The maximum kinetic energy of photoelectron depends only on the light frequency and the work function. function If the light intensity is doubled, the number of electrons emitted also doubled but the maximum kinetic energy remains unchanged.
PHYSICS Classical predictions
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.
CHAPTER 9 Experimental observation
Photoelectrons are emitted from the surface of the metal almost instantaneously after the surface is illuminated, even at very low light intensities.
Modern theory
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.
PHYSICS Classical predictions
Energy of light depends only on amplitude ( or intensity) intensity and not on frequency. Note:
CHAPTER 9 Experimental observation
Energy of light depends on frequency.
Modern theory
According to Planck’s quantum theory which is
E=hf Energy of light depends on its frequency.
Table 9.3
Experimental observations deviate from classical predictions based on wave theory of light. light 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. photons
PHYSICS
CHAPTER 9
Example 5 : a. Why does the existence of a threshold frequency in the photoelectric effect favor a particle theory for light over a wave theory? b. In the photoelectric effect, explains why the stopping potential depends on the frequency of light but not on the intensity. Solution : a. Wave theory predicts that the photoelectric effect should occur at any frequency, provided the light intensity is high enough. However, as seen in the photoelectric experiments, the light must have a sufficiently high frequency (greater than the threshold frequency) for the effect to occur. b. The stopping voltage measures the kinetic energy of the most energetic photoelectrons. Each of them has gotten its energy from a single photon. According to Planck’s quantum theory the photon energy depends on the frequency of the light. The intensity controls only the number of photons reaching a unit area in a unit time. ,
PHYSICS
CHAPTER 9
Example 6 :
In a photoelectric experiments, a graph of the light frequency f is plotted against the maximum kinetic energy K of the photoelectron as shown in Figure 9.10. m ax
14
f × 10 Hz
4.83
K max (eV)
0
Figure 9.10 Based on the graph, for the light of frequency 7.14× 1014 Hz, calculate a. the threshold wavelength, b. the maximum speed of the photoelectron. (Given c =3.00× 108 m s 1, h =6.63× 10 34 J s, m =9.11× 10 31 kg and −
−
−
e
e=1.60× 10 19 C) −
PHYSICS
CHAPTER 9
Solution : f
= 7.14 × 1014 Hz
a. By rearranging Einstein’s photoelectric equation, 14
f × 10 Hz
hf = K max
+ W 0
4.83
K max (eV)
0 From the graph, f 0
W 0 1 f = K max + h h 1 f = K max + f 0 h
y
m x
=
= 4.83 × 1014 Hz
Therefore the threshold wavelength is given by λ 0
= =
λ
c
f 0 3.00 × 108 4.83 × 1014 6 21 × 10 −7
+
c
PHYSICS
CHAPTER 9
Solution : f
= 7.14 × 1014 Hz
b. By using the Einstein’s photoelectric equation, thus
hf = hf = 1 1 2
2
mvmax
2
1
2 1 2
2
+ W 0
2
+ hf 0
mvmax mvmax
= h( f − f 0 )
(9.11 × 10 −31 )vmax 2 = 6.63 × 10 −34 (7.14 × 1014 − 4.83 × 1014 ) vmax
= 5.80 × 10 5 m s −1
PHYSICS
CHAPTER 9
Exercise 9.2 : Given c =3.00× 108 m s−1, h =6.63× 10−34 J s, me=9.11× 10−31 kg and
e=1.60× 10− C 19
1.
A photocell with cathode and anode made of the same metal connected in a circuit as shown in the Figure 9.11a. 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 Figure 9.11b.
I ( nA )
365 nm
5 G V
−1 Figure 9.11a
0
V ( V ) Figure 9.11b
PHYSICS
CHAPTER 9
Exercise 9.2 : 1.
a. Calculate the maximum kinetic energy of 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.
ANS. : 1.60 10 19 J, 3.85 10 19 J; 1.57 V 2. When EM radiation falls on a metal surface, electrons may be emitted. This is photoelectric effect. a. Write Einstein’s photoelectric equation, explaining the meaning of each term. b. Explain why for a particular metal, electrons are emitted only when the frequency of the incident radiation is greater than a certain value? c. Explain why the maximum speed of the emitted electrons is independent of the intensity of the incident radiation? (Advanced Level Physics, 7th edition, Nelkon&Parker, Q6, p.835)
