Physics Form 5 KSSM_Neat

Problem Solving Using Einstein's Photoelectric Equation CHAPTER 7


LET’S ANSWER
Example 1 LET’S ANSWER Quantum Physics

A blue light with a frequency of 6.67 × 10 Hz is shone on a clean
14
caesium metal surface. What is the maximum kinetic energy of
KEMENTERIAN PENDIDIKAN MALAYSIA
photoelectrons emitted?
–19
–34
[Work function of caesium = 3.43 × 10 J, Planck’s constant = 6.63 × 10 J s] https://bit.
ly/2YwWUsj
Solution

Step 1: Step 2: Step 3: Step 4:
Identify the Identify the Identify the Solve the problem
problem information given formula to be used numerically


1 Maximum kinetic energy of the 3 hf = W + K
max
photoelectron, K
max
–34
4 (6.63 × 10 )(6.67 × 10 ) = 3.43 × 10 + K
14
–19
14
2 Frequency, f = 6.67 × 10 Hz max
–19
–19
–19
Work function, W = 3.43 × 10 J K max = 4.42 ×10 − 3.43 × 10
–20
Planck's constant, h = 6.63 × 10 J s = 9.92 × 10 J
–34
Example 2


Figure 7.14 shows the change in kinetic energy of K / eV
photoelectrons released from lithium for different light
frequencies. Determine the threshold frequency from 2.0
the graph and calculate the work function of lithium. 1.8
[Planck's constant, h = 6.63 × 10 J s] 1.6
–34
1.4
Solution 1.2
1.0
Threshold frequency, f = 5.6 × 10 Hz 0.8
14
0
Planck’s constant, h = 6.63 × 10 J s 0.6
–34
Work function, W = hf 0 0.4
= (6.63 × 10 ) (5.6 × 10 ) 0.2
14
–34
= 3.71 × 10 J 0
–19
1 2 3 4 5 6 7 8 9 10
f / 10 Hz
14
Figure 7.14

LS 7.3.4 241

Example 3

What is the maximum velocity of the photoelectron emitted when a monochromatic light
(l = 550 nm) is shone on a metal which has a work function of 2.00 eV?
[Given hc = 1.243 × 10 eV nm, 1 eV = 1.60 × 10 J, mass of electron, m = 9.11 × 10 kg]
3
–19
–31
Solution
KEMENTERIAN PENDIDIKAN MALAYSIA
Wavelength, l = 550 nm
3
hc = 1.243 × 10 eV nm
–19
1 eV = 1.60 × 10 J
Mass of electron, m = 9.11 × 10 kg
–31
E = hc Then 1 mv 2 = 2.26 – 2.00
l 2 max = 0.26 eV
3
= 1.243 × 10 eV nm 1
550 nm mv 2 = 0.26 × 1.60 × 10 J
–19
= 2.26 eV 2 max = 4.16 × 10 J
–20
=
v max ! 2 × 4.16 × 10 –20
–31
9.11 × 10
= 3.02 × 10 m s –1
5
Example 4
When a photocell is shone on with a red light (l = 750 nm) and then with a blue light
1
(l = 460 nm), the maximum kinetic energy of the photoelectron emitted by the blue
2
light is two times that of the red light.
(a) What is the work function of the photoelectric Info GALLERY
material in the photocell? The maximum K max
(b) What is the threshold wavelength of the wavelength of light
photoelectric material? needed for a metal
to emit electrons is
Solution known as threshold 0 1
–
wavelength, l for 1 λ
Wavelength of the red light, l = 750 × 10 m the metal. 0 – λ
–9
1
Wavelength of the blue light, l = 460 × 10 m 0
–9
2
Maximum kinetic energy of the photoelectron of the red light, K 1
Maximum kinetic energy of the photoelectron of the blue light, K = 2K 1
2
Work function = W
–34
Planck’s constant, h = 6.63 × 10 J s
Speed of light in vacuum, c = 3.00 × 10 m s –1
8
(a) hf = W+ K (b) hc = W,
hc = W+ K l
l l = 6.63 × 10 × ( 3.00 × 10 –20)
0
8
-34
8
–34
6.63 × 10 × ( 3.00 × 10 –9) = W + K 1 0 = 2.03 × 10 m 9.80 × 10
–6
750 × 10
2.65 × 10 = W + K ------ (1)
–19
1
8
6.63 × 10 × ( 3.00 × 10 –9) = W + K 2
–34
460 × 10
4.32 × 10 = W + 2K ------ (2)
–19
1
(1) × 2 – (2):
2.65 × 10 × 2 – 4.32 × 10 = (2W – W) + (2K – 2K )
-19
–19
1
1
W = 9.80 × 10 J
-20
242 LS 7.3.4

Generating Photoelectric Current in a CHAPTER 7
Photocell Circuit Glass tube Light
Figure 7.15 shows a photocell circuit consisting of a glass Cathode _ + Anode Quantum Physics
vacuum tube. The semi-cylindrical cathode is coated with a Vacuum
light-sensitive metal and connected to the negative potential.
The anode is a metal rod fixed at the axis of the semi-cylindrical
KEMENTERIAN PENDIDIKAN MALAYSIA
cathode and connected to the positive potential. When the
photocell is illuminated by light, the production of photoelectric _ + mA
current is produced in the circuit. Table 7.5 are two examples of
common photocells. Figure 7.15 A photocell circuit

Table 7.5 Production of photoelectric current by photocells coated with caesium and lithium
Caesium Lithium

Caesium Lithium

Light Light
Cathode Cathode


Anode Anode

µA µA

Work function of caesium, W = 2.14 eV Work function of lithium, W = 2.50 eV

14
Threshold frequency, f = 5.16 × 10 Hz Threshold frequency, f = 6.03 × 10 Hz
14
0 0
Maximum wavelength to produce Maximum wavelength to produce
photoelectric current, l = 579 nm photoelectric current, l = 496 nm
On the whole, the higher the work function, the shorter the maximum wavelength required
to produce photoelectric current. As the light intensity increases, the photoelectric current in the
photocell circuit also increases.



Activity 7.9 ISS ICS


Aim: To observe how photoelectric current is produced in a photocells coated with caesium
through computer animations
Instructions:
1. Carry out this activity in pairs.
SCAN ME
2. Scan the QR code to view a simulation related to the SCAN ME
Animation video
production of photoelectric current in photocells coated on the production
with caesium. of photoelectric
3. Based on the simulation, explain how the photoelectric current current
is produced in a photocell coated with caesium. https://bit.ly/3hriJRr
4. Present your findings.


LS 7.3.5 243

Photoelectric Effect Applications
Figure 7.16 shows some examples of applications of photoelectric effect.






KEMENTERIAN PENDIDIKAN MALAYSIA
LED lamps along the road which
are powered by solar cells are
energy efficient and environmentally
friendly. In daylight, the
photoelectric effect of solar cells
enables electrical energy to be stored
in the battery. At night, the LED
lamps will light up with the power
from the battery.








The Noor Complex Solar Power
Plant located in the Sahara Desert
is the world’s largest concentrated
solar power plant. This station is
expected to be completed in 2020 Applications of Photoelectric E ect
and is capable of producing 580
MW capacity for use by 1 million
residents.











Light detectors at the automatic
doors use infrared beam and
photocells as switches. When the
light path is disturbed, photoelectric
current in the photocell circuit will
be disconnected and the door will
remain open.






Figure 7.16 Examples of applications of photoelectric effect

244 LS 7.3.6

The image sensor is a main component in high- CHAPTER 7
resolution cameras. This component is used to
convert light into electrical signals which can be Quantum Physics
processed to form digital images.








The operation of the ISS (International Space Station)
depends on the source of electrical energy generated
from solar panels. The ISS has 16 wings of solar panels
and each wing which measures 35 m × 12 m has
33 thousand solar cells. These panels are capable of
generating 84 – 120 kW of electricity. ISS ICS



Applications of Photoelectric E ect KEMENTERIAN PENDIDIKAN MALAYSIA
Activity 7.10

Aim: To gather information on the applications of photoelectric effect
Instructions:
1. Carry out a Round Table activity.
2. You can obtain information from reading materials or website about other applications of
photoelectric effect.
3. Present your findings in a mind map.



Formative Practice 7.3



1. (a) State Einstein's Photoelectric Equation.
(b) State the meaning of:
(i) work function
(ii) threshold frequency
(iii) the relationship between work function and threshold frequency
2. (a) Sketch a graph to show the relationship between the maximum kinetic energy of
photoelectrons and the frequency of light shone on a metal.
(b) What are the physical quantities represented by the gradient and the intercepts of the
graph sketched in 2(a)?
3. When a metal with a work function of 4.32 × 10 J is shone on by a violet light
–19
(l = 4 × 10 m), what is the maximum kinetic energy of an emitted photoelectron?
–7
–1
8
[Planck’s constant, h = 6.63 × 10 J s, speed of light in vacuum, c = 3.00 × 10 m s ]
–34
LS 7.3.6 245

Games https://bit. ly/2En68AG l applications


Interactive Interactive Einstein's Photoelectric Theory Einstein's Photoelectric Equation, 1 mv 2 hf = W + max 2 Work function, W = hf 0 = hc Photoelectric current production Solar cells • Light detector of • automatic door Image sensor • ISS solar panel •
KEMENTERIAN PENDIDIKAN MALAYSIA




explained by • •









The higher the light photon frequency, The kinetic energy of photoelectrons is independent of the intensity of the light Photon energy, E = hc l Power, P = nhf



Quantum Physics Photoelectric Effect the higher the kinetic energy of the emitted photoelectrons. The minimum frequency to emit a photoelectron is the threshold The emission of photoelectrons is • • Electron microscope













• • frequency, f 0 • • instantaneous

Quantum Theory of Light Discrete energy packet, E = hf de Broglie’s h hypothesis, l = p










waves with particle properties particles with wave properties




oncept Chain Explanation of black body spectrum Max Planck




C

246

Self-Reflection
Self-reflection SCAN ME CHAPTER 7
SCAN ME
Download and print
Self-Reflection
1. New things I have learnt in the chapter on ‘Quantum Quantum Physics
Physics’ are ✎ . https://bit.ly/2QiuNZA

2. The most interesting thing I have learnt in this chapter
KEMENTERIAN PENDIDIKAN MALAYSIA
is ✎ .

3. The things I still do not fully understand are ✎ .

4. My performance in this chapter.
Poor 1 2 3 4 5 Very good

5. I need to ✎ to improve my performance in this chapter.












Summative Practice https://bit.ly/3ht8T1t














1. State the meaning of the following terms:
(a) black body
(b) quantum of energy

2. The minimum energy required for the photoelectron to escape from the sodium metal
surface is 2.28 eV.
(a) Will sodium show photoelectric effect for a red light with a wavelength of 680 nm shone
on it?
(b) What is the threshold wavelength of sodium?










247

3. Wavelength of the yellow line of the sodium spectrum is 590 nm. How much kinetic energy
does one electron have when its de Broglie wavelength is equal to the yellow line of the
sodium spectrum?


4. A laser light beam with a wavelength of 555 nm and a power of 5.00 mW is aimed at an
object without any light reflected. Calculate:
KEMENTERIAN PENDIDIKAN MALAYSIA
(a) the momentum of a photon in the laser beam
(b) the number of photons per second in the laser light beam hitting the object

5. The de Broglie wavelength of an electron is 1.00 nm.
(a) State Louis de Broglie's hypothesis of the wave properties of electrons.
(b) Calculate the momentum of the electron.
(c) Calculate the velocity of the electron.
(d) Calculate the kinetic energy of the electron.


6. (a) Why is a large cavity with a small hole able to act as a black body?
(b) The temperature of a black body is 4 500 K and it looks orange-yellow. Describe
the colour changes in the black body as the body is heated to a temperature of
9 000 K.

7. Photograph 1 shows a communication satellite in outer space. A quantum communication
attempt was performed with a laser pulse of 60 mW and a wavelength of 800 nm.




















Photograph 1
(a) What is the momentum of one photon from the laser pulse?
(b) How much energy does one photon carry?
(c) What is the number of photons per second?
(d) What is the total momentum transferred by the laser pulse per second?







248

8. Complete Table 1 with information on the wavelength and photon energy for several CHAPTER 7
components of waves in the electromagnetic spectrum.
Table 1 Quantum Physics

Wavelength Photon energy Region in the electromagnetic spectrum
500 nm
KEMENTERIAN PENDIDIKAN MALAYSIA
50 eV
5.0 × 10 J
−21

9. Figure 1 shows a photocell constructed using semiconductor material that can be activated
by a light with a maximum wavelength of 1 110 nm.

Glass tube Light
Cathode _ Anode
Semiconductor +
material
Vacuum


_ +
mA


Figure 1
(a) What is the threshold frequency and work function of the semiconductor?
(b) Why does the semiconductor look opaque at room temperature?


10. Muthu conducted an experiment on a grain of sand falling through a small hole. Given
−10
the mass of the grain of sand is 5 × 10 kg, the diameter of the sand is 0.07 mm, the
−1
velocity of the sand falling through the hole is 0.4 m s and the size of the hole is 1 mm:
(a) Estimate the de Broglie wavelength of the sand.

(b) Will the falling sand produce a diffraction pattern when passing through the small
hole? Explain your answer.


11. When a photodiode is shone on with a red light (l = 700 nm) and a blue light
(l = 400 nm), the maximum kinetic energy of the photoelectrons emitted by the
blue light is two times that of the red light.
(a) What is the work function of the photodiode?
(b) What is the threshold wavelength of the photodiode?
(c) What is the de Broglie wavelength of the photoelectron emitted by UV light
(l = 131 nm) from the photodiode?





249

21st Century Challenge


12. Amin conducted an experiment to determine the work function and threshold wavelength
for a material X. The arrangement of the apparatus is as shown in Figure 2.

KEMENTERIAN PENDIDIKAN MALAYSIA
Glass tube Light
Cathode Anode
(material X) Vacuum






+ _ mA
P
Figure 2
When the cathode coated with material X is illuminated by a light beam of wavelength, l,
the emitted photoelectrons will move towards the anode and give a reading in milliammeter.
If the connection to the power supply is reversed, the potential difference at the anode is set
to negative and that will prevent the arrival of the negatively charged photoelectrons. If the
potential divider, P is adjusted until the stopping potential, V results in a zero milliammeter
s
reading, then V is a measure of the maximum kinetic energy, K max of the photoelectrons
s
emitted, of which K max = eV . Table 2 shows the experimental results for the values of l and
s
the corresponding values of V .
s
Table 2
l / nm V / V
s
135 7.53
172 5.59

227 3.98
278 2.92
333 2.06
400 1.43

(a) Based on Einstein's Photoelectric Equation, derive an equation that relates l and V .
s
(b) Plot a suitable graph to determine the Planck’s constant, work function and threshold
wavelength for material X.
(c) Calculate the wavelength of light for the production of a 10.0 eV photoelectron using
the work function in (b).

(d) What is the de Broglie wavelength for the 10.0 eV photoelectron?
(e) Why is material X a critical component in a night vision device?



250

Answers Scan the QR code for
Answers


complete answers

https://bit.ly/34WEgOA
ONLY SELECTED ANSWERS ARE PROVIDED HERE
KEMENTERIAN PENDIDIKAN MALAYSIA
6. (a) Horizontal component = 9.83 N
Vertical component = 6.88 N
(b) The horizontal component moves the knife
Summative Practice
forward.
The vertical component pushes the knife
1. Therefore, worker Y has to apply a force that downward.
makes an angle of 88.58° with the direction of
the force from worker X. 7. T = 4.0 N
S = 6.93 N
2. (a) F = 188 N at an angle of 33° with the
direction of the force applied by P.
(b) – Advantages: The tree will fall in the
direction of the resultant force. A larger
angle will ensure that there is a large space Summative Practice
between P and Q. The tree will fall on to
the ground without endangering P and Q.
2. (a) A and B are at the same level in a stationary
– Disadvantage: The large angle between the liquid.
directions of the forces produces a resultant (b) 972 kg m –3
force with a smaller magnitude.
(c) The direction of the resultant force makes 4. (a) Pressure at point X = atmospheric pressure
a smaller angle with the direction of the Pressure at point Y = 0
force by P. The tree will fall nearer to P. (b) Since point X and point Z are at the same level,
Therefore, P has to be more careful. Pressure at point X = pressure at point Z
Pressure at point X = atmospheric pressure, and
3. 5 493.6 N m –1
Pressure at point Z = pressure due to mercury
column + 0
4. The resultant force of the two forces has the
largest magnitude when the forces act on an Atmospheric pressure = pressure due to
object in the same direction. mercury column
If the force 17 N and the force 13 N are in the Therefore, the height of the mercury column, h
same direction, resultant force = 17 + 13 is a measure of atmospheric pressure.
= 30 N (c) 100 862 Pa
The resultant force of the two forces has the 6. (a) 800 N cm –2
smallest magnitude when the forces are in (b) Pascal’s principle
opposite directions.
(c) Cross-sectional area of slave cylinder
If the force 17 N and the force 13 N are in 2
opposite directions, resultant force = 17 + (–13) = π × 2.5
4
= 4 N = 4.91 cm 2
Therefore, the resultant forces of 17 N and 13 N Braking force = 3 928 N
has magnitude between 4 N and 30 N.
7. Mass of wooden block = 2.98 kg
5. Stage I: Resultant force = 0 N Weight of wooden block = 29.23 N
Stage II: Resultant force = 450 N to the East Buoyant force = 31.78 N
Stage III: Resultant force = 0 N Buoyant force . weight of block
There is a resultant force upwards
The block moves up with an acceleration


251

(c) Figure (a): Direction of motion of magnet to
the left
Figure (b): Direction of motion of magnet to
Summative Practice
the right
1. – The filament lamps require high resistance to (d) Increase the number turns of the solenoid
produce light. Increase the speed of motion of the magnet
– The coiled filament causes the wire length to
increase. 4. I = 7.2 A
KEMENTERIAN PENDIDIKAN MALAYSIA
S
– The resistance is directly proportional to the The loss of energy from the transformer can be
length of the wire. neglected, that is the transformer is ideal.
– The longer the filament wire, the higher the 7. (a) 2.5 A
resistance.
– The higher the resistance, the brighter the lamp. (b) The transformer is ideal
8. 60.00 %
2. (a) (i) 4.5 Ω
– Use laminated soft iron core
(ii) 1.33 A
(iii) 3.99 V – The secondary coil is wound on top of the
primary coil
(b) Bulb X is the brightest compared to bulb Y
and bulb Z. Bulb Y and bulb Z have the same
brightness.
(c) (i) 6 Ω
(ii) 1.0 A Summative Practice
(iii) 3 V
1. (a)
(d) Bulb X and bulb Y glow with equal brightness.
Bulb Z does not light up.
3. (a) The electromotive force, e.m.f. is the energy
transferred or work done by a source of
electrical energy to move one coulomb of
charge in a complete circuit.

(b) The bulb does not light up because the diode is
in reverse biased state.
Summative Practice
2. (a)
2. Fore finger: Direction of magnetic field
Middle finger: Direction of current
Thumb: Direction of force
Force/ Motion
(b)

Magnetic
field


3. (a)
Current

D D
4 1
Key:
Positive cycle
3. (a) Induced current is the current produced in Negative cycle
a conductor when there is relative motion
between the conductor and a magnet that causes D D
the conductor to cut magnetic field lines. 2 3
(b) X: north pole
Y: south pole (c) Half-wave rectification will occur
252

4. (b) Under bright conditions, LDR resistance (b) Heat energy boils the cold water. The high
becomes low. Therefore, the voltage across pressure steam produced is capable of rotating
LDR decreases but the voltage across R is a turbine at extremely high speed.
increased. The I is low and the transistor is (c) The rotation of a turbine switch on a dynamo
B
turned off. The I will be low and the LED will
C which will generate electrical energy by
hot light up. electromagnetic induction.
(c) the LED with an alarm, the resistor with a
KEMENTERIAN PENDIDIKAN MALAYSIA
thermistor and the LDR with a resistor.

Summative Practice

Summative Practice 1. (a) A black body is an ideal body that is able to
absorb all the electromagnetic rays that fall
1. (a) A radioactive decay is a random and on it.
spontaneous process by which an unstable
nucleus will decay by emitting radioactive (b) Quantum of energy is a discrete packet of
energy and not a continuous energy.
radiation to become a more stable nucleus.
(b) The half life, T1, is the time taken for a sample 2. (a) Work function of sodium metal
2
of radioactive nuclei to decay to half of its = 3.65 × 10 J
–19
initial number. Photon energy of the red light
–19
(c) Nuclear energy is the energy produced by = 2.93 × 10 J
reactions in atomic nuclei. Photoelectric effect does not occur because of
the photon energy of the red light is lower than
2. (a) X is the helium nucleus or α-particle, Y is g-ray. work function of sodium metal.
(b) 3 a-particles and 2 b-particles are released. (b) 545 nm
3. (a) 11.2 s 3. 6.93 × 10 J
–25
(b) n = 5
–27
4. (a) 1.19 × 10 kg m s
–1
so after 5T1, only 3.125% of the sample remains.
(b) 1.40 × 10 s
2 16 –1
4. (a) A is the older sample. The ratio of uranium-238
to plumbum-206 is smaller. 5. (a) Louis de Broglie hypothesized that particles
such as electrons could have wave properties.
(b) Suppose that during the rock formation, only
uranium-238 was trapped. The oldest rock de Broglie wavelength, l = h p
e
formed on Earth is about 4.28 billion years. The p is the electron momentum
half-life of uranium-238 is 4.5 billion years. (d) 2.41 × 10 J
–19
Therefore, the decay process of uranium-238
in a rock sample has gone through less than 6. (a) The rays of light that enter the large cavity will
one half-life. Hence, less than half of the undergo repeated reflections on the inner walls
uranium-238 nuclei in the sample of rock had of the cavity. At each reflection, part of the rays
decayed to form lead-206 nuclei. So the number are absorbed by the inner walls of the cavity.
of lead-206 nuclei cannot be more than the Reflections continue to occur until all the rays
remaining uranium-238 nuclei. are absorbed and none of them can leave the
cavity. Thus, the cavity acts like a black body.
6. (a) Nuclear fusion
1
4
3
–28
2 1 H + H ˜ He + n + energy 7. (a) 8.29 × 10 kg m s –1
1
2
0
–19
(b) 2.82 × 10 J (b) 2.49 × 10 J
–12
17
(c) 2.41 × 10 s –1
8. (a) The chain reaction resulting from neutron
bombardment on the uranium-235 nuclei
produces a large amount of nuclear energy in
the reactor.
253

Glossary
Glossary




A black body Half-life
An idealised body that is able to absorb all The time taken for a sample of a radioactive
electromagnetic radiation that falls on it nuclei to decay to half of its initial number
KEMENTERIAN PENDIDIKAN MALAYSIA
An electric field Ideal transformer
The region around a charged particle Transformer that does not experience any loss
where any electric charge in the region will of energy, that is the efficiency, h is 100%
experience an electric force
Nuclear energy
Atmospheric pressure Atomic energy released during nuclear reactions
Pressure due to the weight of the layer of air such as radioactive decay, nuclear fission and
acting on the surface of the earth. nuclear fusion.
Nuclear fission
Buoyant force Nuclear reaction when a heavy nucleus splits
The force acting upwards on an object into two or more lighter nuclei while releasing a
immersed in a liquid when there is pressure large amount of energy
difference between the lower surface and upper
surface of the object Nuclear fusion
Nuclear reaction in which small and light nuclei
Catapult field fuse to form a heavier nucleus while releasing a
Resultant magnetic field produced by the large amount of energy
interaction between the magnetic field from a
current-carrying conductor and the magnetic Resolution of forces
field from a permanent magnet The process of resolving a force into two
components
Elasticity
The property of material that enables an object Resultant force
The single force that represents the vector sum
to return to its original shape and size after the
force applied on it is removed of two or more forces acting on an object
Semiconductor diode
Electromagnetic induction An electronic component which allows electric
Production of an e.m.f. in a conductor when current to flow in one direction only
there is relative motion between the conductor
and a magnetic field or when the conductor is Thermionic emission
in a changing magnetic field The emission of free electrons from a heated
metal surface
Electromotive force (e.m.f.)
The energy transferred or work done by an Threshold frequency
electrical source to move one coulomb of The minimum frequency for a light photon to
charge in a complete circuit produce photoelectric effect
Work function
Forces in equilibrium The minimum energy required for a
When the forces acting on an object to
produce a zero resultant force photoelectron to be emitted from a metal surface



254

References
References




Anderson, M., Berwald, J., Bolzan, J. F., Clark, R., Craig, P., Frost, R., & Zorn, M. (2012).
Integrated iScience Glencoe. United State of America: McGraw-Hill Education.
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Charles, C., Leong, S. C., & Chow, S. F. (2001). Physics A Course for ‘O’ Level (2nd ed.). Times
Media Private Limited.

Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2018). Physics (11th ed.). United State:
John Wiley & Sons, Inc.
Hewitt, P. G. (2015). Conceptual Physics (12th edu.). England: Pearson Education Limited.

Heyworth, R. M. (2010). New Science Discovery Volume 2 (2nd ed.). Singapore: Pearson
Education South Asia Pte Ltd.
Johnson, K. (2016). Physics For You (5th ed.). United Kingdom: Oxford University Press.

Hamper, C. (2009). Higher Level Physics Developed Specifically For The IB Diploma. England:
Pearson Education Limited.
Ho, P. L. (2010). In Science Volume 1. Singapore: Star Publishing Pte Ltd.

Ho, P. L. (2010). In Science Volume 2. Singapore: Star Publishing Pte Ltd.

Honeysett, I., Lees, D., Macdonald, A., & Bibby, S. (2006). OCR Additional Science for GCSE.
United Kingdom: Heinemann.

Lau, L., & Fong, J. (2013). All You Need To Know: Science (Physics) For GCE ‘O’ Level. Singapore:
Alston Publishing House Pte Ltd.
Loo, W. Y., & Loo, K. W. (2013). All About Physics ‘O’ Level: Pearson Education South Asia Pte Ltd.

Pople, S. (2014). Complete Physics for Cambridge IGCSE (3rd edu.). United Kingdom: Oxford
University Press.
Rickard, G., Phillips, G., Johnstone, K., & Roberson, P. (2010). Science Dimensions 2.
Australia: Pearson.

Sang, D. (2014). Cambridge IGCSE Physics Coursebook (2nd edu.). United Kingdom: Cambridge
University Press.

Stannard, P. & Williamson, K. (2006). Science World 7 (3rd ed.). Australia: MacMillan Education
Australia Pte Ltd.
Tho, L. H., Tho, M. Y., & Fong, J. (2008). Interactive Science For Inquiring Minds Lower Secondary
Volume B. Singapore: Panpac Education Private Limited.

Tong, S. S., Ip, H. W., Lam, W. L., & Wong, T. P. (2012). Interactive Science 3B (2nd ed.). Hong
Kong: Pearson Education Asia Limited.


255

Index
Index




Acceleration 9-12, 15-18 Electrical potential energy Nuclear fusion 208-209, 212-213
Aerofoil 82-83 179-180 Nuclear reactor 213-216
Alpha decay 200 Electrical power 122, 124-126 Ohm 100-102, 109, 117, 125
KEMENTERIAN PENDIDIKAN MALAYSIA
Alternating current generator Electromotive force (e.m.f.) Ohm’s law 100, 125
159-160 152-160
Amplifier 189-190 Electrons 174-176, 179-180, 226- Parallel circuit 103, 120
Aneroid barometer 51-52 227, 231-232, 234, 236-237 Pascal’s principle 60, 62-65
Angle of attack 82-83 Energy 99, 114, 117, 122 Photocell 234, 243-244
Archimedes’ principle 66-67, Energy quantum 227-229, 238 Photoelectric 227, 234, 236-238,
71, 74, 77 Equilibrium of force 18-23 240-245
Atmospheric pressure 45-52, Faraday’s law 148, 155 Photoelectron 234, 236-242
54-56, 58 Fleming’s left hand rule 139 Photon 227, 229-232, 237-239
Automatic switch 190-193 Fleming’s right hand rule 157 pnp transistor 187-188
Bernoulli’s principle 78, 80, Fortin barometer 51-52 Potential difference 92, 95,
98-110, 115-120, 124-126

83-84 Forward bias 181-182, 184 Pressure 40-62, 66-67, 73-75,
Beta decay 201 Gamma ray 200-201 77-82
Black body 224-226 Gas pressure 56-59
Brushed motor 146-148 Quantum of light 224, 227, 238
Brushless motor 146-148 Half-life 202-203, 205-206 Radioactive decay 200-201,
Buoyant force 66-71, 74-77 Hooke’s law 27 203-205, 207, 211-212
Hot air balloon 75
Capacitor 182, 185-186 Hydraulic system 61, 62, 64 Resistance 100, 102-113, 116-118
Cartesian diver 73, 74 Hydrometer 71-72 Resolution of force 13-15, 21
Cathode rays 174-179 Resultant force 2-12, 15-18, 23
Charged particle 94-96 Ideal transformer 163-164 Reverse bias 181, 182, 184
Conductor 98, 111-112, Induced current 153-154, Rotation 142-148, 160-161
136-141, 152-156 157-159 Semiconductor diode 181-185
Continuous energy 229, 237 Internal resistance 103, 114, Series circuit 103, 120
Critical temperature 112 116-121 Solenoid 149-151, 153-156,
Current 92, 98-112, 116-126, Lenz’s law 159 158-159
138-148, 151-154, 157-165 Lift force 80-83 Submarine 55, 72-75
Decay series 202, 205 Liquid pressure 40-47, 66 Superconductor 111-112
Direct current generator Magnetic field 136-146, 157-158, Thermionic emission 174-175
159-160 162-165 Threshold frequency 236-237,
Direct current motor 143-146 Magnitude 3-4, 6-16, 18, 142-144, 239-241
Discrete energy 229 155-156, 162-163 Transformer 162-167

Elasticity 24-25 Manometer 56-59 Transistor 187-193
Electric charge 94-96, 98 Mass 7-8, 10-12, 15-17, 20, 26 Velocity 9-11, 19
Electric current 98, 109, Mass defect 210-212
140-141, 151 Mercury manometer 58-59 Water manometer 56-57
Electric field 92-99 npn transistor 187-188 Wave-particle 230, 232
Electric field strength 95 Nuclear energy 208-209, Weight of an object 69, 76
Electrical energy 143, 163, 211-212, 216-217 Work function 239-243
166-167, 181, 213-214, Nuclear fission 208-209, 211,
216-217, 244-245 214-216

256

KEMENTERIAN PENDIDIKAN MALAYSIA























































RM 13.50
ISBN 978-983-092-447-2




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