Physics_Anyflip Sem 2

PENERBIT
ILMU
BAKTI SDN. BHD.

Contents

PENERBITChapter 1 Electrostatics 1
ILMU 7
BAKTI SDN. BHD.1.1 Coulomb’s Law 14
1.2 Electric Field 20
1.3 Electric Potential 28
1.4 Charge in a Uniform Electric Field
Summative Practice 1

Chapter 2 Capacitors and Dielectrics 30
39
2.1 Capacitance and Capacitors in Series and Parallel 43
2.2 Charging and Discharging of Capacitors 50
2.3 Capacitors with Dielectrics
Summative Practice 2

Chapter 3 Electric Current and Direct-Current Circuits

3.1 Electrical Current 52

3.2 Ohm’s Law and Resistivity 55

3.3 Variation of Resistance with Temperature 58

3.4 Electromotive Force (E.M.F), Internal Resistance and Potential Difference 61

3.5 Resistors in Series and Parallel 64

3.6 Kirchhoff’s Rules 69

3.7 Electrical Energy and Power 79

3.8 Potential Divider 82

3.9 Potentiometer 86

Summative Practice 3 93

Chapter 4 Magnetism 95
102
4.1 Magnetic Field 111
4.2 Resultant Magnetic Field Produced By Current-Carrying Conductor 116
4.3 Force on a Moving Charged Particle in a Uniform Magnetic Field 120
4.4 Force on a Current-Carrying Conductor in a Uniform Magnetic Field 122
4.5 Forces between Two Parallel Current-Carrying Conductors 125
4.6 Application of Motion of Charged Particle
Summative Practice 4

Chapter 5 Electromagnetic Induction

5.1 Magnetic Flux 127

5.2 Induced E.M.F. 130

5.3 Self-Inductance 137
PENERBIT
ILMU5.4 Energy Stored in Inductor 140
BAKTI SDN. BHD.
5.5 Mutual Inductance 143

Summative Practice 5 146

Chapter 6 Alternating Current 149
152
6.1 Alternating Current 155
6.2 Root Mean Square (R.M.S.) 165
6.3 Resistance, Reactance and Impedance 170
6.4 Power and Power Factor
Summative Practice 6

Chapter 7 Optics 173
180
7.1 Reflection at a Spherical Surface 183
7.2 Refraction at a Spherical Surface 192
7.3 Thin Lenses 194
7.4 Huygens’s Principle 197
7.5 Constructive and Destructive Interferences 203
7.6 Interference of Transmitted Light through Double-Slits 211
7.7 Interference of Reflected Light in Thin Films 216
7.8 Diffraction by a Single-Slit 220
7.9 Diffraction Grating
Summative Practice 7

Chapter 8 Wave Properties of Particle 223
228
8.1 de Broglie Wavelength 234
8.2 Electron Diffraction
Summative Practice 8

Chapter 9 Nuclear and Particle Physics

9.1 Binding Energy and Mass Defect 237

9.2 Radioactivity 241

9.3 Particle Accelerator 246

9.4 Fundamental Particle 252

Summative Practice 9 258

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• Summative Assessment
Test (UPS) 1, 2, & 3

• Semester Examination
• Complete Answers

PENERBIT1CHAPTER
ILMU
BAKTI SDN. BHD.Electrostatics

1.1 Coulomb’s Law

 Learning Outcomes

You should be able to: Qq kQq
4πεor2 = r2
● State Coulomb’s Law, F =

● Sketch the electric force diagram.

● Apply Coulomb’s Law for a system of point charges.
*Simple configuration of charges with a maximum four charges in 2D

Electrostatic Force between Charges

 1 This chapter delves into the study of electric charges at rest (point charge), the
electric force law that describes the interactions between electric charges, and
the effects caused by the distribution of charges in a region or space.

 2 There are two types of charge in nature which are positive charge and
negative charge.

F F

+ +
Qq
Like charges repel

+ F F −

Q q

Unlike charges attract

Figure 1.1  Attraction and repulsion of charges

(a) Like charges (positive-positive or negative-negative) repel each other.
(b) Unlike charges (positive-negative or negative-positive) attract each other.
  3 Coulomb’s Law states that the magnitude of electrostatic force, F that exists
between two point charges is directly proportional to the product of the

1

 Electrostatics

Smart Tips! magnitudes of the point charges and inversely proportional to the square of
the distance between the point charges. In equation form,

F = k Qq F = k Qq
r2 r2

Notes: Qq can be QQ or where BAKTI SDN. BHD.
qq or qQ

F = electrostatic force

Q, q = magnitudes of point charges

r = distance between point charges

k = electric force constant, electrostatic constant, or Coulomb’s constant

= 9 × 109 N m2 C‒2

Chapter 1 *k = 1 , where εO (permittivity of free space) = 8.85 × 10‒12 F m‒1.
4πεO
 4 Electrostatic force, F is also known as electric force or Coulomb force.

Graph of electrostatic force against the distance between two

charges

 1 Figure 1.2(a) shows the graph of electrostatic force, F against the distance, r

between two charges. Figure 1.2(b) shows the graph of F against 1 .
r2
Mid Semester Test 1
FF

Gradient,
m = kQq

1
O r O r2

(a) (b)
Figure 1.2  Graphs for electrostatic force
ILMU
Example 1

AnswersPENERBIT Smart Tips! State Coulomb’s Law.

• State the relationship Solution
between electrostatic
force and two charges. Coulomb’s Law states that the magnitude of electrostatic force that exists

• State the relationship between two charges is directly proportional to the product of the magnitudes
between electrostatic
force and the distance of the two charges and inversely proportional to the square of the distance
between two charges.
between them.
• W rite the equation for
Coulomb force. Electrostatic force, F = k Qq
r2

2

Example 2 Electrostatics 

Sketch the electrostatic force exerted on Smart Tips!
(a) charge Q, and
(b) charge q • The word ‘on’ means
as shown below. target charge will move
away or towards the
+ other charge.

Q • Like charges repel.
PENERBIT + • T he force must be
ILMU
BAKTI SDN. BHD. q acting on the target
charge.
Solution • T he subscripts for
electrostatic force must
(a) +FQq Q + in the standard format: Chapter 1
‘on’, then ‘by’ (F ‘on’ Q
q ‘by’ q is FQq).

Both are like charges and thus repel. So, the electrostatic force on Q is to Common Error

the left, or away from q. • T he arrow was drawn
on the incorrect charge.
(b) + +Q q FqQ
• T he subscript for F in
the wrong order.
Electrostatic force on q is to the right, or away from Q.
Smart Tips!
Example 3 Mid Semester Test 1
Unlike charges attract.
Sketch the electrostatic force exerted on −
(a) charge Q, and 3
(b) charge q. q

+

Q

Solution

(a) + −

Q FQq q

Both are unlike charges and thus attract. So, the electrostatic force on Q is Answers

to the right, or towards q.

(b) + −Q FqQ q


Electrostatic force on q is to the left, or towards Q.

Example 4

Two charges, Q and q are 30 μC and 50 μC respectively.

+−

Qq

If charges Q and q are 2 cm apart, determine the magnitude of the electrostatic
force exerted on charge Q and its direction.

 Electrostatics

Smart Tips! Solution

Use Coulomb’s Law FQq = k Qq
equation to calculate r2
the magnitude of force
(ignore sign of charge). FQq = (9 × 109)  (30 × 10−6)(50 × 10−6)
(2 × 10−2)2
Common Error = 3.38 × 104 N BAKTI SDN. BHD.

Students consider sign of Unlike charges attract. So, the direction of the electrostatic force exerted on Q
charges in calculation.
is towards q.

Chapter 1 + −

Q FQq q

Mid Semester Test 1 Example 5

Three point charges, Q1 = 9 μC, Q2 = 1 μC and Q3 = 2 μC are separated by the
distances shown in the figure below.

− Q2

3 cm

+Q1 4 cm + Q3

Determine the electrostatic force exerted on Q3. F32
ILMU
Smart Tips! Solution

• Sketch free body F32 = k Q3Q2
(r32)2

diagram showing all the 109) (2 × 10−6)(1 × 10−6)
ownhQile3.Q2 F32 = (9 × (3 × 10−2)2
forces acting
• a Qt1trraecptselQs Q3. 3, = 20 N

AnswersPENERBIT F31 = k Q3Q1 F
(r31)2 31

F31 = (9 × 109) (2 × 10−6)(9 × 10−6)
(4 × 10−2)2

= 101.25 N

Resolving F32 into x-and y-components,
F32x = 0 N, F32y= 20 N

Resolving F31 into x-and y-components,
F31x = 101.25 N, F31y = 0 N

Summing all the forces in
the x-direction,
F3x = F32x + F31x
F3x = 0 + 101.25
= +101.25 N

4

and y-direction, Electrostatics 
F3y = F32y + F31y
F3y = 20 + 0 Common Error
F3y = +20 N
Students do not resolve
Magnitude of resultant force, the forces into x-and
PENERBIT y-components.
ILMUF3 = (F3x)2 + (F3y)2
BAKTI SDN. BHD. 5
F3 = (+101.25)2 + (+20)2
F3 = 103.21 N

Direction of resultant force, ∑F3y ∑F3
θ ∑F3x
  tan θ = F3y Chapter 1
F3x

 +20

tan θ = +101.25

θ = 11.17° above the positive x-axis

Therefore, the net electrostatic force exerted on Q3 is 103.21 N at an angle of
11.17° above the positive x-axis.

Example 6 Mid Semester Test 1

Determine the electric force exerted on a proton, Q1 by two electrons, Q2 and
Q3 as shown in the figure below.

− Q2

2 cm 2 cm

+Q1 2 cm − Q3

[Charge of proton = +1.60 × 10−19 C, charge of electron = −1.60 × 10−19 C]

Solution Answers

− Q2

2 cm

θ

+Q1 1 cm

Determining angle θ, F12
θ
cos θ = 1
2 F13

θ = 60°

F12 = k Q1Q2
(r12)2

F12 = (9 × 109)  (1.60 × 10−19)(1.60 × 10−19) = 5.76 × 10−25 N
(2 × 10−2)2

 Electrostatics F13 = k Q1Q3
(r13)2
6
F13 = (9 × 109)  (1.60 × 10−19)(1.60 × 10−19)
(2 × 10−2)2
F13 = 5.76 × 10−25 N BAKTI SDN. BHD.

Chapter 1 Resolving F12 into x-and y-components,
F12x = +(5.76 × 10−25) cos 60°
F12x = +2.88 × 10−25 N
F12y = +(5.76 × 10−25) sin 60°
F12y = +4.99 × 10−25 N

Resolving F13 into x-and y-components,
F13x = +5.76 × 10−25 N
F13y = 0 N

Summing all the forces in
the x-direction,

Mid Semester Test 1 F1x = F12x + F13x
F1x = (2.88 × 10−25) + (5.76 × 10−25)
F1x = +8.64 × 10−25 N

and y-direction,
F1y = F12y + F13y
F1y = (4.99 × 10−25) + 0
F1y = +4.99 × 10−25 N

Magnitude of resultant force,
ILMU
F1 = (F1x)2 + (F1y)2
F1 = (+8.64 × 10−25)2 + (+4.99 × 10−25)2
F1 = 9.98 × 10−25 N

Direction of resultant force,

∑F1y

AnswersPENERBIT ∑F
1

θ ∑F1x

  tan θ = F1y
F1x

 +4.99 × 10−25

tan θ = +8.64 × 10−25

θ = 30° above the positive x-axis

Therefore, the net electrostatic force exerted on Q1 is 9.98 × 10−25 N at an angle
of 30° above the positive x-axis.

Quick Check 1.1 Electrostatics 

 1 Two point charges, Q1 = 85 μC and Q2 = ‒50 μC are separated by a distance of 7
3.5 cm. Determine the magnitude and direction of the electric force that
(a) Q2 exerts on Q1,
(b) Q1 exerts on Q2.

[Coulomb’s constant, k = 9.0 × 109 N m2 C−2]

 2 Two identical point charges which are 30 mm apart experience a repulsive
force of 980 N. What is the magnitude of each charge?

 3 If the repulsive force between two identical point charges of – 4 µC is 200 N,
what is the separation between them?

 4 Calculate the magnitude of the electrostatic force exerted on a point charge
of 10 μC by another point charge of 5 μC that is located 60 cm away.

 5 Three charges, q1 = q2 = q3 = 10 μC, are separated by the distances shown in
the figure below.

q2
PENERBIT Chapter 1
ILMU
BAKTI SDN. BHD.3 cm Mid Semester Test 1

q1 4 cm q3

Calculate the magnitude of the electrostatic force exerted on charge q1.

1.2 Electric Field

 Learning Outcomes

You should be able to: F Answers
● Define and use electric field strength, E = qo
kQ
● Use E = r2 for point charge.

● Sketch the electric field strength diagram.
*Simple configuration of charges with a maximum four charges in 2D
● Determine the electric field strength, E for a system of charges.
*Simple configuration of charges with a maximum four charges in 2D

Electric Field

  1 Electric field is the region around a point charge where electrostatic force is
experienced by a positive test charge when it is placed in the region.

 2 Electric field can be visualised by sketching electric field lines which consist of
lines and arrows. The arrowhead indicates the direction of electric field, while
the length of arrow indicates the magnitude of electric field.

 Electrostatics  3 A stronger electric field means a larger magnitude of electric field. Therefore,
the electric field lines must be closer together.
8
 4 The tangent to the electric field line at a point corresponds to the electric field
at that point. The direction of electric field lines depends on the type of point
charge. If the point charge is
(a) positive: The direction of electric field lines around the point charge is
radially outwards.
BAKTI SDN. BHD.
+ +C +

Chapter 1 Single positive charge Two identical positive charges
Point C is neutral point, EC = 0

Figure 1.3(a)  Electric field lines for positive charges

(b) negative: The direction of electric field lines around the point charge is
radially inwards.

Mid Semester Test 1 − −C−

Single negative charge Two identical negative charges
Point C is neutral point, EC = 0
ILMU
Figure 1.3(b)  Electric field lines for negative charges

 5 (a) Figure 1.4(a) shows the electric field lines for unlike charges of different
magnitudes.

AnswersPENERBIT +−

4Q −Q

Figure 1.4(a)  Unlike charges of different magnitudes

(b) Figure 1.4(b) shows the electric field lines for unlike charges of the same
magnitude.

+−

Figure 1.4(b)  Unlike charges of the same magnitude

Electric Field Strength Electrostatics 

  1 Electric field strength, E is defined as 9

E = Fe
qo
PENERBIT
ILMU where Chapter 1
BAKTI SDN. BHD.
E = electric field strength

Fe = electrostatic force
qo = positive test charge
Electric field strength is a vector quantity with both magnitude and direction.
The unit for electric field strength is N C−1 or V m−1.

 2 Figure 1.5 shows a test charge, q at a distance, r from a point charge, Q.

Q+ +q

r

Figure 1.5  Two charges at a distance

Electric field strength at distance, r from Q,

kQqo

E= Fe = r2
qo qo
Mid Semester Test 1
So, E= kQ
r2

where

E = electric field strength
k = Coulomb′s constant (9 × 109 N m2 C−2)

Q = point charge

r = distance between test charge and point charge

 3 The magnitude of electric field strength depends on two factors:

(a) Magnitude of a point charge, Q

E  Q

The greater the magnitude of a point charge, the greater the electric field

strength. Answers

(b) Distance of a point or test charge from a point charge, r
1
E  r2

The greater the distance from a point charge, the smaller the electric field

strength.

 4 If a system contains more than one point charge, the electric field strength at

a point of the system can be calculated by determining the total electric field

strength or resultant electric field strength at that point.

Example 1

A point, P is 10 cm from a negative point charge, Q of 20 μC as shown in the
figure below.

−Q P

Calculate the electric field strength and its direction at point P.

 Electrostatics

Smart Tips! Solution

The direction of electric Electric field strength at P,
field around negative kQ
point charge is radially E = r2
inwards.
E = (9 × 109)(20 × BAKTI SDN. BHD.10−6)
Common Error (0.1)2

Students substitute = 1.80 × 107 N C−1
negative sign into the
Chapter 1 equation. −Q EP
P

The direction of electric field strength at point P is to the left.

Example 2

Mid Semester Test 1 The figure below shows two point charges, Q1 and Q2 separated by a distance
of 5 cm. Point P is 2 cm from Q2.

2 cm

+ P−
Q2 = 5 μC
Q1 = 2 μC

ILMU 5 cm

Determine the electric field strength at point P.

Smart Tips! Solution

• Observe the figure + E1P E2P Q2 E1P is to the right (+E1P)
carefully. Q2 is a E2P is to the right (+E2P)
negative charge. Q1 P

• S ketch the electric field
AnswersPENERBIT P Q1 r1P = 5 – 2 = 3 cm
line at point due to
and Q2.
Magnitude of E1P and E2P ,

Common Error E1P = kQ1 E2P = kQ2
r1P2 r2P2

• Students forget that E E1P = (9 × 109)(2 × 10−6) E2P = (9 × 109)(5 × 10−6)
is a vector quantity and (3 × 10−2)2 (2 × 10−2)2
the direction of E must E1P = 2.0 × 107 N C−1 E2P = 1.13 × 108 N C−1
also be determined.
So, total magnitude at point P,
• Students do not sketch
the electric field lines, EP = E1P + E2P
making mistakes EP = (2.0 × 107) + (1.13 × 108)
when determining the EP = 1.33 × 108 N C−1
direction of the electric The direction of EP is positive, that is to the right.
field. Therefore, the electric field strength at point P is 1.33 × 108 N C‒1 to the right.

• Students do not convert
the unit for distance to
S.I. unit.

10

Example 3 Electrostatics 

The figure below shows two point charges, Q1 and Q2 separated by a distance Smart Tips!
of 6 cm.
Net electric field strength
+ P + is zero means E due to
the two charges are the
Q1 = 10 μC Q2 = 7 μC same but in opposite
PENERBIT directions.
ILMU 6 cm
BAKTI SDN. BHD. Common Error
If the net electric field strength at point P is zero, determine the distance of
point P from point charge Q1. • Students cannot solve
quadratic equation.
Solution Chapter 1
• Students do not convert
Let the distance between Q1 and P = r the unit for distance to
S.I. unit.
r
• Students difficult to
+ +E2P P E1P decide the final answer
Q1 = 10 μC Q2 = 7 μC because there are two
values of r.
6 cm
11
Magnitude of E1P and E2P.
kQ1
E1P = r1P2 Mid Semester Test 1

E1P = k(10 × 10−6)
r2

E2P = kQ2
r2P2

E2P = k(7 × 10−6)
(0.06 – r)2

Given EP = 0,

(+E1P) + (–E2P) = 0

E1P = E2P

k(10 × 10−6) = k(7 × 10−6)
r2 (0.06 – r)2
Answers
10 × 10−6 = 7 × 10−6
r2 (0.06 – r)2

(0.06 – r)2 = 7 × 10−6
r2 10 × 10−6

(3.6 × 10−3) – 0.12r + r2 = 0.7
r2

(3.6 × 10−3) – 0.12r + r2 = 0.7r2

0.3r2 – 0.12r + (3.6 × 10−3) = 0

r = 0.37 m (r > 6 cm is not accepted and ignored).
OR
r = 0.03 m

∴ The distance of point P from point charge Q1 is 0.03 m.

 Electrostatics Example 4

12 A system is made up of two point charges, Q1 and Q2 of 6 μC and 5 μC
respectively as shown in the figure below. Point P is 5 cm vertically below Q2.

10 cm

− +Q1 Q2

5 cm
BAKTI SDN. BHD.
Chapter 1 P

Determine the electric field strength at point P.

Solution

Let the angle between line P to Q1 = θ, and the distance between P and
Q1 = r1P

− +10 cm

Q1 Q2

Mid Semester Test 1 r1P 5 cm

E1P θ

P

E2P

tan θ = 10
ILMU 5

θ = 63.43°

r1P = (10)2 + (5)2
r1P = 11.18 cm = 11.18 × 10−2 m

Magnitude of E1P and E2P.

AnswersPENERBIT E1P = kQ1 E2P = kQ2
r1P2 r2P2

E1P = (9 × 109)(6 × 10−6) E2P = (9 × 109)(5 × 10−6)
(11.18 × 10−2)2 (5 × 10−2)2
E1P = 4.32 × 106 N C−1 E2P = 1.80 × 107 N C−1

Resolving E1P and E2P into x- and y-components,

E (N C−1) x-component (N C−1) y-component (N C−1)

E1P = 4.32 × 106 E1Px = – (4.32 × 106) E1Py = + (4.32 × 106)
sin 63.43° cos 63.43°

E1Px = –3.86 × 106 E1Py = 1.93 × 106

E2P = 1.80 × 107 E2Px = 0 E2Py = –1.80 × 107

Sum EPx = –3.86 × 106 EPy = –1.61 × 107

Let angle between resultant force and x-axis = θ Electrostatics 

∑EPx Common Error

θ Students make mistakes
when resolving x-and
PENERBIT ∑EP y-components.
ILMU
BAKTI SDN. BHD. ∑EPy

Magnitude of EP.

EP = (EPx)2 + (EPy)2

EP = (–3.86 × 106)2 + (–1.61 × 107)2 Chapter 1
EP = 1.66 × 107 N C−1

Direction of EP.

  tan θ =EPy
EPx

 –1.61 × 107

tan θ = –3.86 × 106

θ = 76.52° below the negative x-axis Mid Semester Test 1

Therefore, the electric field strength at point P is 1.66 × 107 N C‒1 at an angle
of 76.52° below the negative x-axis.

Quick Check 1.2

 1 Two point charges, Q1 = –5 × 10−8 C and Q2 = 7 × 10−8 C as shown in the figure
below are separated by a distance of 5 cm.

2 cm

Q C Q
1 2

5 cm

Determine the electric field strength at point C. Answers

 2 A test charge of –6 μC is placed 3 cm from a point charge Q of 10 μC. Calculate
the magnitude of electric field strength on the test charge at that point.

 3 Two charges of +12 μC and ‒10 μC are placed near point C as shown in the
figure below.

C
5 cm

+ −

Q1 = 12 μC Q2 = 10 μC

3 cm

Determine the net electric field strength at point C.

13

Summative Practice 1

Chapter 1 Objective Questions BAKTI SDN. BHD. 5 Which of the following statements is true about
electric field?
 1 Which of the following is not true about A Direction of electric field depends on the mass
electrostatic force? of the particle
A Electrostatic force is experienced by a moving B Magnitude of electric field is directly
charge when it enters a region of electric field proportional to the magnitude of a point charge
B Electrostatic force is directly proportional to C Magnitude of electric field is directly
the distance between two point charges proportional to the distance between the point
C Electrostatic force is due to the interaction charge and the test charge
between two charges D If a positive test charge is placed in a region of
D Electrostatic force is a vector quantity an electric field, it will experience an electric
force that is opposite in direction to the electric
 2 Electric field strength depends on all of the field.
following except ……
Mid Semester Test 1 A the distance between the test charge and pointILMU  6 Which of the following statements is true about
charge. electric potential?
B the magnitude of the point charge. A Electric potential is a vector quantity
C the shape of the point charge. B Electric potential can be positive or negative
D the type of charge. C Electric potential does not depend on the
position of charge
 3 Coulomb’s Law states that …… D Electric potential is inversely proportional to
A the force of attraction or repulsion between the magnitude of charge
two charged bodies is directly proportional
AnswersPENERBIT to the product of their charges and inversely  7 Electric potential difference will be zero if ……
proportional to the square of the distance A two points are on the same equipotential surface.
between them. B the magnitude of point charge is less than test
B the force of attraction or repulsion between charge.
two charged bodies is inversely proportional C the mass of point charge is larger than the mass
to the product of their charges and directly of test charge.
proportional to the square of the distance D the maximum work is done to move from one
between them. point to another point.
C the force of attraction or repulsion between
two charged bodies is directly proportional  8 Which of the following pairs is correct?
to the product of their charges and inversely
proportional to the distance between them. Quantity Unit
D the force of attraction or repulsion between
two charged bodies is directly proportional to A Electric field strength N C−1
the product of their charges and the square of
the distance between them. B Electrostatic force kg m

 4 Two unlike charges are placed near each other. C Electric force m s−2
What will most likely happen?
A They will move in circular paths D Electric field V
B They will attract each other
C They will repel each other  9 Which of the following quantities is a scalar
D Nothing will happen quantity?
A Electric potential difference
28 B Electrostatic force
C Electric field
D Acceleration

Electrostatics 

10 Which of the following statements is not true B Direction of electric field lines is away from
about electric field lines? positive charge
A Electric field lines must be straight lines that
cannot be curved C Direction of electric field lines is towards
negative charge

D Electric field lines cannot cross each other

PENERBITStructured Questions
ILMU
BAKTI SDN. BHD. 1 Two point charges of 10 μC and 5 μC are separated by a distance of 3 cm. Determine the
electrostatic force exerted on the 10 μC charge.

 2 A test charge of 15 C is placed 2 cm from a point charge, Q. If the electric field strength felt by Chapter 1
Q is 20 N C−1, calculate the magnitude of charge Q.

 3 The potential difference between two parallel metal plates separated by a distance of 4.0 cm is
9 kV. Calculate the electric field strength.

 4 Calculate the electric potential at a point 30 cm from a point charge of 50 μC.

 5 An electron with an initial speed of 20 km s−1 perpendicularly enters a region of the uniform Mid Semester Test 1
electric field of 55 N C−1. If the electron moves in a parabolic path in the electric field, calculate

the speed of the electron 2.0 ns after entering the field.
[Charge of electron = ‒1.60 × 10−19 C, mass of electron = 9.11 × 10−31 kg]

 6 A charged particle of −25 μC is placed in a region of uniform electric field between two
charged plates separated by a distance of 15 cm. If the potential difference between the plates is
10 V, calculate the magnitude of the electric field.

 7 The electric field strength at a distance, d from a point charge of 5 nC is 300 N C−1. Calculate
the magnitude of the electric field at a distance, 3d from the charge.

 8 Figure 1 shows two point charges, Q1 = 3 μC and Q2 = ‒5 μC, B
placed 4 cm apart. Point B is 3 cm above point C. 3 cm

Calculate the electric potential at point B.

+Q1 4 cm − Q2 Answers
A Figure 1
C

 9 Figure 2 shows a system made up of two point charges, Q1 = ‒10 μC −A 5 cm
Q1 30°
and Q2 = 15 μC. ∠BAC is a right angle. C
Determine the electric field strength at point C.

B +Q
2
Figure 2

10 A system is made up of two point charges, Q1 = 6 μC and Q2 = 7 μC separated by a distance
of 80 cm. If a test charge, q experiences zero net electrostatic force when placed on the line
joining Q1 and Q2, determine the position of the test charge.

29

PENERBIT6CHAPTER
ILMU
BAKTI SDN. BHD.Alternating Current

6.1 Alternating Current

 Learning Outcomes

You should be able to:

● Define alternating current (AC).

● Sketch and interpret sinusoidal AC waveforms.

● Use sinusoidal voltage and current equations:
VI ==IOVsOinsinωtωt
®
®

Definition

 1 A current is referred to as an alternating current (AC) if the current reverses
direction at regular intervals. AC can be represented by sinusoidal waveforms.
Figure 6.1 shows the symbol for an AC power source.

Figure 6.1  Symbol for AC power source 149

AC Waveform and Equation

 1 Figure 6.2 shows the graph of current against time for an AC.

Current

Io
0 Time
−Io

Figure 6.2  Sinusoidal alternating current

 Alternating Current

Chapter 6 The peak current, IO is the magnitude of the maximum current. The equation BAKTI SDN. BHD.
for the AC is written as
Mid Semester Test 1
I = IO sin ωt
where IO = peak current
ωt = phase
 2 Figure 6.3 shows the graph of voltage against time for an AC.

Voltage

Vo
0 Time

−Vo

Figure 6.3  Sinusoidal alternating voltage

The equation for the alternating potential difference is written as

V = VO sin ωt

where VO = peak voltage
There will be an alternating potential difference (voltage) across a resistor

when an AC passes through it.
 3 If the voltage is zero when the current is zero, and the current and voltage reach

their peaks at the same time, then the current and voltage are said to be in phase.

Example 1

The figure below shows the current against time graph of a sinusoidal AC.

I (A)
ILMU

AnswersPENERBIT 2

0 t (ms)
10 20 30 40

−2

Smart Tips! Write the equation for the alternating current.

Determine the period for Solution
one cycle.
General equation, I = IO sin ωt

Recap! From the graph, IO = 2 A and T = 20 ms

2π ∴ ω = 2π = 2π = 100π rad s−1
T T 20 × 10−3
Angular speed, ω =
Hence, I = 2 sin 100πt (A)
T = Time period

150

Alternating Current 

Example 2

Given the AC voltage, V = VO sin 50πt, where V is in volts and t is in s. Sketch
a graph to show the variation of voltage, V with time.

Solution
PENERBIT
ILMUComparing V = VO sin ωt with V = VO sin 50πt,
BAKTI SDN. BHD.
ω = 2π = 50π rad s−1
T

∴ Period, T = 2π = 0.04 s
50π

V (V) Chapter 6

Vo

0 0.02 0.04 t (s)
−Vo

Example 3 Mid Semester Test 1

The figure below shows the sinusoidal waveform of an AC.

V (V)

8

0 t (s)
0.2 0.4 0.6 0.8 1.0 1.2

−8

Determine Answers
(a) the peak voltage,
(b) the period and frequency,
(c) the angular speed.

Solution Recap!

(a) As seen from the graph, the peak voltage = 8 V Frequency, f = 1
T
(b) Period, T = 0.4 s T = Time period

Frequency, f = 1 = 1 = 2.5 Hz
T 0.4

(c) Angular speed,

ω = 2π = 2π = 5π rad s−1
T 0.4

151

 Alternating Current

Quick Check 6.1

 1 If an AC has a peak voltage of 5 V and a frequency of 50 Hz, write the
equation for the alternating voltage in terms of time.

 2 Write the equation for an alternating voltage with a frequency of 100 Hz and
a peak voltage of 230 V. Then, sketch the voltage versus time graph.

 3 A sinusoidal alternating voltage is represented by
V = 3 sin 10πt
where V is in volts and t is in seconds. Determine its angular frequency,

frequency, period and peak voltage.

 4 An AC has a peak current of 15 A and a frequency of 60 Hz. Find the angular
frequency and write the current equation of the AC.

 5 The equation for an AC is
I = 10 sin 5πt
where I is in A and t is in s. If a 5 Ω resistor is connected to the AC source,

what is the peak current, peak voltage and frequency?
Chapter 6 BAKTI SDN. BHD.

Mid Semester Test 1 6.2 Root Mean Square (R.M.S.)

 Learning Outcomes

ILMU
You should be able to:

● Define root mean square (r.m.s.) current and voltage for AC source.
IO VO
● Use Irms = 2 , Vrms = 2

Root Mean Square

AnswersPENERBIT R.m.s. current

 1 Referring to Figure 6.2 in the previous section, it is seen that in one complete

cycle of the AC, any positive value, +I of the current will have a corresponding

negative value, ‒I. This will result in a mean current value of zero.

 2 However, when an AC flows through a resistor, heat is still dissipated from the

resistor. This indicates that the mean value of current is not the effective value

of the AC.

 3 Therefore, in the discussions on P

AC, the root-mean-square (r.m.s.) IO2R
value is usually used. It is derived

as below. Pav = 1 IO2R
At any given time, the 2

instantaneous power dissipated

from a resistor, R is t
P = I2R = (IO sin ωt)2R 0
Figure 6.4  Graph of power, P against time, t
= IO2R sin2 ωt

152

Alternating Current 

The current is squared, so the power is always positive as shown in the graph.

Thus, the average power is IO2R
2
Pav (mean value of I2)
= Irms2 R =    Irms =
I2
∴ Irms2 = IO2 Irms =
2
PENERBIT
ILMU Irms = IO
BAKTI SDN. BHD. 2

The root-mean-square current, Irms of an AC is the equivalent steady direct
current, when flowing through a resistor for a given time, will dissipate the

same amount of heat as the AC does in the same resistor at the same time. Chapter 6

R.m.s. voltage

 1 Similarly, the root-mean-square voltage, Vrms of an AC is the equivalent steady
direct voltage, when applied across a resistor for a given time, will dissipate the

same amount of heat as the AC does in the same resistor at the same time.

The average power, Pav = V2 = 1 VRO2   Vrms = (mean value of V 2)
rms 2
R Vrms = V2

∴ Vrms2 = VO2 Mid Semester Test 1
2

Vrms = VO
2

Example 1

An AC source is represented by the equation Answers
V = 110 sin 120t

where V is in volts and t is in seconds.
Determine
(a) the r.m.s. voltage,
(b) the frequency,
(c) the instantaneous voltage value when t = 2 ms.

Solution

From the equation, VO = 110 V, ω = 120 rad s−1
VO
(a) R.m.s. voltage, Vrms = 2

Vrms = 110
2
= 77.78 V Common Error

(b) Angular speed, ω = 2πf Students forget that ωt is
in radians. The calculator
Frequency, f = ω = 120 = 19.1 Hz mode should be set
2π 2π accordingly.

(c) When t = 2 ms,
V = 110 sin [(120)(2 × 10−3)] = 26.15 V

153

 Alternating Current

Example 2

An alternating current, I is represented by the equation
I = 2 × 102 sin 115πt

where I is in amperes and t is in seconds. Determine

(a) the r.m.s. current,

(b) the instantaneous current when t = 5 s.
BAKTI SDN. BHD.
Solution

From the equation, IO = 2 × 102 A, ω = 115π rad s−1
(a) R.m.s. current, Irms = IO
2
Chapter 6 2 × 102
Irms = = 141.4 A
2

(b) Instantaneous current when t = 5 s,
I = 2 × 102 sin [(115π)(5)] = 0 A

Mid Semester Test 1 Example 3

The figure below shows the graph of potential difference, V against time, t of
an AC.

V (V)

25

ILMU 0 0.5 1.0 t (s)

−25

AnswersPENERBIT Find
(a) the r.m.s. voltage,
(b) the frequency,
(c) the angular speed.

Common Error Solution

Students forget that ω is From the equation, VO = 25 V, T = 0.5 s
in rad s‒1, not m s‒1. VO
(a) R.m.s. voltage, Vrms = 2
154
Vrms = 25
2

= 17.7 V

(b) Frequency, f = 1 = 1 = 2 Hz
T 0.5

(c) Angular speed, ω = 2πf

ω = 2π(2)

= 4π rad s−1

Quick Check 6.2 Alternating Current 

 1 A generator generates an alternating voltage represented by 155

V = 230 sin 100πt

where V is in volts and t is in seconds. Determine the r.m.s. voltage and
frequency of the AC.

 2 The output of an AC voltage source is given by V = 2.8 × 102 sin 2πft, where
V is in volts and t is in seconds. A 150 Ω resistor is connected to the source.
Determine
(a) the r.m.s. voltage across the resistor,
(b) the r.m.s. current flowing through the resistor,
(c) the maximum current in the circuit.

 3 An AC supply has a peak voltage of 440 V and r.m.s. current of 15 A. Find
(a) the r.m.s. voltage,
(b) the peak current.

 4 The peak current of an AC flowing through a 2 500 W device is 15 A. What is
the r.m.s. voltage across the device?

 5 The potential difference, V across a 5 Ω pure resistor in an AC circuit is
represented by

V = 230 sin ωt
where V is in volts and t is in seconds. Calculate
(a) the r.m.s. voltage,
(b) the r.m.s. current flowing through the resistor.
PENERBIT Chapter 6
ILMU
BAKTI SDN. BHD. Mid Semester Test 1
6.3 Resistance, Reactance and Impedance

 Learning Outcomes

You should be able to:

● Sketch and use phasor diagram and sinusoidal waveform to show the

phase relationship between current and voltage for a single component Answers

circuit of:
® resistor, R
® capacitor, C
® inductor, L

● Use phasor diagram to analyse voltage, current and impedance of series
circuit of RL, RC and RLC.

● Define and use:

® capacitive reactance, XC = 1
2πfC
® inductive reactance, XL = 2πfL

® impedance, Z = R2 + (XL − XC)2
(XL − XC)
® phase angle, Ø = tan−1 R

● Explain graphically the dependence of R, XC, XL and Z on f and relate it to

resonance.

 Alternating Current

Phasor Diagrams

 1 A phasor is a vector that rotates anticlockwise about the origin with constant
angular velocity. It is used to represent sinusoidally varying quantities such as
alternating current and alternating voltage.

 2 A phasor diagram is used to analyse alternating currents most simply.
Figure 6.5 shows a phasor diagram. The angle between two phasors is known
as phase angle, Ø.

V
ω

Ø
I
Chapter 6 BAKTI SDN. BHD.
Figure 6.5  Phasor diagram. Both phasors V and I have angular velocity, ω.
Mid Semester Test 1
Single-Component AC Circuits

Pure resistor

 1 Figure 6.6 shows a pure resistor, R connected to an AC source. The alternating
current is represented by equation I = IO sin ωt.

R

VR

AC source
ILMU
Figure 6.6  AC circuit with a pure resistor

 2 By using Ohm’s Law,

V = IR

= (IO sin ωt) R
AnswersPENERBIT
= IOR sin ωt

We know that, VO = IOR

Thus, V = VO sin ωt

 3 Figure 6.7 shows the phasor diagram and the graphs of variation with time for

V and I. It is seen that the current and voltage reach their maximum, minimum

and zero values at the same time. The current and voltage are thus in phase.

ΔV, ΔI

ω VO
IO

IV T 2T t

0

−IO
−VO
(a) (b)

Figure 6.7  (a) Phasor diagram and (b) graphs of V and I against time for a pure resistor

156

Since V and I have the same phase angle, ωt, the phase angle between V and I Alternating Current 
is ΔØ = ωt − ωt = 0°.
157
 4 By using Ohm’s Law, the r.m.s. voltage across resistor, R is related to r.m.s.
current by
VR = Vrms = IrmsR

Pure capacitor

 1 Figure 6.8 shows a pure capacitor, C connected to an AC source. The
alternating voltage is represented by equation V = VO sin ωt.

C
PENERBIT
ILMU VC Chapter 6
BAKTI SDN. BHD. AC source

Figure 6.8  AC circuit with a pure capacitor

 2 Current in the circuit,

I= dQ …… 1 Mid Semester Test 1
dt
and charge in the capacitor,
Q = CV

Since V = VO sin ωt, Q = CVO sin ωt …… 2

Substituting 2 into 1 , I = d(CVO sin ωt)
dt


I = CVO ω cos ωt with CVO ω = IO
Hence,
 or ωt π
I = IO cos ωt I = IO sin + 2 Answers

 3 Figure 6.9 shows the phasor diagram and the graphs of variation with time for

V and I. It is seen that the voltage across the capacitor reaches its maximum

value of 90° after current reaches its maximum value. The voltage thus lags

current by π rad or 90°, or current leads voltage by π rad or 90°.
2 2

ΔV, ΔI

ω VO
I IO

V T 2T t
0

−IO
−VO
(a) (b)

Figure 6.9  (a) Phasor diagram and (b) graphs of V and I against time for a pure capacitor

Summative Practice 6

Chapter 6 Objective Questions BAKTI SDN. BHD. 6 What is the equation for an alternating current
with r.m.s. value of 10 A and a frequency of 60 Hz?
 1 If a resistor, a capacitor and an inductor are
connected in series to an AC source, the current in [Assume I = 0 A at time, t = 0 s]
the resistor would lag behind … A I = 10 sin (120πt) (A)
A the current in the inductor. B I = 10 cos (120πt) (A)
B the current in the capacitor. C I = 10 2 sin (120πt) (A)
C the voltage across the inductor. D I = 10 2 cos (120πt) (A)
D the voltage across the capacitor.

 2 Which of the following does not depend on the  7 In an LC series circuit, the current is I = IO sin ωt
frequency of the AC source? and the supply voltage is V = VO cos ωt. What is
A Resistance the power dissipation of the circuit?
B Impedance
C Inductive reactance A 0
D Capacitive reactance
B IOεO
 3 Which electrical component in an AC series 2
circuit has both voltage and current in phase? C IOεO
Mid Semester Test 1 A Inductor D IOεO sin ωt
B Resistor
C Capacitor  8 The current flowing through an inductor is I. What
D All the above
is the power dissipated by the inductor?
 4 Figure 1 shows a resistor, R and an inductor, L
connected in series to an AC source. A 0

RL B LI

ILMU C LI 2

D 1 LI 2
2

 9 The maximum alternating current flowing through

VR VL a pure capacitor of capacitance, C is IO and the
AC source maximum voltage across it is εO. If the frequency
AnswersPENERBIT of the AC is 50 Hz, what is the average power

dissipated from the capacitor?

A 0

B εOIO
2
Figure 1 C εOIO

Which of the following is correct about RL series D I2
100πC
circuit shown?

A V = VR C V = VL + VR 10 The current flowing in an AC circuit is given as

B V < VL + VR D V > VL + VR I = 5 cos ωt, while the voltage is given as

 5 What is the condition for the alternating current in V = 7 cos ωt. What is the power factor of the

an RLC series circuit to reach its peak? circuit?

A XL = 0 C XL = XC A 0 C 1

B XC = 0 D XL2 + XC2 = 1 B 5 D ∞
7

170

Alternating Current 

PENERBIT Structured Questions Chapter 6
ILMU
BAKTI SDN. BHD. 1 The instantaneous current in an AC circuit is given by I = 8 sin (ωt + Ø), where I is in A and Mid Semester Test 1
t is in seconds. What is the r.m.s. value of the current?
Answers
 2 A 15 µF capacitor is connected to a 220 V, 70 Hz AC power supply. Find the maximum current
of the circuit.

 3 Sketch the graph of R, XC, XL and Z against frequency,  f for an RLC series circuit. Relate it to
the resonance frequency.

 4 If an inductor, L of 28 mH, a resistor, R of 8.7 kΩ and a capacitor, C of 6 250 pF are connected
in series to a 725 V (r.m.s.), 10.0 kHz AC source, find
(a) the total impedance,
(b) the phase angle,
(c) the r.m.s. current.

 5 An RLC series circuit is supplied with an AC voltage, V = 80 sin 800t, where V is in volts and
t is in seconds. If the resistance is 250 Ω, the capacitance is 5.5 µF and the inductance is
0.60 H, determine the average power delivered to the circuit.

 6 An AC generator with r.m.s. output of 45.0 V at a frequency of 70.0 Hz supplies power to a
60.0 Ω resistor and a 50.0 µF capacitor connected in series. Find
(a) the r.m.s. current,
(b) the r.m.s. voltage drop across the resistor,
(c) the r.m.s. voltage drop across the capacitor,
(d) the phase angle.

 7 The output voltage of an AC source is given by V = (2.8 × 102) sin 2πft, where V is in volts and
t is in seconds. A 1.5 × 102 Ω resistor is connected to this source. Find
(a) the r.m.s. voltage across the resistor,
(b) the r.m.s. current.

 8 A 200 Ω resistor and a 15.0 µF capacitor are connected in series to a 220 V, 50 Hz AC source.
Calculate
(a) the current,
(b) the r.m.s. voltages across the resistor and capacitor.

 9 An RLC series circuit with R = 5 Ω, L = 26 mH and C = 800 µF is supplied with a sinusoidal
voltage of peak value 240 V and frequency 50 Hz. Find
(a) the impedance,
(b) the phase angle between the voltage of the power source and the current,
(c) the power dissipated in the system,
(d) the power factor.

10 An RLC series circuit with R = 2.5 Ω, L = 35 mH and C = 795 µF is supplied with a sinusoidal
voltage of peak value 280 V.
(a) At what source frequency will resonance occur?
(b) Determine the impedance, current and power dissipated of the circuit when resonance
occurs.

171

 Alternating Current

11 An 8.8 µF capacitor is connected to an AC generator with a frequency of 75.0 Hz and a r.m.s.
voltage of 1.6 × 102 V. Determine

(a) the capacitive reactance,

(b) the r.m.s. current.

12 An RLC series circuit has a resistance of 2.9 × 102 Ω, an inductance of 0.75 H and a capacitance
of 3.5 μF. The AC source has a frequency of 50.0 Hz and a peak voltage of 1.70 × 102 V.
Determine
(a) the impedance,
(b) the maximum current,
(c) the phase angle,
(d) the maximum voltages across the components,
(e) the average power delivered.
Chapter 6 BAKTI SDN. BHD.
13 Consider an RLC series circuit with R = 1 200 Ω, L = 28 mH, Vrms = 24.0 V and f = 650 Hz.
Determine

(a) the capacitance value for which the r.m.s. current is maximum,

(b) the maximum r.m.s. current.

Mid Semester Test 1 14 (a) Figure 1 shows a 125 Ω resistor connected to an AC source. 125 Ω
The AC voltage is given by V = 220 sin ωt, where V is in
volts, t is in seconds. V = 220 sin ωt
Figure 1
Find
(i) the r.m.s. voltage,
(ii) the r.m.s. current,
(iii) the average power delivered to the circuit.

(b) An RLC series circuit consisting of a 25 mH inductor, a
30 µF capacitor and a 96 Ω resistor is connected to an AC
generator supplying a voltage of 230 V at a frequency of
65 Hz. Calculate EXAM CLONE
(i) the capacitive reactance,
(ii) the inductive reactance,
(iii) the impedance,
(iv) the phase angle.
ILMU

AnswersPENERBIT 15 (a) An AC source with a peak voltage of 110 V supplies power to an RLC series circuit with
a power factor of 0.82. The current is given by I = 2.5 sin 3πt, where I is in A and t is in s.
Determine
(i) the r.m.s. current,
(ii) the impedance,
(iii) the resistance.

(b) An inductor is connected to an AC source with a frequency of 50 Hz and a r.m.s. voltage of
40 V. The reactance of the inductor is 112 Ω. Determine EXAM CLONE
(i) the inductance,
(ii) the peak current.

172

PENERBIT9CHAPTER
ILMU
BAKTI SDN. BHD.Nuclear and Particle Physics

9.1 Binding Energy and Mass Defect

 Learning Outcomes

You should be able to:
● Define and use mass defect, Δm = (Zmp + Nmn) − mnucleus
● Define and use binding energy, EB = Δmc2
EB
● Determine binding energy per nucleon, A

● Sketch and describe graph of binding energy per nucleon against nucleon

number.

Mass Defect

  1 Mass defect, Δm is the difference between the sum of the individual masses of
nucleons and the actual mass of the nucleus, that is,

Δm = (total mass of nucleons) – (actual mass of the nucleus)

Δm = (Zmp + Nmn) − mnucleus

Δm = [Zmp + (A − Z)mn] − mnucleus

where

Z = atomic number or proton number (number of protons in the nucleus)

N = neutron number (number of neutrons in the nucleus)

A = mass number or nucleon number (number of nucleons* in the nucleus)

* Nucleon is a generic term used to refer to either a proton or a neutron.

Mass defect is measured in kilograms (kg) or atomic mass units (u), where

1 u = 1.66 × 10–27 kg

 2 As the nucleons combine to form the nucleus, some of their mass is released as

energy, resulting in a reduction in mass.

237

 Nuclear and Particle Physics

 3 Conversely, in any attempt to separate the nucleons, this same amount of
energy must be provided. This energy is referred to as the binding energy of
the nucleus.

BAKTI SDN. BHD.+ Binding energy =

Nucleus Separated nucleons
(smaller mass) (greater mass)

Figure 9.1  Separation of a nucleus

Chapter 9 Binding Energy

 1 Binding energy, EB is the amount of energy required to separate a nucleus into
its constituent nucleons.

Binding energy, EB = Δmc2

where EB is in J, ∆m is in kg and c is the speed of light (3 × 108 m s‒1).
 2 Binding energy can also be expressed as

 EB
Mid Semester Test 1 = Δm 931.5 MeV
u
where EB is in eV and Δm is in atomic mass unit (u).

 3 It is seen that the binding energy of a nucleus is equal to the energy equivalent

of the mass defect. The unit for binding energy is joules (J) or electron volt

(eV), where

ILMU 1 eV = 1.60 × 10‒19 J

Binding Energy per Nucleon

 1 The binding energy per nucleon is calculated by dividing the total binding

energy, EB of a nucleus by the number of nucleons in that nucleus, A. It is a

measure of the stability of the nucleus.
Binding energy, EB
Binding energy per nucleon = Nucleon number, A
AnswersPENERBIT
= Δmc2
A

 2 The nuclear force and electrostatic force which exist inside the nucleus are the

only two factors that affect its stability. Nuclear force is the strongest force

(attraction force) that holds all of the nucleons in the nucleus together, whereas

electrostatic force acts to repel charged particles (protons) in the nucleus.

 3 When nuclear force exceeds electrostatic force, the nucleus is regarded as

stable. The binding energy per nucleon is measured in electron-volts per

nucleon (eV/nucleon) or joules per nucleon (J/nucleon).

 4 The nuclear force is referred to as ‘strong’ nuclear force. The reason is, only

13.6 eV is required to ionise one electron in the ground state of hydrogen,

whereas several million electron volts are required to separate a nucleon from

its nucleus held together by the nuclear force.

238

Nuclear and Particle Physics 

Graph of binding energy per nucleon against nucleon number

 1 Figure 9.2 shows the graph of binding energy per nucleon against atomic
number, A.

9 34S 58Fe 84Kr 119Sn BAKTI SDN. BHD.
16O
Binding energy per nucleon (MeV) Region of very 205Ti 235U
8 12C stable nuclides 238U
14N
Chapter 9
7 4He

6 7Li
5 6Li

4

3 3H
2 3He

1 2H

0 20 40 60 80 100 120 140 160 180 200 220 240 260
Mass number, A

Figure 9.2  Graph of binding energy per nucleon against nucleon number

The typical binding energy per nucleon ranges from 6 MeV to 10 MeV, with an Mid Semester Test 1

average of 8 MeV.

 2 For light nuclei, EB rises sharply as A rises. Except for the lightest nuclei, most
A
EB are between 7 MeV/nucleon and 9 MeV/nucleon. The graph shows that the
A
ILMU
highest EB and most stable nuclei are those with nucleon numbers 50 to 80.
A
Iron-56 (Fe) is the most stable nucleus with a maximum EB of 8.8 MeV per
A
nucleon.

Example 1

PENERBITCalculate Answers
(a) the mass defect,
(b) the binding energy per nucleon Smart Tips!
(i) in MeV,
(ii) in J Number of electrons =
of boron (150B). Number of protons
[mn = 1.008665 u, mp = 1.007276 u, me = 5.48 × 10−4 u, atomic mass of boron,
mB = 10.01294 u, speed of light in vaccum = 3.0 × 108 m s−1]

Solution

(a) mnucleus = matomic − Zme
mnucleus = 10.01294 u − 5(5.48 × 10−4 u)
= 10.01020 u
Δm = (Zmp + Nmn) − mnucleus
Δm = [5(1.007276 u) + 5(1.008665 u)] − 10.01020 u
= 0.069505 u

239

 Nuclear and Particle Physics

 (b) (i)
EB = Δm 931.5 MeV
u

  = (0.069505 u) 931.5 MeV
u
= 64.7 MeV BAKTI SDN. BHD.
∴ EAB =
64.7 MeV
10 nucleon

= 6.47 MeV/nucleon

(ii) EB = 64.7 MeV
= 64.7 × 106 × 1.60 × 10‒19

Chapter 9 = 10.35 × 10–12 J
EAB = 10.35 × 10−12 J
10 nucleon
= 10.35 × 10−13 J/nucleon

Example 2

Gold, 17997Au is bombarded with slow neutrons, and electrons are released.

Mid Semester Test 1 Calculate, in MeV, the energy released in the reaction.

[Atomic mass of 19797Au = 196.966552 u, 198 Hg = 197.966752 u,
80
mn = 1.008665 u, me = 5.48 × 10−4 u]

Solution

17997Au + 1 n → 19808Hg + 0 e
0 –1

Δm = (mAu + mn) − (mHg + me)
ILMU
Δm = (196.966552 u + 1.008665 u) − (197.966752 u + 5.48 × 10−4 u)

= 7.917 × 10−3 u
 EB
= (7.917 × 10−3 u) 931.5 MeV
u
= 7.37 MeV

AnswersPENERBIT Quick Check 9.1

 1 The binding energy of an element is 28.6 MeV and its binding energy per
nucleon is 7.14 MeV/nucleon. Calculate the number of nucleons in the nucleus.

 2 42He has a binding energy of 28.8 MeV. Calculate the binding energy per nucleon.
 3 Determine the total binding energy and binding energy per nucleon for 5266Fe.
[ mn = 1.008665 u, mp = 1.007276 u, me = 5.48 × 10−4 u, atomic mass of

5266Fe = 55.93494 u]
 4 Calculate the binding energy per nucleon of 2100Ne. Given atomic mass

2100Ne = 19.992 u.

240

Nuclear and Particle Physics 

9.2 Radioactivity

 Learning Outcomes

PENERBITYou should be able to:
ILMU
BAKTI SDN. BHD.● Explain α, β+, βˉ and γ decays.
*Radioactive decay as spontaneous and random process
**Introduce neutrino and antineutrino
dN
● State and use decay law, dt = –λN

● Define and determine activity, A and decay constant, λ.
*Consider decay curve Chapter 9
● Use N = Noe−λt or A = Aoe−λt

● Define and use half-life, T1 = ln 2
λ
2

Radioactive Decay Mid Semester Test 1

 1 A radioactive atom will emit particles and rays to become stable. In other Answers
words, if an atom is unstable, it will emit particles and the nucleus will reach
a lower energy state. This is the disintegration (decay) process of radioactive
atoms.

 2 This radioactive decay is a spontaneous process that is unplanned and
unpredictable. It is also a random process, therefore it is impossible to foresee
which nucleus will disintegrate. However, the number of nuclei that will decay
in a specific amount of time may be predicted. Every nucleus decays with the
same probability per unit time.

 3 The rate of decay is affected by the type of atom and the number of atoms. It
is not affected by physical conditions (such as pressure or temperature) and
chemical bonding.

 4 Radioactive substances emit three types of radiation which are alpha, beta and
gamma. These emissions cause
(a) the photographic plate to become black,
(b) gases to ionise,
(c) damage to cells or molecules of living systems.

Alpha decay

 1 In the nuclear decay process, the nucleus emits alpha (α)-particles to gain

stability. α-particles are helium nuclei with the symbol 4 He or 42α (two protons
2

and two neutrons). A nucleus thus loses four nucleons during alpha decay, two

of which are protons. The two protons have a total charge of +2e.

 2 Uranium-238 is one substance that experiences decay process.

U238 → 4 He + 23940Th + Q
92 2

During this radioactive decay, the uranium atom (23982U) emits an α-particle and
decays into a thorium atom (23940Th). Another α-particle emitter is thorium-230.

23900Th → 42He + 22868Ra + Q

241

 Nuclear and Particle Physics

Beta decay

 1 A neutrino is a subatomic particle that is very similar to an electron but has no

electrical charge and a very small mass.

 2 An antiparticle is a particle that has the same mass as a particle but opposite

charge (refer to section 9.4). In beta decay, neutrino and its antiparticle,
BAKTI SDN. BHD.
antineutrino are emitted to comply with the laws of conservation.

 3 There are two types of beta decay:
(a) Positron (beta-plus, β+) decay

The antiparticle of an electron, known as a positron, has the same mass as

an electron but is positively charged, +1e. During positron decay, a proton

is removed from the nucleus, while a positron and a neutrino are emitted.

Chapter 9 For example,

172N → 162C + 01e + ν + Q
172N = parent nucleus
162C = daughter nucleus
01e = beta-plus particle
v = neutrino

Mid Semester Test 1 Q = heat release

(b) Electron (beta-minus, β–) decay

A beta-minus particle (–01e) has a mass equal to an electron and is negatively
charged, ‒1e. During beta-minus decay, the nucleus obtains an additional

proton, while an electron and an antineutrino are emitted. The nucleon

number, A of the parent and daughter nuclei remain the same as one of the

neutrons in the parent nucleus changes into a proton and an electron. The

electron is then emitted.

For example, 164C → 174N + −01e + ¯ν + Q

where
ILMU
164C = parent nucleus
174C = daughter nucleus
−10e = beta-minus particle

¯ν = antineutrino

AnswersPENERBIT Q = heat released

Gamma rays

 1 Gamma rays (γ) are high-energy photons (electromagnetic radiation) with no
mass and no charge (neutral). Emission of gamma rays does not change the
parent nucleus into a different nuclide because neither the charge nor nucleon
number changes. The nuclei in an excited state emit gamma ray photons when
they transition to a ground state.

Radioactive Decay

 1 The Law of Radioactive Decay states that the number of nuclei, N undergoing
radioactive decay (α, β, or γ) per unit time, is proportional to the total number
of nuclei in the sample material. Hence, the rate of change of N is

dN = –λN
dt


242

Nuclear and Particle Physics 

or dN = –λ dt
N

where

ddNt = rate of decay
PENERBIT
ILMU λ = decay constant
BAKTI SDN. BHD.
N = number of nuclei

The negative sign indicates decreasing number of nuclei over time.

 2 Let the decay start at t = 0 when N = No (initial number of radioactive nuclei in

the sample).

∫ ∫∴ N dN = – λ t Chapter 9
No N
dt

0

[ln N]NNo = −[λt]0t
N
ln No = −λt

N = N0e−λt

 3 Figure 9.3 shows the plot of N against t for a radioactive decay process. This Mid Semester Test 1
curve is called decay curve.

N
N

0

N = N e−λt
0

1 N0
2

1 N0
4
1
8 N0 t

0 T½ 2T½ 3T½

Figure 9.3  Graph of number of radioactive nuclei, N present at any time, t

 4 The decay rate of a radioactive sample, dN is also called activity, A. The S.I.
dt

unit for activity is Becquerel (Bq), with 1 Bq = 1 s‒1. Another unit for activity is

Curie (Ci), with 1 Ci = 3.7 × 1010 s‒1. Answers

 5 Since A = dN = –λN and N = N0e−λt,
dt

A = −λN0e−λt and Ao = –λNo

Hence, A = Aoe−λt

where

A = activity at a specific time

Ao = initial activity at t = 0
  6 Decay constant, λ is defined as the probability of a radioactive nucleus

decaying within a second. It can be expressed as

dN

λ = –  dt
N

= decay rate
number of remaining radioactive nuclei

Its unit is per second (s−1).

243

 Nuclear and Particle Physics

 7 Half-life, T1 is the time taken for half of a specified number of radioactive

2

nuclei to decay. Any given radioactive nuclide has a fixed half-life which is

independent of the number of remaining nuclei.
No
From the equation N = Noe−λt, let N = 2 , t = T1

2
BAKTI SDN. BHD.No Noe−λT 1
2 =
2

1 = −λT 1
2 2
e

2 = λT 1
2
e

ln 2 = λT1

Chapter 9 2

T 1 = ln 2
λ
2

 8 Also, referring to Figure 9.3, when t = 0, N = No,

t = T1, N = 1 No,
2
2
1
t = 2T 1 , N = 4 No,

2 1
8
Mid Semester Test 1 t = 3T 1 , N = No, ...

  Therefore, N = 1 2
2 n

No, where n = number of half-lives

Example 1

ILMU
Carbon dating tests on an old wooden box show that only 6.0% of the wood is

fresh with 164C atoms. The half-life of carbon-14 is 5 730 years. What is the age
of the box?

Smart Tips! Solution

Only 6.0% of the wood   N = No 1 n
is still fresh, the rest has 2
decayed.
AnswersPENERBIT   N = 1n
No 2
 
0.06 = 1n
2

−n ln 2 = ln 0.06

n = 4.06

Number of half-lives, n = 4.06

Age, t = nT1

2

= (4.06)(5 730)
= 2.3 × 104 years

244

Nuclear and Particle Physics 

Example 2

Given the half-life of 238U is 1.89 × 1019 s, calculate the decay constant.

Solution

T1 = 1.89 × 1019 s
PENERBIT
ILMU2 ln 2
BAKTI SDN. BHD.λ
T 1 =
ln 2
2 λ

1.89 ×1019 =

λ = ln 2 Chapter 9
1.89 × 1019

= 3.67 × 10−20 s−1

Example 3

The half-life of a radioactive substance is 3.2 hours. How much radioactive
substance is lost from 1 g of the substance in 1 hour?

Solution Smart Tips!

−λT N is the amount of
T1 = 3.2 hours N = Noe− substance left after
= Noe ln 2 (1) 1 hour. Mid Semester Test 1
2 3.2

T 1 = ln 2 = N e−0.3125(ln 2)
λ o
2   = No 1 0.3125
2
3.2 hr = ln 2
λ
N
λ = ln 2 hr‒1 No = 0.805
3.2

Decayed substance

= 1 ‒ 0.805

= 0.195 g

Quick Check 9.2

 1 231Pa has a half-life of 3.25 × 104 years. How much of a 6 µg original sample Answers
of 231Pa is still present 1.2 × 102 years later?

 2 The half-life of tritium is 12.5 years. How much of a sample will still be intact
after 30 years? Give the answer in terms of No.

 3 If a radioactive sample consists of 40 g of substance, how much substance
will still be in the sample after 2.6 × 105 s? Given that the sample has a half-
life of 1.3 × 105 s.

 4 A certain radioactive sample has a half-life of 8.02 hours. If the initial activity of
the sample is 1.8 × 105 Bq, find its activity after being left on the shelf for 3 hours.

 5 226 Ra has a half-life of 1.9 ×103 years. If a sample initially has the 226 Ra nuclei
88 88
of 3.5 ×1018, calculate

(a) the initial activity,

(b) the number of radium nuclei left after 5.2 × 103 years,

(c) the activity at this later time.

245

 Nuclear and Particle Physics

9.3 Particle Accelerator

 Learning Outcomes

You should be able to:
● State the thermionic emission.
● Explain the acceleration of particle by electric and magnetic fields.
● State the role of electric and magnetic fields in particle accelerators (linac and

cyclotron) and detectors (general principles of ionisation and deflection only).
● State the need for high energies to investigate the structure of nucleon.
BAKTI SDN. BHD.
Chapter 9 Thermionic Emission

Mid Semester Test 1   1 Thermionic emission is defined as the phenomenon where metal surfaces emit
electrons when heat energy is supplied. It is a simple technique for creating an
electron beam.

 2 Electrons are emitted when a metal is heated to a point where the free electrons
inside the metal have enough kinetic energy to break free and move away from
the surface of the metal.

 3 The rate of thermionic emission can be increased by
(a) changing the metal type
(b) raising the cathode temperature
– More electrons will be released if the emitter metal has a higher temperature.
(c) increasing the metal (filament) surface area
– More electrons can be emitted due to the larger surface area.

ILMU
Particle Acceleration

 1 Table 9.1 shows the motion and explains the acceleration of a particle in
electric and magnetic fields.

Table 9.1  A particle in electric and magnetic fields

Motion of Diagram Description
Particle/Charge
AnswersPENERBIT
Charge moving Acceleration along the y-axis
perpendicular
to the electric due to electrostatic force.
field
+++++++++++++ v F = Fe = ma
− − and
q0
E Fe = qoE
−−−− −−−−−−−−−
Hence, a = qoE
m
where

Fe = electrostatic force
a = acceleration of particle

qo = charge of particle
E = electric field strength

m = mass of particle

246

Nuclear and Particle Physics 

Motion of Diagram Description
Particle/Charge

Stationary Positive charge Due to electrostatic force
charge and
charge moving ++++++++++ +++ acting in parallel to the
parallel to the
electric field + electric field, the particles
a
E Fe
PENERBIT move in a straight line.
ILMU
BAKTI SDN. BHD. F = Fe = ma
and

−−−− −−−−−−−−− Fe = qoE

Negative charge Hence, a = qoE
m
++++++++++ +++ Chapter 9
Fe

a
E−

−−−− −−−−−−−−−

Charge moving Since the direction of the
perpendicular
to the magnetic velocity of particle and the
field
direction of magnetic field

Bin are at an angle, the particles Mid Semester Test 1
v
move in a circular path.
q + FB
r FC = FB = mac
and

v FB + q FB = qovB

FB Hence, ac = qovB
m
where
+v
FC = centripetal force
q FB = magnetic force
ac = centripetal acceleration
v = velocity of particle

B = magnetic flux density

Particle Accelerators Answers

Linac

 1 Linac stands for linear particle accelerator. Figure 9.4 shows the basic setup
of a linac. It consists of a particle source and alternating tubes at an alternating
polarity that causes the particle to accelerate.

Radio frequency
AC source

Particle AB Acceleration only occurs within the gaps
source CD

Alternate tubes at alternate polarity

Drift tubes

Figure 9.4  Linac

247

Summative Practice 9

Objective Questions BAKTI SDN. BHD. 6 Which of the following is not a fundamental

 1 Which of the following statements is incorrect? particle?
A Mass number is the sum of all protons and
electrons in an atom A Proton C Electron
B Mass defect is the amount of matter that would
be converted into energy if a nucleus is formed B Lepton D Tau
from initially separated protons and neutrons
Chapter 9 C Nuclear binding energy is the energy released  7 Which of the following statements is false about
in the formation of an atom from subatomic cyclotron?
particles A The speed of an ion does not change inside a
D Nuclei with the highest binding energies are dee
the most stable B The evacuated chamber of cyclotron reduces
energy loss due to ions colliding with air
 2 A positron has a mass number of ,a molecules
C A constant electric field limits the ions to
charge of , and a mass equal to that the two dees and keeps them travelling in a
circular pattern
of a(an) . D The ions spend the same amount of time in
each dee as their speed increases
Mid Semester Test 1 A 0, +1e, proton

B 1, +2e, proton

C 0, +1e, electron

D 1, +2e, electron  8 Which of the following best describes the
relationship between the number of neutron, N and
 3 Which of the following emissions leaves both the proton, Z in an atom with a stable nucleus?
atomic number and mass number unchanged? A N  Z
A Positron emission B N  Z
B Neutron emission C N  Z
C Alpha particle emission D N  Z
D Gamma radiation emission
ILMU
 4 The nucleus of an atom is made up of ……  9 Which of the following pairs is a particle and its
A neutrons and protons. antiparticle?
B electrons and protons. A Proton/positron
C electrons, protons and neutrons. B Electron/positron
D electrons and neutrons. C Neutron/proton
AnswersPENERBIT D Neutron/neutrino

 5 The helium atom does not contain …… 10 Which of the following does not belong to the
A two protons. family of leptons?
B two electrons. A Electron
C two neutrons. B Muon
D six nucleons. C Proton
D Neutrino

258

Nuclear and Particle Physics 

Structured Questions

[mp = 1.007276 u, mn = 1.008664 u, me = 5.48 × 10−4 u]

 1 The helium (42He) nucleus has a mass defect of 0.0303 u. What is the binding energy, in MeV, of
helium?
PENERBIT
ILMU 2 Determine the binding energy, in MeV, of the 8346Kr nucleus.
BAKTI SDN. BHD. [Atomic mass of 8346Kr = 83.913 u]

 3 Find the binding energy and binding energy per nucleon of 5266Fe.
56
[Atomic mass of 26 Fe = 55.935 u]

 4 Calculate the total binding energy and binding energy per nucleon of 17985Pt. Chapter 9
[Atomic mass of 17985Pt = 194.964774 u]

 5 Calculate the binding energy per nucleon of a 14 N nucleus.
7
[Mass of nucleus of 174N = 14.003074 u]

 6 The decay constant of a nucleus is 7.5 × 10−3 s−1. Determine its half-life.

 7 Radon has a half-life of 3.7 days. Determine the amount of time required for 75% of a radon Mid Semester Test 1
sample to decay.

 8 If the disintegration constant of a radioactive substance is 0.005 per minute, determine the half-
life of the substance.

 9 The activity of a radioactive sample is 1 550 per minute. The activity reduces to 28 per minute
after 14 hours. Determine its half-life.

10 Figure 1 shows the decay curve of an isotope.

Activity

80

70

60 Answers

50

40

30

20

10

0 t (min)
5 10 15 20 25

Figure 1

How many days would it take for a 10.0 mg sample of the isotope to decay until only 1.25 mg
remains?

11 A radioactive sample contains 2.6 μg of pure 173N which has a half-life of 10 minutes.
(a) How many nuclei are present initially?

(b) What is the initial activity?

(c) What is the activity, in Ci, after 2 hours?

259

 Nuclear and Particle Physics

12 The number of nuclei in a sample of U238 is 6.50 × 1019.
92

(a) Find the decay constant.

(b) How many disintegrations take place on average every minute?

[Half−life of U238 = 1.59 × 105 years]
92

13 The half-life of carbon-11 is 21.04 minutes. The initial mass of a sample of carbon-11 is 3 mg.
(a) Determine the initial number of carbon-11 atoms.
(b) Find the maximum activity of the carbon-11 sample.
(c) What is the activity of the carbon-11 sample after 70 minutes?
BAKTI SDN. BHD.
Chapter 9 14 (a) Calculate the binding energy, in MeV, of 73Li.
[Mass of 73Li = 7.016003 u]
(b) The activity of a radioisotope sample is 620 s‒1 nine days after it is prepared. The activity
reduces to 250 s‒1 fifteen days after the date of preparation. Calculate

(i) the decay constant, and

(ii) the half-life of the radioisotope. EXAM CLONE

Mid Semester Test 1 15 (a) Determine the binding energy per nucleon, in MeV/nucleon, of copper, 6249Cu.
[Atomic mass of Cu = 63.546 u]

(b) The half-life of certain radioactive nuclei is 0.85 s. How long does it take for 35% of the

nuclei to decay? EXAM CLONE

ILMU

AnswersPENERBIT

260

PENERBIT
ILMU
BAKTI SDN. BHD.