CHAPTER 9: NUCLEAR & PARTICLE PHYSICS

CHAPTER 9

NUCLEAR
&

PARTICLE PHYSICS

KMM

CHAPTER 9: NUCLEAR &
PARTICLE PHYSICS

ATOM RADIOACTIVE
ATOM

THERMIONIC
EMISSION

1

SUBTOPICS….

9.1 BINDING ENERGY AND MASS
DEFECT

9.2 RADIOACTIVITY
9.3 PARTICLE ACCELERATOR
9.4 FUNDAMENTAL PARTICLE

2

PARTICLE PROPERTIES

Particle Symbol Mass

Proton 1 p mp =1.672 x 10-27 kg @ 1.007277 u
1
Electron
Neutron 0 e me =9.11 x 10-31 kg @ 0.000549 u
Hidrogen 1

1 n mn=1.674 x 10-27 kg @ 1.008665 u
0

1 H mH=1.673 x 10-27 kg @ 1.007825 u
1

UNIT CONVERSION RELATION

1 u = 1.66 x 10-27 kg c2 = 931.5MeV
u
1eV = 1.6 x 10-19 J
c= 3.00 x 108 m s -1

3

9.1 Binding Energy & Mass Defect

At the end of this topic, students should be able to:

a) Define and use mass defect,

 m  Zmp  Nmn  mnucleus

b) Define and use binding energy,
EB  mc2

c) Determine binding energy per nucleon, EB .
A

d) Sketch and describe graph of binding energy per nucleon
against nucleon number.

4

Properties of Nucleus

• An atom is consists of electron and nucleus where the
electron orbiting the nucleus. (Figure 11.1.1)

• A nucleus of an atom is made up of protons and
neutrons that is also known as nucleons. (Figure 11.1.2)

Figure 11.1.1

Figure 11.1.2 5

9.1 (a) Mass Defect, Δm

Definition: Formula:

The difference  m  Zmp  Nmn  mnucleus
between the sum of
OR
the masses of
individual nucleons m  ZmH  Nmn  matom

that form an mp = mass of a proton
atomic nucleus and mH = mass of hydrogen atom
mmnatuocmleu=s =mmasasssoof fana nucleus
the mass of the atom
nucleus. mn = mass of a neutron
Z = number of protons
** Its unit is a.m.u (u) N = number of neutrons
@ kg. 6

Example 9.1.1:

1.00728 amu per proton

1.00866 amu per neutron

Mass of nucleus
= 7.016005 amu

From example above, can you determine the value of
mass defect ?

(Ans: 0.040475 a.m.u @ 0.040475u)

7

9.1 (b) Binding Energy, EB

Definition: Formula:

Energy required to = ∆ 2
separate a nucleus into
its individual protons ΔmEB : Binding energy
and neutrons. : Mass defect
c : speed of light
OR = 3.00 x 108 m s-1
Energy released when
nucleus is formed from ** Its unit is MeV @ J.
its individual nucleons.

8

Method to find EB

EB ( in unit J ) EB ( in unit MeV )
Δm ( in unit kg ) Δm ( in unit u )

c = 3.00 x 108 m s-1 c2  931.5 MeV

Example: u

Let Δm = 1 u = 1.66 x 10-27 kg

EB = Δmc2 EB = Δmc2

=(1.66×10-27kg)(3.00×108ms-1)2 @ =(1 u)  931.5 MeV 
 u 
=1.4904×10-10kgm2s-2

=1.4904×10-10 J = 931.5 MeV

Note : 1eV = 1.6 x 10 -19J 9

Example 9.1.2:

Calculate binding energy of the Helium , 4 He in SI unit.
2

(Given mass of helium atom = 4.002603 u, mH= 1.007825 u
& mn= 1.008665 u)

∆ = + −

= 2 1.007825 + 2 1.008665 − 4.002603
= 0.030377

10

Example 9.1.2: 2 METHODS TO FIND EB

 Method (i):

Find EB in unit MeV then convert to Joule.

= ∆ 2 931.5

= 0.030377

= 28.30

= 28.30 × 106 × 1.6 × 10−19 = . × −

 Method (ii):

Convert Δm to unit kg then find EB in unit Joule.
∆ = 0.030377 × 1.66 × 10−27 = 5.0426 × 10−29

= ∆ 2
= 5.0426 × 10−29 3.00 × 108 2

= . × − 11

Example 9.1.3:

Calculate
a) mass defect and
b) binding energy of the deuterium, 2 .
1 H

(Given mass nucleus deuterium 2.013553 u, mp= 1.007277 u
& mn=1.008665 u)

Solution:

a) ∆ = + −
= 1 1.007277 + 1 1.008665 − 2.013553

= 0.002389

b) = ∆ 2 931.5

= 0.002389
= 2.23
12

9.1 (c) Binding Energy per Nucleon,



Definition: Formula:

Binding energy per = ∆ 2
nucleon number.
*to measure the
stability of the nucleus.

*The larger the binding energy per nucleon, the greater
the work must be done to remove the nucleon from
the nucleus, the more stable the nucleus.

** Its unit is MeV/nucleon @ J/nucleon.

13

Example 9.1.4:

Calculate the average binding energy per nucleon of the
 nucleus iron-56 .
56 Fe
26

Given Mass nucleus 2566 = 55.93494
mp= 1.007277 u
mn= 1.008665 u

∆ = + − − 55.93494

= 26 1.007277 + 30 1.008665
= 0.514212

14

Binding energy per nucleon = ∆ 2



0.514212 ((931.5 )/ )
= 56
= 8.55 /

15

Example 9.1.5:

23 Mg has an atomic mass of 22.994127 u. Calculate its
12
binding energy per nucleon.

Given mH= 1.007825 u
mn= 1.008665 u

∆ = + −
= 12 1.007825 + 11 1.008665 − 22.994127
= 0.195088

16

Binding energy per nucleon = ∆ 2



0.195088 ((931.5 )/ )
= 23
= 7.90 /

17

9.1 (d) Graph of Binding Energy per
Nucleon against Nucleon Number

From the graph:
 Hydrogen-1 has zero binding energy because it

has only one nucleon.
 The EB /A increases rapidly with increasing A for

light nuclei and reaches a maximum at iron-56
(Fe).
 The Iron-56 (Fe) is the most stable nucleus
(maximum EB /A: about 8.8 MeV / nucleon).

18

 Nuclei with nucleon number 50 ≤ A ≤ 80 have
the highest EB /A are most stable (almost 9.0 MeV).

 Value of EB /A decreases slowly for A > 62
(nucleus less tightly bound) as the nuclei get
heavier.

 For heavy nuclei (A = 200 to 240) have EB /A
which are smaller ranges between 7 – 7.5 MeV.
These nuclei are unstable and radioactive.

19

Greatest stability Binding energy per nucleon as a
function of mass number,A

Binding energy per nucleon (MeV/nucleon) Area of maximum
stability ( 50 ≤ A ≤ 80 )

EB

/A

Mass number, A 20

21

Exercise 9.1:

1) The binding energy of the neon 20 Ne is 160.64 MeV.
Find its nucleus mass. 10

Given mp= 1.007277 u (Ans: 19.99 u)
mn= 1.008665 u

2) Determine the total binding energy and the binding
energy per nucleon for the nitrogen -14 nucleus
Given

147 = 14.003074 u
11 = 1.007825 u
10 = 1.008665 u

(Ans:104.66 MeV, 7.48 MeV/nucleon)

22

3) Calculate the binding energy of an aluminum nucleus, 27 Al in
13
MeV.

(Given mass of neutron, mn=1.008665 u ; mass of hidrogen,
mH=1.007825 u ; speed of light in vacuum, c=3.00108 m s1
and atomic mass of aluminum, MAl=26.98154 u)

(Ans: 224.95 MeV)

4) Calculate the binding energy per nucleon of a

boron, 11 Bo nucleus in J/nucleon.
5
(Given mass of neutron, mn=1.008665 u ; mass of proton,
mp=1.007277 u ; speed of light in vacuum, c=3.00108 m s1
and nucleus mass of boron, MB=10.01294 u)

(E = 1.46x10 -11 J/nucleon)

23

5) Why is the uranium-238 nucleus is less stable than
carbon-12 nucleus? Give an explanation by referring to
the binding energy per nucleon.

(Given mass of neutron, mn=1.008665 u ; mass of
hidrogen, mH=1.007825 u ; speed of light in vacuum,
c=3.00108 m s1; atomic mass of carbon-12,
MC=12.00000 u and atomic mass of uranium-238,
MU=238.05079 u )

(Ans: U think)

24

9.2 Radioactivity

At the end of this topic, the student should be able to :

a) Explain , +, - and  decays.

b) State and use decay law, dN  N .

dt

c) Define and determine activity, A and decay

constant, .
d) Use N  Noet or A  Aoet .

e) Define and use half-life, T1  ln 2 .

2 

25

Properties of , , 

Three kinds of Alpha particle () Beta particle () Gamma rays ()
radioactive
emissions Helium Electron E.W
( 4 ) ( 10e @ 0 ) ()
The nature of 2 He 1 e 0 (uncharged)
radioactive Yes
emissions +2e 1e @ +1e
Charge
Ability to affect a Yes Yes
photographic plate
Ability to produce Yes Yes Yes
fluorescence
Strong Moderate Weak
Ionization power  have a lowest
 have a relatively  have a lower ionizing power
because they have
large mass & charge ionizing power than no mass & charge.
produce the ion particles because
pairs in medium, of their smaller mass 26
hence they have & charge.
large ionizing power.

Properties of , , 

-Weak -Moderate -Strong

 lowest penetrating  have strong penetrating  have the highest
power & can stopped power & can pass through a penetrating power &
by a sheet of paper. sheet of paper but can be it is the most difficult
stopped by a few mm of to be stopped.
aluminum sheet.
27

Properties of , , 

Three kinds Alpha particle () Beta particle () Gamma
of rays ()

radioactive
emissions

Deflection Yes Yes No
by electric  rays
and  particles are  negatively charged
magnetic which are
fields positively charged & are particles are attracted to neutral
attracted towards the the positive plate. are not
negative plate. deflected.

  and particles are deflected in opposite direction. 28
The deflection of particles is less than particles
because particles are much heavier than particles.

9.2 (a) ALPHA DECAY ()  Alpha decay usually happens to the heavier unstable nucleus.
 In Alpha decay, the nucleus of heavy radioactive elements emits an

particle.
 The nuclear equation for an decay can be written as :

A X  YA4  4 He
Z 2
Z 2

 In decay, the proton number is reduced by 2 while the nucleon

number is reduced by 4 (consists of 2 protons & 2 neutrons).

Example: Proton 
Neutron

10A 6 B  4 He  Q
4 2
6

Parent Daughter  particle

U238  23940Th + 4 He  Q
92 2

226 Ra  222 Rn+ 4 He+Q 29
88 86 2

 The decay shown can be represented by the equation :

9.2 (a) ALPHA DECAY () U238  23940Th  4 He
2
92

U238 : Parent

92

23940Th : Daughter

4 He :  particle
2

 It is positively charged particle and its value is +2e with mass of

4.001506 u.

 It is produced when a heavy nucleus (Z > 82) decays.

 Alpha particles can penetrate a sheet of paper.

 The energy released appears in the form of kinetic energy in the

daughter nucleus and the alpha particle which is given by

  Q  mx  mY  m c2 If Q < 0, decay
could not occur
30

9.2 (a) BETA DECAY ()

 In  decay, the nucleus emits a  particle (+ and ) that has high velocity (v ~ c).

 It has the same mass as electron or 0.000549 u.

 There are two kinds of particles : Beta minus () & Beta plus (+ ).
 Beta minus () known as Beta particle & Beta plus (+ ) known as positron.

Beta minus , β Beta plus , β+

 Also called as Negatron OR Electron.  Also called as Positron OR

 Symbol :  OR 01β OR 0 e AntiElectron. 10e
1
 Symbol : + OR 01β OR

 Negatron decay Positron decay
no. protons > no. neutron
no. neutrons > no. protons

 It is produced when one of the  It is produced when one of the
protons in the parent nucleus decays
neutrons in the parent nucleus decays into a neutron, a positron and a
into a proton, an electron and an neutrino.
antineutrino. 31

Beta minus , β

• Example

23940Th  234 Pa  -10e Q
91

234 Pa  U234  0 e  Q
91 -1
92

Beta plus , β+

• Example

12 N 162C  10e  v  Q
7

32

9.2 (a) BETA DECAY ()

Beta minus , β Beta plus , β+

 Electron is emitted,  Positron is emitted,
 the mass number does not change  the change of the parent nucleus

but the change of the parent decreases.
nucleus increases.

 General equation : antineutrino  General equation : neutrino

A X  Z+A1Y+ 01β v A X  ZA1Y+ 01β  v
Z Z

parent Daughter  particle parent Daughter  particle

 v is called antineutrino  v is called neutrino

 an elementary particle that exist to  an elementary particle that exist to

account the missing energy in account the missing energy in
negatron decay. positron decay.

33

Beta minus , β
 When an unstable nucleus has too many neutrons, it will decay with β

emission.
 During a β emission, a neutron in the nucleus breaks up into a proton

and an electron.
 The electron is emitted from the nucleus as a β particle, whereas the
proton is retained in the nucleus, as shown in the following equation.

Massless,
neutral

34

Beta plus , β+

 When an unstable nucleus has too many protons, it will decay with β+

emission.

 During a β+ emission, a proton, 11H in the nucleus breaks up into a

neutron 01n and a positron 0 e .
1

 The positron is emitted from the nucleus as a β+ particle, and the neutron

is retained in the nucleus, as shown in the following equation.

Massless,
neutral

35

Conclusion

Negatron decay, β- Example: Proton
Neutron
deNcaeygawtirlol nha(p-poenr 10e )
when the number of 7A7 B10 e v
neutrons are more than 3
the number of protons 2
in a nucleus.
 0 e  v

1

It is produced when Neutron  proton
one of the neutrons in +
the parent nucleus
decays into a proton, electron
an electron and an +
antineutrino.
v

36

Conclusion

Positron decay, β+ Example: Proton
Neutron

Positron h(a+ppoern10 e ) 7 A47 B 10e 
decay will 5

when the number of   0 e v
protons is more than 1
the number of neutrons
in a nucleus.

It is produced when one Proton  neutron
of the protons in the +
parent nucleus decays into
a neutron, a positron and positron
a neutrino. +

v

37

9.2 (a) GAMMA DECAY ()

 Gamma decay occurs when an unstable nucleus releases its
energy in the form of high frequency electromagnetic waves

called rays.
 In -decay, a photon (-ray) is emitted when the excited nucleus

changes from a higher level energy state to a lower level. The
wavelengths of the electromagnetic radiation are shorter than
1010 m.
 Gamma rays photon are emitted when an excited nucleus in an
excited state makes a transition to a ground state. This will
happen when the nucleus decays into alpha or beta particles.

(A nucleus that undergoes  or  decay may also emit rays).

 Thus gamma-ray emission often associates with other type of
decays.

38

9.2 (a) GAMMA DECAY ()

 During gamma decay, the mass number and the atomic number of the
nucleus remains the same.

 The nuclide does not change after a gamma decay. (There no change
in the proton number and nucleon number).

 Gamma decay equation : A*A
Z
X  X  Z

39

 General equation for gamma decay:

A X*  A X  
Z Z
where X* = excited nucleus

X = stable nucleus

 The asterisk (*) indicates that the nucleus is in an excited
state.

 Examples of -decay;

218 Po 28124Pb 24He  γ Gamma ray
84

234 Pa 234 U  10e  γ
91 92

20881Ti 20881Ti  γ
 -ray is uncharged (neutral) ray and zero mass.

40

Example 9.2.1:

Write equations to represent the following radioactive decay.

a) 29p328hUotodenc.ays by emitting an alpha-particle and a gamma

b) 32 P decays by beta-emission.
15

c) 64 Cu decays by positron-emission.
29

You may use X, Y and Z to represent the daughter nuclides.

Solution 9.2.1:
a)
238 U  234 X+ 4 He+γ
92 90 2

b) P32  3126Y+ 0 e
1
15

c) 64 Cu  64 Z+ 0 e
29 28 1

41

Example 9.2.2:

2w8131hBerie decays to Po according to the equation 211 Bi  Po  β  γ
 is a -particle and  is a -ray photon. 83

proton number (atomic number) of Po? What is the

A. 82 B. 84 C. 207 D. 212
Ans; B : In - emission, proton number increases by 1.

Example 9.2.3:

Complete the following radioactive decays and identify the
radiations emitted.
234 e0
a) 23940Th  91 Pa   (-particle)
-1

b) 14 C 147 N -01 e + γ (-particle & -ray)
6 Ne +01 e (positron)

c) 22 Na 2120 42
11

Exercise 9.2:

1. The nucleus of uranium 23 8 undergoes successive decays,

emitting an α-particle, a β-particle and a γ-photon. What is
the resultant nucleus?

A. 23 8 B. 23 4 C. 23 3 D. 23 4

2. A radioactive isotope undergoes decay with the emission of

β-particle. The decay produces 11418 . The isotope is:

A. 1 1 1 B. 1 1 3 C. 1 1 1 D. 1 1 2

43

9.2 (b) DECAY LAW

State that: Formula:

ddeirceacytlyrapterop odrdtNtioniasl  dN   N dN  N  At
to the number of  dt  dt
radioactive nuclei /
Negative sign means
atom N remaining in the number of nuclei
present decreases
the source. with time

44

9.2 (c) ACTIVITY (A)

Definition: Formula:

the disintegrations Rate of dN  At
per second by the Decay dt
radioactive nucleus. known as
(1 Bq = 1 decay/s)
(1 Ci (curie) =3.7 × 1010 Bq) Activity, A

of the
sample

Unit is the Becquerel, (Bq)
 1 Bq is the rate of decay of 1 nucleus

per second

45

9.2 (c) DECAY CONSTANT ()

Definition: Formula:

Decay constant of a dN  N  At
nuclide is define as dt
the probability that
a radioactive atom
will decay in one decay constant
second.
dN

   dt

N
• The larger , the greater rate of
decay.

• Different nuclides, different values ** Its unit is s-1.

for .

46

9.2 (d) USE = −λ & = −λ

Formula: Formula:

dN
 N t
NN0   A  N and N  N0et
dt
0  A   N0et

ln N N  t t   N0 et
N0
0 and A0  N0

ln N  λt A  A0eλt
N0

N  N0eλt

Nuclei remain Nuclei at Activity at time t Activity at

at time t time, t =0 time, t =0

9.2 (e) HALF LIFE ( ൗ )

Definition: Formula:

the time required for the At t = 1Τ2and N = N0 / 2
number of radioactive
nuclei to decrease to half from N  Noeλt
of the original number of
nuclei. N  N e1  λT1/2
o
**The units of the ൗ are 2o
second (s), minute (min), hour
(hr), day and year (yr).  e1 λT1/2
**Its depend on the unit of the
decay constant. 2

e λT1/2 2

λT1/2  ln 2

T1  ln 2

2  48

 The half-life of any given radioactive nuclide is
constant, it does not depend on the number of
nuclei present.

Table 11.1 Isotope Half-life
shows the 4.5  109 years
value of half- 238 U 1.6  103 years
life for several 92
isotopes. 138 days
226 Ra 24 days
88 3.8 days
20 minutes
210 P o
884

23940Th

222 Rn
86

214 Bi
83

49