### 26.1 Properties of nucleus

```UNIT 26 : NUCLEUS
(2 HOURS)
is defined as the
central core of an
atom that is
positively charged
and contains
protons and
neutrons.
26.1 Properties of nucleus
26.2 Binding energy and mass defect.
1
At the end of this topic, students should be able to:



State the properties of proton and neutron
Define
◦ Proton number
◦ Nucleon number
◦ Isotopes
Use
to represent a nuclide
2
26.1 Properties of nucleus
• A nucleus of an atom is made up of protons and
neutrons that is also known as nucleons.
Figure 26.1.1( atom)
Figure 26.1.2 (nucleus)
3
Proton
 Particle with positive charge of the nucleus
 Charge : +1.60 x 10-19 C
 Mass : 1.672 x 10-27 kg / 1.007276 u
Neutron
 Particle with no charge of the nucleus
 Charge :  Mass : 1.675 x 10-27 kg / 1.008665 u
4
Proton number
 Definition: the number of protons in the nucleus.
 Also called as atomic number
 Symbol : Z
Nucleon number
 Definition : the total number of neutrons and protons in
the nucleus.
 Also called as atomic mass number
 Symbol : A
Isotope
 Definition : the atoms of the same element whose nuclei
contain the same number of protons (Z) but different
number of neutrons (N).
1
2
3
 Example : 1 H , 1 H , 1 H
(Hydrogen, deuterium, tritium)
5
 The atomic nucleus can be represented as
A
Z
where
X
X = symbol for the element
Z = atomic number (number of protons)
A = atomic mass number
= total number of protons and neutrons
Example :
56
26
Fe
Element : Iron-56
Proton no, Z = 26
Nucleon no, A = 56
Neutron = 56-26 = 30
A-Z=N
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Example 26.1
Complete the table below:
Element
nuclide
1
1
9
4
Number of
neutrons
Number of
electrons
H
Be
14
7
N
16
8
O
23
11
59
27
Na
31
16
133
55
Number
of protons
8
8
8
Co
S
Cs
238
92
U
7
At the end of this topic, students should be able to:



Define and determine mass defect
Define and determine binding energy,
Identify the average value of binding energy per nucleon
of stable nuclei from the graph of binding energy per
nucleon against nucleon number.
8
Definition
the difference between the sum of the masses of individual
nucleons that form an atomic nucleus and the mass of the
nucleus.
Formula
Δ m  Zm p  Nm n   M
A
m p : mass of a proton
M
A
 mass of a nucleus
m n : mass of a neutron
Z  number
of protons
N  number
of neutrons
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Example 26.2
From example above, can you determine the value of
mass defect ?
(Ans : 0.040475 a.m.u)
10
Definition
Energy required to separate a nucleus into its individual
protons and neutrons.
@ Energy released when nucleus is formed from its
individual nucleons.
Formula
Where
E : Binding energy
Δm : Mass defect
c : speed of light = 3.00 x 108ms-1
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
There are 2 methods to determine the value of Binding
Energy, EB
EB ( in unit MeV )
Δm ( in unit u )
EB ( in unit J )
Δm ( in unit kg )
c = 3.00 x 108ms-1
2
c =
931.5 M eV
u

Example :
Let Δm = 1 u = 1.66 x 10-27kg
E B = Δmc
2
= (1.66× 10
E B = Δmc
-27
8
-1 2
kgm s
 9 3 1 .5 M eV 
= (1 u ) 

u


J
= 9 3 1 .5 M eV
kg)(3.00× 10 m s )
= 1.4904 × 10
-10
= 1.4904 × 10
-10
2
2 -2
=
Note : 1eV = 1.6 x 10-19J
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Example 26.3
Calculate
a) mass defect and
b) binding energy of the deuterium.
2
Given
H m ass  2 .0 1 3 5 5 3 u
1
Solution:
13
Example 26.4
4
2
Calculate binding energy of the Helium nucleus, H e
in SI unit.
Given mass of helium atom = 4.002603 u
Solution:
14
EB
N
Definition
mean (average) binding energy of a nucleus
Binding energy per nucleon 
Binding energy ( EB )
Nucleon number( A)
Binding energy per nucleon 
mc
2
A
Binding energy per nucleon is measure the stability of of the
nucleus.
 The greater the binding energy per nucleon, the more
stable the nucleus is.
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Binding energy per nucleon (MeV/nucleon)
Greatest stability
Binding energy per nucleon as
a function of mass number,A
Mass number A
16
From the graph:





For light nuclei the value of EB/A rises rapidly from 1
MeV/nucleon to 8 MeV/nucleon with increasing mass
number A.
For the nuclei with A between 50 and 80, the value of
EB/A ranges between 8.0 and 8.9 Mev/nucleon. The
nuclei in these range are very stable.
62
The nuclide 28 Ni has the largest binding energy per
nucleon (8.7945 MeV/nucleon).
For nuclei with A > 62, the values of EB/A decreases
slowly, indicating that the nucleons are on average, less
tightly bound.
For heavy nuclei with A between 200 to 240, the binding
energy is between 7.5 and 8.0 MeV/nucleon.These
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Example 26.5
Calculate the average binding energy per nucleon of the
iron-56 56 Fe .

Given
26
56
26
1
1

Fe mass  55.93494
u
H  1 p mass  1.00782 u
1
0
1
n mass  1.00867 u
Solution:
18
19
Exercise
20
1) The binding energy of the neon 10
Ne is160.64 MeV.
Find its atomic mass.
Given 1
p mass  1.007825 u
1
1
0
n mass  1.008665
u
(Ans: 19.992u)
2) Determine the total binding energy and the binding
energy per nucleon for the nitrogen -14 nucleus 14 N
7
Given 14

7
1
1
N mass  14.003074
u
H  1 p mass  1.007825
1
0
1
n mass  1.008665

u
u
(Ans:104.6 MeV,7.47 MeV/nucleon)
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3) Calculate the binding energy of an aluminum nucleus
in MeV.
(Given mass of neutron, mn=1.00867 u ; mass of proton,
mp=1.00782 u ; speed of light in vacuum, c=3.00108 m
s1 and atomic mass of aluminum, MAl=26.98154 u)
(Ans: 225 MeV)
4)
Calculate the binding energy per nucleon of a boron
nucleus in J/nucleon.
(Given mass of neutron, mn=1.00867 u ; mass of proton,
mp=1.00782 u ; speed of light in vacuum, c=3.00108 m
s1 and atomic mass of boron, MB=10.01294 u)
(E = 1.04x10 -12 J/nucleon)
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5) Why is the uranium-238 nucleus is less stable than carbon12 nucleus? Give an explanation by referring to the binding
energy per nucleon.
(Given mass of neutron, mn=1.00867 u ; mass of proton,
mp=1.00782 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)
The end…..
Next chapter : nuclear reaction
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