is defined as the
central core of an
atom that is
positively
charged and
contains protons
and neutrons.
CHAPTER 7: Nucleus
(3 Hours)
Dr Ahmad Taufek Abdul RahmanSchool of Physics & Material Studies, Faculty of Applied Sciences, Universiti Teknologi MARA Malaysia, Campus of Negeri Sembilan
1
At the end of this chapter, students should be able to:
State the properties of proton and neutron.
Define
proton number
nucleon number
isotopes
Use
to represent a nucleus.
Explain the working principle and the use of mass
spectrometer to identify isotopes.
Learning Outcome:
7.1 Properties of nucleus (1 hour)
XAZ
2
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
7.1.1 Nuclear structure
A nucleus of an atom is made up of protons and neutrons that
known as nucleons (is defined as the particles found inside
the nucleus) as shown in Figure 7.1.
7.1 Properties of nucleus
Proton
Neutron
Electron
Figure 7.1
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Proton and neutron are characterised by the following properties
in Table 7.1.
For a neutral atom,
The number of protons inside the nucleus
= the number of electrons orbiting the nucleus
This is because the magnitude of an electron charge
equals to the magnitude of a proton charge but opposite
in sign.
Proton (p) Neutron (n)
Charge (C)
Mass (kg)
+e 0
2710672.1 2710675.1
)1060.1( 19 )uncharged(
Table 7.1
4
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Nuclei are characterised by the number and type of nucleons they contain as shown in Table 7.2.
Any nucleus of elements in the periodic table called a nuclide is
characterised by its atomic number Z and its mass number A.
The nuclide of an element can be represented as in the Figure 7.2.
Table 7.2
Number Symbol Definition
Atomic number Z The number of protons in a nucleus
Neutron number N The number of neutrons in a nucleus
Mass (nucleon)
numberA The number of nucleons in a nucleus
Relationship : (7.1)
5
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
The number of protons Z is not necessary equal to the number
of neutrons N.e.g. :
Figure 7.2
Atomic number
Mass number
Element X
Mg24
12 S32
16; Pt195
78;
21Z12 ZAN 6
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Since a nucleus can be modeled as tightly packed sphere
where each sphere is a nucleon, thus the average radius of
the nucleus is given by
31
0 ARR (7.2)
where nucleus of radius average :Rfm 1.2 OR m102.1constant : 15
0
Rnumber (nucleon) mass :A
femtometre (fermi)
m 101fm 1 15
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Based on the periodic table of element, Write down the symbol of nuclide for following cases:
a. Z =20 ; A =40
b. Z =17 ; A =35
c. 50 nucleons ; 24 electrons
d. 106 nucleons ; 48 protons
e. 214 nucleons ; 131 protons
Solution :
a. Given Z =20 ; A =40
c. Given A=50 and Z=number of protons = number of electrons =24
Example 1 :
XA
ZCa40
20
XA
ZCr50
24
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
What is meant by the following symbols?
State the mass number and sign of the charge for each entity above.
Solution :
Example 2 :
n1
0 p1
1; e0
1;
n1
0
p1
1
e0
1
Neutron ; A=1
Charge : neutral (uncharged)
Proton ; A=1
Charge : positively charged
Electron ; A=0
Charge : negatively charged
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Complete the Table 7.3.
Example 3 :
Table 7.3
Element
nuclide
Number of
protons
Number of
neutrons
Total charge
in nucleus
Number of
electrons
H1
1
N14
7
Na23
11
Co59
27
Be9
4
O16
8
S31
16
Cs133
55
U238
92
16 15 16e 16
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
is defined as the nuclides/elements/atoms that have the
same atomic number Z but different in mass number A.
From the definition of isotope, thus the number of protons or electrons are equal but different in the number of neutrons
N for two isotopes from the same element.
For example :
Hydrogen isotopes:
Oxygen isotopes:
7.1.2 Isotope
H1
1
H2
1
H3
1
: Z=1, A=1, N=0
: Z=1, A=2, N=1
: Z=1, A=3, N=2
proton )p(1
1
deuterium )D(2
1
tritium )T(3
1
O16
8
O17
8
O18
8
: Z=8, A=16, N=8
: Z=8, A=17, N=9
: Z=8, A=18, N=10
equal
equal not equal
not equal
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Mass spectrometer is a device that detect the presence of
isotopes and determines the mass of the isotope from known
mass of the common or stable isotope.
Figure 7.3 shows a schematic diagram of a Bainbridge mass
spectrometer.
29.1.3 Bainbridge mass spectrometer
Figure 7.3
+-
++++
++
----
--
2B
E
1B
1r2r
S1S2
S3
Photographic plate
Plate P1 Plate P2
Ion source
Ions beam
Evacuated
chamber
1m 2m
d
Separation
between isotopes
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Plate P1 Plate P2
Working principle
Ions from an ion source such as a discharge tube are narrowed to a fine beam by the slits S1 and S2.
The ions beam then passes through a velocity selector (plates
P1 and P2) which uses a uniform magnetic field B1 and a uniform
electric field E that are perpendicular to each other.
The beam with selected velocity v passes through the velocity
selector without deflection and emerge from the slit S3. Hence,
the force on an ion due to the magnetic field B1 and the electric
field E are equal in magnitude but opposite in direction (Figure
7.4).
v
EF
BF
Using Fleming’s
left hand rule.Figure 7.4 13
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Thus the selected velocity is
The ions beam emerging from the slit S3 enter an evacuated
chamber of uniform magnetic field B2 which is perpendicular to
the selected velocity v. The force due to the magnetic field B2
causes an ion to move in a semicircle path of radius r given by
EB FF
qEqvB 90sin1
1B
Ev
cB FF
r
mvqvB
2
2 90sin
qB
mvr
2
and
1B
Ev
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Since the magnetic fields B1 and B2 and the electric field E are
constants and every ion entering the spectrometer contains the
same amount of charge q, therefore
If ions of masses m1 and m2 strike the photographic plate with
radii r1 and r2 respectively as shown in Figure 7.3 then the ratio
of their masses is given by
kmr and constant21
qBB
Ek
mr
2
1
2
1
r
r
m
m (7.4)
qBB
mEr
21
(7.3)
15
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
A beam of singly charged ions of isotopes Ne-20 and Ne-22 travels
straight through the velocity selector of a Bainbridge mass
spectrometer. The mutually perpendicular electric and magnetic
fields in the velocity selector are 0.4 MV m1 and 0.7 T respectively.
These ions then enter a chamber of uniform magnetic flux density
1.0 T. Calculate
a. the selected velocity of the ions,
b. the separation between two isotopes on the photographic plate.
(Given the mass of Ne-20 = 3.32 1026 kg; mass of Ne-22 =
3.65 1026 kg and charge of the beam is 1.60 1019 C)
Solution :
a. The selected velocity of the ions is
Example 4 :
T 0.1 T; 7.0;m V 104.0 2116 BBE
1B
Ev
7.0
104.0 6v
15 s m 1071.5 v16
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Solution :
b. The radius of the circular path made by isotope Ne-20 is
and the radius of the circular path made by isotope Ne-22 is
Therefore the separation between the isotope of Ne is given by
T 0.1 T; 7.0;m V 104.0 2116 BBE
qBB
Emr
21
11
m 119.0
19
626
11060.10.17.0
104.01032.3
r
m 130.0
19
626
21060.10.17.0
104.01065.3
r
12 ddd 12 22 rrd
122 rr
119.0130.02
m 102.2 2d
Figure 7.3
17
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
At the end of this chapter, students should be able to:
Define mass defect and binding energy.
Use formulae
Identify the average value of binding energy per
nucleon of stable nuclei from the graph of binding
energy per nucleon against nucleon number.
Learning Outcome:
7.2 Binding energy and mass defect (2 hours)
2mcE
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
7.2.1 Einstein mass-energy relation
From the theory of relativity leads to the idea that mass is a
form of energy.
Mass and energy can be related by the following relation:
e.g. The energy for 1 kg of substance is
7.2 Binding energy and mass defect
2mcE (7.5)
where
massrest : m)s m10003( in vacuumlight of speed: 18 .c
energy ofamount : E
J1000.9 16E
2mcE 28)1000.3)(1(
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Unit conversion of mass and energy
The electron-volt (eV)
is a unit of energy.
is defined as the kinetic energy gained by an electron in being accelerated by a potential difference (voltage) of 1 volt.
The atomic mass unit (u)
is a unit of mass.
is defined as exactly the mass of a neutral carbon-12 atom.
J1060.1eV 1 19
J 10601eV 10Me 1 136 .V
12
1
12
Cof massu 1
126
kg10661u 1 27 .
20
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
1 atomic mass unit (u) can be converted into the unit of
energy by using the mass-energy relation (eq. 7.5).
in joule,
in eV/c2 or MeV/c2,
J 1049.1 10E
2mcE 2827 )1000.3)(1066.1(
J 1049.1u 1 10
26
19
10
eV/ 105.9311060.1
1049.1 cE
26 eV/ 105.931u 1 c
2MeV/ 5.931u 1 c
OR
21
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
The mass of a nucleus (MA) is always less than the total mass of its constituent nucleons (Zmp+Nmn) i.e.
Thus the difference in this mass is given by
where m is called mass defect and is defined as the mass difference between the total mass of the constituent nucleons and the mass of a nucleus.
The reduction in mass arises because the act of combining the nucleons to form the nucleus causes some of their mass to be released as energy.
Any attempt to separate the nucleons would involve them being given this same amount of energy. This energy is called the binding energy of the nucleus.
7.2.2 Mass defect
np NmZmM A where
neutron a of mass: nmproton a of mass: pm
AMNmZmm np(7.6)
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
The binding energy of a nucleus is defined as the energy
required to separate completely all the nucleons in the
nucleus.
The binding energy of the nucleus is equal to the energy
equivalent of the mass defect. Hence
Since the nucleus is viewed as a closed packed of nucleons,
thus its stability depends only on the forces exist inside it.
The forces involve inside the nucleus are
repulsive electrostatic (Coulomb) forces between protons and
attractive forces that bind all nucleons together in the nucleus.
7.2.3 Binding energy
2B cmE (7.7)Binding energy
in jouleSpeed of light in
vacuumMass defect in kg
7.2.4 Nucleus stability
23
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
These attractive force is called nuclear force and is responsible
for nucleus stability.
The general properties of the nuclear force are summarized
as follow :
The nuclear force is attractive and is the strongest force
in nature.
It is a short range force . It means that a nucleon is
attracted only to its nearest neighbours in the nucleus.
It does not depend on charge; neutrons as well as protons
are bound and the nuclear force is same for both.
e.g.
The nuclear force depends on the binding energy per
nucleon.
proton-proton (p-p)
neutron-neutron (n-n)
proton-neutron (p-n)
The magnitude of nuclear
forces are same.
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Note that a nucleus is stable if the nuclear force greater than
the Coulomb force and vice versa.
The binding energy per nucleon of a nucleus is a measure of
the nucleus stability where
Figure 7.5 shows a graph of the binding energy per nucleon as
a function of mass (nucleon) number A.
)number(Nucleon
)(energy Bindingnucleonper energy Binding B
A
E
A
mc 2
nucleonper energy Binding
(7.8)
25
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Mass number A
Bin
din
g e
ne
rgy p
er
nu
cle
on
(M
eV
/nu
cle
on
)
Greatest stability
Figure 7.526
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
From Figure 7.5,
The value of EB/A rises rapidly from 1 MeV/nucleon to 8
MeV/nucleon with increasing mass number A for light nuclei.
For the nuclei with A between 50 and 80, the value of EB/Aranges between 8.0 and 8.9 Mev/nucleon. The nuclei in
these range are very stable. The maximum value of the
curve occurs in the vicinity of nickel, which has the most
stable nucleus.
For 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 nuclei
are unstable and radioactive.
Figure 7.6 shows a graph of neutron number N against atomic
number Z for a number of stable nuclei.
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Atomic number Z
Ne
utr
on
nu
mb
er,
N
N=Z
Line of
stability
Figure 7.628
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
From Figure 7.6,
The stable nuclei are represented by the blue dots, which lie
in a narrow range called the line of stability.
The dashed line corresponds to the condition N=Z.
The light stable nuclei contain an equal number of
protons and neutrons (N=Z) but in heavy stable nuclei
the number of neutrons always greater than the number
of protons (above Z =20) hence the line of stability
deviates upward from the line of N=Z.
This means as the number of protons increase, the
strength of repulsive coulomb force increases which
tends to break the nucleus apart.
As a result, more neutrons are needed to keep the
nucleus stable because neutrons experience only the
attractive nuclear force.
29
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
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=3.00108 m s1 and
atomic mass of aluminum, MAl=26.98154 u)
Solution :
The mass defect of the aluminum nucleus is
Example 5 :
Al2713
Al2713 13Z and 1327N
14N
u 2415.0m
98154.2600867.11400782.113
Alnp MNmZmm
30
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Solution :
The binding energy of the aluminum nucleus can be calculated by
using two method.
1st method:
Thus the binding energy in MeV is
271066.12415.0 m
kg 100089.4 28
kg 10661u 1 27 .
2828
B 1000.3100089.4 E
J 10608.3 11
B
E
2
B cmE
in kg
MeV 226B E
13
11
B1060.1
10608.3
E
J 10601MeV 1 13 .
31
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Solution :
2nd method: 2
B cmE
MeV 225B E
22
u 1
MeV/ 5.931c
cm
2MeV/ 5.931u 1 c
in u
22
u 1
MeV/ 5.931u 2415.0 c
c
32
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
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=3.00108 m s1 and
atomic mass of boron, MB=10.01294 u)
Solution :
The mass defect of the boron nucleus is
Example 6 :
B105
B10
5 5Z and 510N5N
u 06951.0m
01294.1000867.1500782.15
Bnp MNmZmm
33
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Solution :
The binding energy of the boron nucleus is given by
Hence the binding energy per nucleon is
2827 1000.31066.106951.0
J 1004.1 11
B
E
2
B cmE
J/nucleon 1004.1 12B A
E10
1004.1 11
B
A
E
34
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Why is the uranium-238 nucleus less stable than carbon-12
nucleus ? Give an explanation by referring to the repulsive
coulomb force and 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=3.00108 m s1; atomic
mass of carbon-12, MC=12.00000 u and atomic mass of uranium-
238, MU=238.05079 u )
Solution :
From the aspect of repulsive coulomb force :
Uranium-238 nucleus has 92 protons but the carbon-12 nucleus has only 6 protons.
Therefore the coulomb force inside uranium-238 nucleus
is or 15.3 times the coulomb force inside carbon-12
nucleus.
Example 7 :
C126
U23892
6
92
35
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Solution :
From the aspect of binding energy per nucleon:
Carbon-12 :
The mass defect :
The binding energy per nucleon:
C12
66Z and 6N
Cnp MNmZmm
u 09894.0m
00000.1200867.1600782.16
A
cm
A
E 2
C
B
12
u 1
MeV/ 5.931u 09894.0 2
2
cc
nMeV/nucleo 68.7C
B
A
E
36
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Uranium-238 :
The mass defect :
The binding energy per nucleon:
Since the binding energy of uranium-238 nucleus less than the binding energy of carbon-12 and the coulomb force inside uranium-238 nucleus greater than the coulomb force inside carbon-12 nucleus therefore uranium-238 nucleus less stable than carbon-12 nucleus.
U238
9292Z and 146N
u 93447.1m
05079.23800867.114600782.192 m
238
u 1
MeV/ 5.931u 93447.1 2
2
U
B
cc
A
E
nMeV/nucleo 57.7U
B
A
E
37
DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Exercise 7.1 :Given c =3.00108 m s1, mn=1.00867 u, mp=1.00782 u
1. Calculate the binding energy in joule of a deuterium nucleus.
The mass of a deuterium nucleus is 3.34428 1027 kg.
ANS. : 2.781013 J
2. The mass of neon-20 nucleus is 19.99244 u. Calculate
the binding energy per nucleon of neon-20 nucleus in MeV
per nucleon.
ANS. : 8.03 MeV/nucleon
3. Determine the energy required to remove one neutron from an
oxygen-16 . The atomic mass for oxygen-16 is
15.994915 u
(Physics, 3rd edition, James S. Walker, Q39, p.1108)
ANS. : 15.7 MeV
Ne2010
O16
8
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DR.ATAR @ UiTM.NS PHY310 NUCLEUS
Next Chapter…CHAPTER 8 :
RADIOACTIVITY
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