The nuclear atom
• More than 2000 years ago, the Greeks
suggested that matter was made up of very
tiny (small) particles which they called atoms.
What is inside the atom?
What is the structure of
an atom?
19.1 The atomic model
1
Models of an atom
Plum pudding model
19.1 The atomic model
2
Plum pudding model
• The plum pudding model of
the atom by J.J. Thomson,
who discovered the electron
in 1897, was proposed in
1904.
• The atom is composed of
electrons, surrounded by a
soup of positive charge to
balance the electron's
negative charge, like
negatively-charged "plums"
surrounded by positivelycharged "pudding".
• The model was disproved by
the 1909 gold foil experiment,
19.1 The atomic model
3
Geiger-Marsden scattering
experiment
• Experiment:
• a particles are made to hit the thin (super thin)
gold foil.
• Flashes of light will be observed when a particles
hit the zinc sulphide screen.
19.1 The atomic model
4
Geiger-Marsden scattering
experiment
• We expect all the a
particles can pass
through the gold foil.
Results:
Nearly all a particles pass
straight through the gold foil.
Some α-particles (about
1/8000) were scattered by
angles greater than 90∘and
very few even rebounded
back along original paths.
19.1 The atomic model
5
Rutherford’s Remark
• It was quite the most incredible event that has ever
happened to me in my life.
• It was almost as incredible as if you fired a 15-inch
shell at a piece of tissue paper and it came back and
hit you.
• On consideration, I realized that this scattering
backward must be the result of a single collision, and
when I made calculations I saw that it was impossible
to get anything of that order of magnitude unless you
took a system in which the greater part of the mass
of the atom was concentrated in a minute nucleus.
• It was then that I had the idea of an atom with a
minute massive centre, carrying a charge.
19.1 The atomic model
6
How can the heavy a particles
bounce back after hitting the
thin gold foil?
a particles – He2+
Explanation:
All positive charge of the
atom and most of the
mass were concentrated
in a tiny core called
nucleus.
The rest of the atom was
largely empty space.
19.1 The atomic model
7
How can the heavy a
particle bounce back after
hitting the thin gold foil?
• Explanation:
• Most of a particles
passed straight through
the empty space of the
gold atoms.
• Some come close to the
nucleus were repelled by
a strong electrostatic
force, so they were
deflected or bounced
back.
19.1 The atomic model
Bounced
back
deflected
8
9
Alpha
particles
+2 e
P
+Ze
Nucleus
• a-particle lose their K.E. on approaching the +ve charged
nucleus, being repelled by an electrostatic force.
• At P, distance of closest approach K.E. lost = P.E. due to
a-particle location in electric field of nucleus.
• a-particle is then ‘reflected’ away from nucleus and finally
acquires the same K.E. as it had initially. Collision is elastic.
19.1 The atomic model
Estimated Upper Limit of the
Size of a Nucleus

Path for glancing
collision
B
Path for
A
head-on collision
N
P
r
p
Centre
of nucleus
• PE = 2Ze2/40r
• At P, distance of nearest approach K.E. of a’s, ½mv2
= 2Ze2/40r (P.E.)
• hence an estimate of r which is upper limit to size of
nucleus.
19.1 The atomic model
10
1
Nuclear fission
11
When a heavy nucleus splits up,
 huge amount of energy is
released.
Two
typical nuclear fission reactions are:
89
1
+ energy released
U  01n144
Ba

Kr

3
56
36
0n
235
92
94
1
+ energy released
U  01n140
Xe

Sr

2
54
38
0n
235
92
The total mass of the nuclear products is slightly less than the
total mass of the original particles.
Energy is released and can be calculated by using Einstein’s
2)
mass-energy relation
(E
=
mc
19.1 The atomic model
3
Nuclear fusion
A huge amount of energy is also released
when two light nuclei join together to form
a heavy nucleus.
This process is called nuclear fusion.
19.1 The atomic model
12
Example of Nuclear Fusion
13
• Deuterium-Tritium Fusion Reaction.
2
1
H 13H 24He  01n  energy
The total mass of the nuclear products is slightly less than the
total mass of the original particles.
19.1 The atomic model
3
14
Nuclear fusion
2H
1
+ 31 H
4 He + 1 n
2
0
+energy
Since nuclei carry +ve charges, they repel
each other.
For fusion to occur, the 2 hydrogen nuclei
must approach each other with very high
speed.
 hydrogen gas of high temperature
(108 C)!
19.1 The atomic model
3
15
Nuclear fusion
Nuclear fusion occurs in the core of the Sun,
giving out heat and light. The reaction takes
place continuously for billions of years.
(Photo credit:
US NASA)
19.1 The atomic model
Nuclear Power – controlled fission 16
• The schematic diagram of a nuclear reactor is
shown below:
19.1 The atomic model
Fuel
• Enriched uranium is used as the
fuel (uranium dioxide).
• Natural uranium contains only
0.7% uranium-235, which does
not undergo fission in these
circumstances.
• Treatment is required to increase
the concentration to 3%.
19.1 The atomic model
17
Moderator
U  n
235
92
1
0
144
56
Ba  Kr 3 n + energy released
89
36
1
0
• The probability that fission caused by a high energy
neutron (fast moving neutron) is quite low.
• Use materials to slow down neutrons so that the
probability of causing a fission is significantly higher.
• These neutron slowing down materials are the so called
moderators.
19.1 The atomic model
18
19
Moderator
• These neutron slowing down materials are the so called
moderators.
m
u
At rest
M
Before collision
v
V
After collision
By momentum conservation: mu = mv + MV
By energy conservation: ½mu2 = ½mv2 + ½MV2
mM
2m
v
u
V
u
M m
M m
19.1 The atomic model
20
Moderator
v
mM
u
M m
v
V
After collision
• The choice of the moderator
– The atoms of an ideal moderator should have the
same mass as a neutron (M = m). So a neutron
colliding elastically with a moderator atom would
lose almost all its KE to the moderator atom.
– The moderator atoms should not absorb neutrons
but should scatter them instead.
– In practice, heavy water (D2O i.e. 2H2O) is chosen as
the moderator. (Heavy water is different from hard
water)
19.1 The atomic model
Control rods
• The control rods are made
of boron-steel, which
absorbs neutrons.
• They are raised and lowered
to vary the number of
neutrons to control the rate
of fission.
19.1 The atomic model
21
2
Chain reaction
When a uranium nucleus splits, 2 or 3
neutrons are emitted.
neutron
U-235
nucleus
splits
If these neutrons carry on splitting other
uranium nucleus...
19.1 The atomic model
22
2
23
Chain reaction
self-sustaining
 chain reaction
neutron
U-235
nucleus
splits
escapes
19.1 The atomic model
Chain reaction
• The fission neutrons enter the moderator and collide
with moderator atoms, transferring KE to these atoms.
• So the neutrons slow down until the average KE of a
neutron is about the same as that of a moderator
atom.
19.1 The atomic model
24
25
The fission neutrons could be absorbed by the U-238
nuclei without producing further fission.
The fission neutron could escape from the isolated block
of uranium block without causing further fission.
The critical mass of fuel is the minimum mass capable of
producing a self-sustaining chain reaction.
19.1 The atomic model
Two
typical nuclear fission reactions are: 26
89
1
+ energy released
U  01n144
Ba

Kr

3
56
36
0n
235
92
94
1
+ energy released
U  01n140
Xe

Sr

2
54
38
0n
235
92
• At the fission of U-235 on the average 2.5
neutrons are released but not all of these
cause fission.
• multiplication factor (k)
number of neutrons in a neutron " generation "
k
number of neutons in the preceding generation
19.1 The atomic model
number of neutrons in a neutron " generation "
k
number of neutons in the preceding generation
neutron
U-235
nucleus
splits
escapes
19.1 The atomic model
27
number of neutrons in a neutron " generation "
k
number of neutons in the preceding generation
• critical reactor
• The number of neutrons in the system is constant, i.e.
they cause the same number of fissions in every
second.
19.1 The atomic model
28
• k > 1, the system is supercritical
• k < 1, the system is subcritical
k can be varied by lowering or raising the
control rods
19.1 The atomic model
29
1
Nuclear power
b
Potential hazards
30
The nuclear waste remains radioactive
for thousands of years.
 serious handling and storage problems
Nuclear accidents lead to the leakage of
radiation.
 widespread and long-lasting disasters
(Photo credits: BREDL; IAEA)
19.1 The atomic model
31
1
Nuclear power
c
Controlled nuclear fusion
Fuel = H-2, H-3
(plentiful in sea water)
Waste product = He-4
(inert and non-radioactive)
(Photo credit: Princeton Plasma Physics Lab)
Cheaper, abundant and safer,
but not yet in practice
19.1 The atomic model
1
Nuclear power
d
Benefits and disadvantages
contentious
Solve energy
shortage crisis

19.1 The atomic model
Create serious
social and
environmental
problems 
32
1
Nuclear power
d
Benefits and disadvantages
Benefits

Solve energy shortage crisis

No fuel transportation problem

Cheaper than coal/oil for generating
power in most cases

Little environment pollution
19.1 The atomic model
33
1
Nuclear power
d
Benefits and disadvantages
34
Disadvantages

Accident  serious consequence

Expensive in maintaining safety
standards

Unnecessary as alternative energy
sources exist

Lead to widespread of nuclear weapons
19.1 The atomic model
35
19.1 The atomic model
Pre-lesson assignment
Einstein's mass-energy
relation (E = mc2)
1. According to Einstein’s mass-energy
equation E = mc2, what is the amount of
energy produced if 1 kg of a certain
element completely changes into energy?
Solution:
Energy released
= (1)(3 x 108)2
= 9 x 1016 J
19.1 The atomic model
36
37
Q2
90
1
1
U  01n144
Ba

Kr

n

56
36
0
0 n  energy
235
92
• Mass of
U-235:
3.9030 x 10-25 kg
Ba-144:
2.3899 x 10-25 kg
Kr-90:
1.4931 x 10-25 kg
Neutron:
1.6749 x 10-27 kg
• Calculate the nuclear energy released in the nuclear
fission.
Solution:
• Mass difference
= 3.9030 x 10-25 – (2.3899 x 10-25 + 1.4931 x 10-25 +
1.6749 x 10-27)
= 3.251 x 10-28 kg
• Energy released = (3.251 x 10-28 )(3 x 108)2
= 2.93 x 10-11 J
19.1 The atomic model
Nuclear Energy
1. unified atomic mass unit
2. unit of energy: eV
3. Binding energy
19.1 The atomic model
38
39
Atomic mass
Particle or
element
Proton (p)
Atomic mass / kg
1.673 x 10-27
1.00728u
Neutron (n)
1.675 x 10-27
1.00867u
Electron (e)
9.14 x 10-31
0.00055u
Hydrogen (H)
1.674 x 10-27
1.00794u
Helium (He)
6.646 x 10-27
4.00260u
Lithium (Li)
1.165 x 10-26
7.01601u
1 u = unified atomic mass unit
19.1 The atomic model
40
The unified atomic mass unit (u) is defined as one
twelfth of the mass of the carbon-12 atom.
History
Question:
How to express the unified atomic mass unit
(1u) in kg?
Solution:
Given that:
mass of a C-12 atom = 1.9927 x 10-26 kg
∴1u = 1.9927 x 10-26 kg /12
= 1.66 x 10-27 kg
19.1 The atomic model
Example 1
• Given that the atomic mass of a
hydrogen atom is 1.00783u. Find the
mass of a hydrogen atom.
• Solution
Mass of a hydrogen atom
= 1.00783 x 1.66 x 10-27
= 1.67 x 10-27 kg
19.1 The atomic model
41
Units of Energy
Very large unit of energy:
kWh
(electric bill)
S.I. unit: J
Very small unit of energy: eV
(Atomic physics)
Energy = charge Q x voltage V
19.1 The atomic model
42
Energy in eV
43
+
–
1.6 x 10-19 C
–
By W = QV
1 eV = (1.6 x 10-19)(1)
∴ 1 eV = 1.6 x 10-19 J
1V
1 eV is the energy (K.E.) gained by an electron when it is
accelerated through 1 V.
19.1 The atomic model
44
Energy in eV
+
1V
1.6 x 10-19 C +
–
1 eV is the work done in moving a charge of 1.6 x 10-19 C
through 1 V.
19.1 The atomic model
Example 1 page 25
45
• Show that 1 u of mass is equivalent to 931 MeV
by the mass-energy relation.
1 u = 1.66 x 10-27 kg
1 eV = 1.602 x 10-19 J
• Solution:
The energy equivalent to 1 u of mass
= mc2
= (1.66 x 10-27)(2.998 x 108)2
= 1.492 x 10-10 J
= (1.492 x 10-10 / 1.602 x 10-19) eV
= 931 x 106 eV
= 931 MeV
1 u = 931 MeV
19.1 The atomic model
Example 2
46
Consider the following nuclear fission.
235
92
90
1
1
U  01 n 144
Ba

Kr

n

56
36
0
0 n  energy
Given that: 1u = 931 MeV
Mass of
neutron: 1.00866 u
U-235: 235.044 u
Ba-144: 143.923 u
Kr-90: 89.9195 u
(a)
Find the amount of nuclear energy released in nuclear reaction.
Solution:
Mass difference = (235.044 – 143.923 – 89.9195 – 1.00866)u
= 0.19284u
Energy released = 0.19284 x 931 MeV
= 180 MeV
19.1 The atomic model
235
92
(180 MeV)
90
1
1
U  01 n 144
Ba

Kr

n

56
36
0
0 n  energy
(b)
Hence, show that the energy released from 1 kg
of U-235 is about 7.4 x 1013 J which is the energy
released by burning about 3 x 106 tonnes of coal.
Solution:
• No of U-235 atom in 1 kg fuel
= 1/(235.044 x 1.66 x 10-27)
= 2.563 x 1024
• Energy released = (2.563 x 1024) x (180 MeV)
= (2.563 x 1024) x (180 x 106) x (1.6 x 10-19)
= 7.4 x 1013 J
19.1 The atomic model
47
Example 3
Consider the following nuclear fusion.
2
1
H  31 H 42 He  01 n  energy
Given that: 1u = 1.66 x 10-27 kg = 931 MeV
Mass of
neutron: 1.00866 u
H-2: 2.01355 u
H-3: 3.01605 u
He-4: 4.0026 u
(a)Find the amount of nuclear energy released in the reaction.
Solution:
Difference in mass
= (2.01355 + 3.01605 – 4.0026 – 1.00866)u
= 0.01834u
Energy released
= 0.01834 x 931 MeV
= 17.1 MeV
19.1 The atomic model
48
(b)
Hence, show that the energy released from 1 kg of fuel is
about 3.3 x 1014 J.
Solution:
No of H-2 and H-3 in 1 kg fuel
= 1/[(2.01355 + 3.01605 ) x 1.66 x 10-27] = 1.1977 x 1026
Total energy released
= (1.1977 x 1026) x (2.74 x 10-12)
= 3.3 x 1014 J
19.1 The atomic model
49
Nuclear Energy – nucleus 50
• Protons &
neutrons are
collectively called
nucleons.
• A = mass number /
Nucleon number
19.1 The atomic model
Nuclear Energy – nucleus
• Consider a helium nucleus.
Mass of a helium atom
(He – 4)
Mass of components
2 protons: 2 x 1.00728 u
2 neutrons: 2 x 1.00866 u
4.0026 u
2 electrons: 2 x 0.00055 u
Total mass: 4.0330 u
+
+
+
19.1 The atomic model
+
51
• The difference between the mass of an atom and
the total mass of the particles in the atom taken
separately is known as mass defect Dm.
Dm =
+
+
–
+
+
Mass of a helium atom
(He – 4)
Mass of components
4.0026 u
2 protons: 2 x 1.00728 u
2 neutrons: 2 x 1.00866 u
2 electrons: 2 x 0.00055 u
Total mass = 4.0330 u
Mass defect of He = 4.0330u – 4.0026 u
= 0.0304 u
19.1 The atomic model
52
Binding Energy
• mass of a nucleus < total mass of separated nucleons
• The energy is required to separate the nucleons
(to overcome nuclear force binding the nucleons)
• This energy is called the binding energy Eb.
19.1 The atomic model
53
Binding Energy
•
binding energy = Δm c2
where Δm is the mass defect of the nucleus.
19.1 The atomic model
54
Find the binding energy per nucleon
in a He-4 atom.
Dm = 4.0330 u – 4.0026 u = 0.0304 u
(1u = 931 MeV)
 Eb = 0.0304 x 931 MeV = 28.3 MeV
Binding energy per nucleon = 7.08 MeV
19.1 The atomic model
55
Binding Energy
• The greater the binding energy per nucleon,
(Eb/A), the more / less stable the nuclei.
19.1 The atomic model
56
Binding energy Curve
1. Fe – 56 has the largest binding energy per nucleon.
Therefore, it is very stable.
2. Either side of maximum binding energy per nucleon
are less stable.
19.1 The atomic model
57
Binding energy Curve
Fusion
3.
When light nuclei are joined together, the binding
energy per nucleon is also increased and become
more stable.
large binding energy per nucleon ⇔ nucleons are at
low energy state. (energy is released.)
19.1 The atomic model
58
Binding energy Curve
Fission
4.
When a big nucleus disintegrates, the binding energy
per nucleon increases and become more stable
(energy is released)
19.1 The atomic model
59
Example 4
60
• Find the binding energy per nucleon for a Pb-206 nucleus.
Solution:
• Mass defect
= (82 x 1.00728 + 124 x 1.00866 u + 206 x 0.00055 u) –
205.969 u
= 1.8151 u
• Binding energy = 1.8151 x 931 MeV
= 1690 MeV
• Binding energy per nucleon Eb/A
= 1690 MeV / 206
= 8.20 MeV
19.1 The atomic model
61
19.1 The atomic model
Conditions for a Fusion
Reaction (2)
Confinement
The
hot plasma must be well
isolated away from material
surfaces in order to avoid cooling
the plasma and releasing
impurities that would
contaminate and further cool the
plasma.
In
the Tokamak system, the
plasma is isolated by magnetic
fields.
19.1 The atomic model
62
History of unified atomic
mass unit
• The chemist John Dalton was the first to
suggest the mass of one atom of hydrogen as
the atomic mass unit.
• Francis Aston, inventor of the mass
spectrometer, later used 1/16 of the mass of
one atom of oxygen-16 as his unit.
• Oxygen was chosen because it forms
chemical compounds with many other
elements.
19.1 The atomic model
63
The discovery of isotopes
• Oxygen-17 and Oxygen-18 was discovered.
• Chemists: one-sixteenth of the average mass
of the oxygen atoms.
• Physicists: one-sixteenth of the mass of an
atom of oxygen-16.
• Difference in atomic weights: about 275 parts
per million
19.1 The atomic model
64
65
• Since 1961, by definition the unified atomic
mass unit is equal to 1/12 of the mass of a
carbon-12 atom.
• Physicists: carbon-12 was already used as a
standard in mass spectroscopy.
• Chemists: difference in atomic mass: 42 parts
per million which was acceptable
19.1 The atomic model