Chemistry 1B
General
Chemistry
Ch. 24:
Nuclear Reactions and
Their
Applications
Instructor:
Dr. Orlando E. Raola
Santa Rosa Junior College
24-1
Chapter 24
Nuclear Reactions and Their Applications
24-2
Nuclear Reactions and Their Applications
24.1 Radioactive Decay and Nuclear Stability
24.2 The Kinetics of Radioactive Decay
24.3 Nuclear Transmutation: Induced Changes in Nuclei
24.4 Effects of Nuclear Radiation on Matter
24.5 Applications of Radioisotopes
24.6 Interconversion of Mass and Energy
24.7 Applications of Fission and Fusion
24-3
Table 24.1 Comparison of Chemical and Nuclear Reactions
Chemical Reactions
Nuclear Reactions
One substance is converted into another,
but atoms never change identity.
Atoms of one element typically are
converted into atoms of another element.
Electrons in orbitals are involved as bonds
break and form; nuclear particles do not
take part.
Protons, neutrons, and other nuclear
particles are involved; electrons in
orbitals take part much less often.
Reactions are accompanied by relatively
small charges in energy and no
measurable changes in mass.
Reactions are accompanied by relatively
large charges in energy and measurable
changes in mass.
Reaction rates are influenced by
Reaction rates depend on number of
temperature, concentration, catalyst, and
nuclei, but are not affected by
the compound in which an element occurs. temperature, catalysts, or, except on rare
occasions, the compound in which an
element occurs.
24-4
Components of the Nucleus
Most of the mass of the atom is concentrated in the
dense, tiny nucleus.
The nucleus comprises the neutrons and protons,
collectively called nucleons.
The total number of nucleons in a nucleus gives its mass number.
A nuclide is a nucleus with a particular composition.
Each isotope of an element has a different nuclide.
A particular nuclide is often designated by its mass
number; for example, chlorine-35 and chlorine-37.
24-5
Notation for Nuclides
The relative mass and charge of a particle is
described by the notation:
A = mass number
Z = charge of the particle
Example:
electron
proton
neutron
24-6
A
Z
X
0
e
-1
1
p
1
1
n
0
Radioactivity
Many nuclides are unstable and spontaneously emit
radiation, a process termed radioactive decay.
- The intensity of the radiation is not affected by temperature,
pressure, or other physical and chemical conditions.
When a nuclide decays, it emits radiation and usually
changes into a nuclide of a different element.
There are three natural types of radioactive emission:
Alpha particles (α, 42 α , or 42 He ) are identical to helium-4 nuclei.
Beta particles (β, β-, or
0
-1β )
are high-speed electrons.
Gamma rays (γ or 00 γ ) are very high-energy photons.
24-7
Figure 24.1 How the three types of radioactive emissions behave in
an electric field.
The positively charged α particles curve toward the negative plate,
the negatively charged β particles curve towards the positive plate,
and the γ rays are not affected by the electric field.
24-8
Nuclear Equations
When a nuclide decays, it forms a daughter nuclide of
lower energy.
The excess energy is carried off by the emitted radiation and the
recoiling nucleus of the daughter nuclide.
The decay process is represented by a balanced nuclear
equation. Both the total charge and the total mass must be
balanced:
Total A
Reactants
Total Z
24-9
=
Total A
Total Z
Products
Modes of Radioactive Decay
• Alpha (α) decay involves the loss of an α particle from
the nucleus.
– For each α particle emitted, A decreases by 4 and Z decreases
by 2 in the daughter nuclide.
– This is the most common form of decay for a heavy, unstable
nucleus.
228
222
4
Ra
→
Rn
+
88
86
2α
• β- decay involves the ejection of a β- particle from the
nucleus.
– A neutron is converted to a proton, which remains in the nucleus,
and a β- particle is expelled: 1 n → 1 p + 0 β
0
1
-1
– A remains the same in the daughter nuclide but Z increases by 1
63
63
0
unit.
Ni
→
Cu
+
28
29
-1 β
24-10
• Positron (β+) emission is the emission of a β+ particle
from the nucleus.
– The positron is the antiparticle of the electron.
– A proton in the nucleus is converted into a neutron, and a
positron is emitted:
1
1
0
p
→
n
+
β
1
0
1
– A remains the same in the daughter nuclide but Z increases by 1
unit.
11
11
0
C
→
B
+
6
5
1β
• Electron capture occurs when the nucleus interacts
with an electron in a low atomic energy level.
– A proton is transformed into a neutron:
1
p
1
+-10 e →
0
n
1
– The effect on A and Z is the same as for positron emission.
24-11
• Gamma (γ) emission involves the radiation of highenergy γ photons.
– Gamma emission usually occurs together with other forms of
radioactive decay.
– Several γ photons of different energies can be emitted from an
excited nucleus as it returns to the ground state.
– γ emission results in no change in either A or Z since γ rays
have no mass or charge.
24-12
Table 24.2 Modes of Radioactive Decay*
24-13
Table 24.2 Modes of Radioactive Decay*
24-14
Sample Problem 24.1
Writing Equations for Nuclear Reactions
PROBLEM: Write balanced equations for the following nuclear
reactions:
(a) Naturally occurring thorium-232 undergoes α decay.
(b) Zirconium-86 undergoes electron capture.
PLAN: We first write a skeleton equation that includes the mass
numbers, atomic numbers, and symbols of all the particles on
the correct sides of the equation, showing the unknown
A
product particle as Z X. Then, because the total of mass
numbers and the total of charges on the left side and the right
side of the equation must be equal, we solve for A and Z, and
use Z to determine the identity of X from the periodic table.
24-15
Sample Problem 24.1
SOLUTION:
(a) Writing the skeleton equation, with the α particle as a product:
For A, 232 = A + 4, so A= 228.
232
A
4
Th
→
X
+
α
90
Z
2
For Z, 90 = Z + 2, so Z = 88.
The daughter nuclide produced in this reaction is Ra.
232
90 Th
→
228
88 Ra
4
+2α
(b) Writing the skeleton equation, with the captured electron as a
reactant:
86
0
A
For A, 86+ 0 = A so A= 86.
Zr
+
e
→
40
-1
ZX
For Z, 40 -1 = Z so Z = 39.
Element X is yttrium, symbol Y.
86
0
86
Zr
+
e
→
40
-1
39Y
24-16
Nuclear Stability
Two key factors determine the stability of a nuclide:
- the number of neutrons (N), the number of protons (Z), and their
ratio (N/Z), and
- the total mass of the nuclide.
A plot of number of neutrons vs. number of protons for all
stable nuclides produces a band of stability that gradually
curves above the line for N = Z.
- Lighter nuclides are stable when N = Z.
- As Z increases, the N/Z for stable nuclei gradually increases.
- All nuclides with Z > 83 are unstable.
24-17
Figure 24.2
24-18
A plot of number of neutrons vs. number of protons
for the stable nuclides.
Stability and Nuclear Structure
Protons within the nucleus experience electrostatic
repulsive forces, which destabilize the nucleus.
The strong force, which exists between all nucleons,
counteracts the weaker repulsive forces.
Nucleons are found in nucleon energy levels, and pairing
of the spins of like nucleons leads to greater stability.
- Elements with an even Z (number of protons) usually have a larger
number of stable nuclides.
- Over half the stable nuclides have both even N and even Z.
24-19
Table 24.3 Number of Stable Nuclides
for Elements 48 to 54*
*
24-20
Element
Atomic
No. (Z)
No. of
nuclides
Cd
48
8
In
49
2
Sn
50
10
Sb
51
2
Te
52
8
I
53
1
Xe
54
9
Even Z shown in boldface.
Table 24.4 An Even-Odd Breakdown of
the Stable Nuclides
Z
N
No. of
nuclides
Even
Even
157
Even
Odd
53
Odd
Even
50
Odd
Odd
4
TOTAL 264
24-21
Sample Problem 24.2
Predicting Nuclear Stability
PROBLEM: Which of the following nuclides would you predict to be
stable and which radioactive? Explain.
18
(a) 10 Ne
32
(b)16 S
(c)
236
90 Th
(d)
123
56 Ba
PLAN: In order to evaluate the stability of each nuclide, we
determine the N and Z values and the N/Z ratio. We can
then compare these to the values for stable nuclides. We
also note whether Z and N are even or odd.
SOLUTION:
(a)
18Ne
is Radioactive.
This nuclide has N = (18 – 10) = 8 and Z = 10, so the N/Z ratio is
18 – 10
= 0.8, which is too low to be stable.
10
24-22
Sample Problem 24.2
(b)
32S
is stable.
This nuclide has N = Z = 16, so N/Z = 1.0.
With Z < 20 and even values for N and Z, this nuclide is most likely stable.
(c) 236Th is radioactive.
This nuclide has Z = 90, and every nuclide with Z < 83 is radioactive.
(d) 123Ba is radioactive.
This nuclide has N = 67 and Z = 56, so N/Z = 1.20.
For Z values of 55 to 60, Figure 24.2A shows that N/Z ≥ 1.3, so this
nuclide has too few neutrons to be stable
24-23
Predicting the Mode of Decay
Nuclide type
Description
Mode of Decay
neutron-rich
high N/Z
β- decay
neutron → proton, lowers N/Z
proton-rich
low N/Z
β+ emission or e- capture
proton → neutron, increases N/Z
heavy nuclides
Z > 83
α decay
reduces both Z and N
24-24
Sample Problem 24.3
Predicting the Mode of Nuclear Decay
PROBLEM: Predict the nature of the nuclear change(s) each of the
following radioactive nuclides is likely to undergo:
(a)
12
5B
(b)
234
92 U
81
(c) 33 As
127
57
(d)
La
PLAN: If the nuclide is too heavy to be stable (Z > 83), it undergoes α
decay. For other cases, we use the Z value to obtain its
atomic mass from the periodic table.
If the mass number of the nuclide is higher than the atomic
mass, the nuclide has too many neutrons: N is too high and βdecay occurs.
If the mass number is lower than the atomic mass, the nuclide
has too many protons: Z is too high and the nuclide decays by
β+ emission or e- capture.
24-25
Sample Problem 24.3
SOLUTION:
(a) 12B has Z = 5 and its atomic mass is 10.81. The nuclide’s A value
of 12 is significantly higher than its atomic mass, so it is neutron
rich.
It will probably undergo β- decay.
(b)
has Z = 92, so it will undergo α decay and decrease its total
mass.
234U
(c) 81As has Z = 33 and its atomic mass is 74.92. The A value of 81 is
much higher than the atomic mass, so it is neutron rich and will
probably undergo β- decay.
(d)
24-26
127La
has Z = 57 and its atomic mass is 138.9 The A value of 127
is much lower than the atomic mass, so it is proton rich and will
probably undergo β+ emission or e- capture.
Figure 24.3
The 238U decay series.
A parent nuclide may undergo a
series of decay steps before a
stable daughter nuclide is formed.
24-27
Detection and Measurement of
Radioactivity
An ionization counter detects radioactive emissions as
they ionize a gas.
Ionization produces free electrons and gaseous cations, which
are attracted to electrodes and produce an electric current.
A scintillation counter detects radioactive emissions by
their ability to excite atoms and cause them to emit light.
Radioactive particles strike a light-emitting substance, which emits
photons. The photons strike a cathode and produce an electric
current.
24-28
Figure 24.4
24-29
Detection of radioactivity by an ionization counter.
Figure 24.5
24-30
A scinatillation “cocktail” in tubes to be placed in
the counter.
Units of Radioactivity
The SI unit of radioactivity is the becquerel (Bq),
defined as one disintegration per second (d/s).
The curie (Ci) is a more commonly used unit:
1 Ci = 3.70x1010 d/s
24-31
Rate of Radioactive Decay
Radioactive nuclei decay at a characteristic rate,
regardless of the chemical substance in which they occur.
The rate of radioactive decay (A) (also called the activity)
is proportional to the number of nuclei present.
A = kN
Radioactive decay follows first-order kinetics, and the
rate constant k is called the decay constant.
The larger the value of k, the higher the activity of the
substance.
24-32
Half-Life of Radioactive Decay
The half-life of a nuclide is the time taken for half the
nuclei in a sample to decay.
- The number of nuclei remaining is halved after each half-life.
- The mass of the parent nuclide decreases while the mass of
the daughter nuclide increases
- Activity is halved with each succeeding half-life.
ln Nt
N0
24-33
= -kt
or Nt = N0e-kt
and
ln
Nt
= kt
N0
Figure 24.6
Decrease in the number of 14C nuclei over time.
t1/2 =
24-34
ln 2
k
Table 24.5 Decay Constants (k) and Half-Lives
(t1/2) of Beryllium Isotopes
Nuclide
7
4 Be
8
4 Be
9
4 Be
10
4 Be
11
4 Be
24-35
k
t1/2
1.30x10-2/day
53.3 days
1.0x1016/s
6.7x10-17 s
Stable
4.3x10-7/yr
1.6x106 yr
5.02x10-2/s
13.8 s
Sample Problem 24.4
Finding the Number of Radioactive Nuclei
PROBLEM: Strontium-90 is a radioactive by-product of nuclear reactors
that behaves biologically like calcium, the element above it
in Group 2A(2). When 90Sr is ingested by mammals, it is
found in their milk and eventually in the bones of those
drinking the milk. If a sample of 90Sr has an activity of
1.2x1012 d/s, what are the activity and the fraction of nuclei
that have decayed after 59 yr (t1/2 of 90Sr = 29 yr)?
PLAN: The fraction of nuclei that have decayed is the change in the
number of nuclei, expressed as a fraction of the starting
number. The activity of the sample (A) is proportional to the
number of nuclei (N), and we are given A0. We can find At from
the integrated form of the first-order rate equation, in which t is
59 yr. We need the value of k, which we can calculate from the
given t1/2.
24-36
Sample Problem 24.4
SOLUTION:
ln 2
ln 2
0.693
t1/2 =
so k =
= 0.024 yr-1
k
t1/2 = 29 yr
Nt
At
ln
= ln
= kt
or lnA0 – lnAt = kt
N0
A0
so lnAt = -kt + lnA0
= -(0.024 yr-1 x 59 yr) + ln(1.2x1012 d/s)
= -1.4 + 27.81
lnAt = 26.4
At = e26.4 = 2.9x1011 d/s
A 0 – At
1.2x1012 d/s – 2.9x1011 d/s
fraction decayed =
=
A0
1.2x1012 d/s
24-37
= 0.76
Radioisotopic Dating
• Radioisotopes can be used to determine the ages of
certain objects.
• Radiocarbon dating measures the relative amounts of
14C and 12C in materials of biological origin.
– The ratio of 14C/12C remains the same for all living organisms.
– Once the organism dies, the amount of 14C starts to decrease as it
decays to form 14N.
– Since 14C decays at a predictable rate, measuring the amount
present indicates the time that has passed since the organism
died.
•
24-38
40K/40Ar
ratios can be used to determine the age of
certain rocks.
Figure 24.7
Ages of several objects determined by
radiocarbon dating.
t = 1 ln A0
At
k
24-39
Sample Problem 24.5
Applying Radiocarbon Dating
PROBLEM: The charred bones of a sloth in a cave in Chile represent
the earliest evidence of human presence at the southern
tip of South America. A sample of the bone has a specific
activity of 5.22 disintegrations per minute per gram of
carbon (d/min·g). If 12C/14C ratio for living organisms
results in a specific activity of 15.3 d/min·g, how old are
the bones (t1/2 of 14C = 5730 yr)?
PLAN: We calculate k from the given half-life. Then use the first-order
rate equation to find the age of the bones, using the given
activities of the bones and of a living organism.
SOLUTION:
k=
ln 2
t1/2
24-40
=
0.693
5730 yr
= 1.21x10-4 yr-1
Sample Problem 24.5
t = 1 ln A0
At
k
=
1
ln 15.3 d/min·g
1.21x10-4 yr-1
5.22 d/min·g
= 8.89x103 yr
The bones are about 8900 years old.
24-41
Nuclear Transmutation
Nuclear transmutation is the induced conversion of the
nucleus of one element into the nucleus of another.
This is achieved by high-energy bombardment of nuclei in
a particle accelerator.
14
7N
4
1
+ 2α → 1 p +
17
8O
Nuclear transmutation reactions can be described
using a specific short-hand notation:
reactant nucleus (particle in, particle(s) out) product nucleus
The above reaction can be written as: 14N (α, p) 17O.
24-42
Figure 24.8
Schematic diagram of a linear accelerator.
The linear accelerator uses a series of tubes with alternating voltage.
A particle is accelerated from one tube to the next by repulsion.
24-43
Figure 24.9
24-44
Schematic diagram of a cyclotron accelerator.
Table 24.6 Formation of some Transuranium Nuclides*
Reaction
239
94 Pu
+ 2 0n →
239
94 Pu
+
4
2α
+
4
2α
242
96 Cm
+
4
2
253
99 Am
24-45
1
241
95 Am
253
99 Es
*
Half-life of Product
+
+
4
2
18
8
→
→
α →
α
→
O →
241
95 Am
+ -1 β
0
242
96
+
Cm
243
97 Bk
245
98 Cf
256
101Md
256
101 Lr
1
0n
1
+ 2 0n
+
+
432 yr
163 days
4.5 h
1
0n
45 min
1
0
76 min
n
1
+ 50 n
28 s
Like chemical reactions, nuclear reactions may occur in several steps.
Effects of Nuclear Radiation on Matter
Radioactive emissions collide with surrounding matter,
dislodging electrons and causing ionization. Each such
event produces a cation and a free electron.
The number of cation-electron pairs is directly related to
the energy of the incoming ionizing radiation.
Ionizing radiation has a destructive effect on living tissue.
The danger of a particular radionuclide depends on
- the type of radiation,
- its half-life, and
- its biological behavior.
24-46
Units of Radiation
The gray is the SI unit for energy absorption.
1 Gy = 1 J absorbed per kg of body tissue.
The rad is more widely used: 1 rad = 0.01 J/kg or 0.01 Gy.
The rem is the unit of radiation dosage equivalent to a
given amount of tissue damage in a human.
no. of rems = no. of rads x RBE
The RBE is the relative biological effectiveness factor. The rem
allows us to assess actual tissue damage by taking into account
the strength of the radiation, the exposure time, and the type of
tissue.
24-47
Figure 24.10
Penetrating power of radioactive emissions.
The effect of radiation on living tissue
depends on both the penetrating power and
the ionizing ability of the radiation.
Penetrating power is inversely related to
the mass, charge, and energy of the
emission.
24-48
Molecular Interactions with Radiation
The interaction of ionizing radiation with molecules
causes the loss of an electron from a bond or a lone pair.
This results in the formation of free radicals, molecular
or atomic species with one or more unpaired electrons.
Free radicals are unstable and extremely reactive.
Double bonds in membrane lipids are very susceptible to
attack by free-radicals:
24-49
Sources of Ionizing Radiation
• There are several natural sources of background
radiation.
• Cosmic radiation increases with altitude.
• Radon is a radioactive product of uranium and thorium
decay.
– Rn contributes to 15% of annual lung cancer deaths.
• Radioactive 40K is present in water and various food
sources.
• Radioactive 14C occurs in atmospheric CO2.
24-50
Figure 24.11
24-51
US radon distribution.
Table 24.7
Typical Radiation Doses from Natural and Artificial
Sources
Source of Radiation
Average Adult Exposure
Natural
Cosmic radiation
30-50 mrem/yr
Radiation from the ground
From clay soil and rocks
In wooden houses
In brick houses
In concrete (cinder block) houses
~25-170 mrem/yr
10-20 mrem/yr
60-70 mrem/yr
60-160 mrem/yr
Radiation from the air (mainly radon)
Outdoors, average value
In wooden houses
In brick houses
In concrete (cinder block) houses
20 mrem/yr
70 mrem/yr
130 mrem/yr
260 mrem/yr
Internal radiation from minerals in tap
water and daily intake of food.
(40K, 14C, Ra)
~ 40 mrem/yr
24-52
Table 24.7
Typical Radiation Doses from Natural and Artificial
Sources
Source of Radiation
Average Adult Exposure
Artificial
Diagnostic x-ray methods
Lung (local)
Kidney (local)
Dental (does to the skin)
0.04-0.2 rad/film
1.5-3 rad/film
≤ 1 rad/film
Locally ≤ 10,000 rad
Therapeutic radiation treatment
Other Sources
Jet flight (4 h)
Nuclear testing
Nuclear power industry
~1 mrem
< 4 mrem/yr
< 1 mrem/yr
Total average value
100-200 mrem/yr
24-53
Figure 24.12
Two models of radiation risk.
The linear response model proposes that radiation effects
a`ccumulate over time regardless of dose.
The S-shaped response model implies there is a threshold
24-54 above which the effects are more significant.
Table 24.8
24-55
Acute Effects of a Single Dose on Whole-Body
Irradiation
Radioactive Tracers
• The isotopes of an element exhibit very similar chemical
and physical behavior.
– A small amount of radioactive isotope mixed with the stable
isotope will undergo the same chemical reactions, and can act
as a tracer.
• Radioactive tracers are used
–
–
–
–
to study reaction pathways,
to track physiological functions,
to trace material flow,
to identify the components of a substance from a very small
sample, and
– to diagnose a wide variety of medical conditions.
24-56
Figure 24.13 The use of radioisotopes to image the thyroid gland.
This 131I scan shows an
asymmetric image that is
indicative of disease.
24-57
A 99Tc scan of a healthy
thyroid.
Table 24.9
24-58
Some Radioisotopes Used as Medical Tracers
Isotope
Body Part or Process
11C, 18F, 13N, 15O
PET studies of brain, heart
60Co, 192Ir
Cancer therapy
64Cu
Metabolism of copper
59Fe
Blood flow, spleen
67Ga
Tumor imaging
123I, 131I
Thyroid
111In
Brain, colon
42K
Blood flow
81mKr
Lung
99mTc
Heart, thyroid, liver, lung, bone
201Tl
Heart muscle
90Y
Cancer, arthritis
Figure 24.14
PET and brain activity.
These PET scans show brain activity in a normal person (left)
and in a patient with Alzheimer’s disease (right). Red and yellow
indicate relatively high activity within a region.
24-59
Other Applications of Ionizing Radiation
• Radiation therapy
– Cancer cells divide more rapidly than normal cells, and are
therefore susceptible to radioisotopes that interfere with cell
division.
• Destruction of microbes
– Irradiation of food increases its shelf life by killing
microorganisms that cause rotting or spoilage.
• Insect control
• Power for spacecraft instruments
24-60
Figure 24.15
24-61
The increased shelf life of irradiated food.
The Interconversion of Mass and Energy
The total quantity of mass-energy in the universe is
constant.
Any reaction that releases or absorbs energy also loses or gains
mass.
E = mc2 or ΔE = Δmc2
so Δm = ΔE
c2
In a chemical reaction, the energy changes in breaking
or forming bonds is relatively small, so mass changes
are negligible.
In a nuclear reaction, the energy changes are enormous
and the mass changes are easily measurable.
24-62
Nuclear Binding Energy
The mass of the nucleus is less than the combined
masses of its nucleons.
- Mass always decreases when nucleons form a nucleus, and the
“lost” mass is released as energy.
- Energy is required to break a nucleus into individual nucleons.
The nuclear binding energy is the energy required to
break 1 mol of nuclei into individual nucleons.
- Binding energy is expressed using the electron volt (eV).
Nucleus + nuclear binding energy → nucleons
1 u = 931.5 x 106 eV = 931.5 MeV
24-63
Sample Problem 24.6
Calculating the Binding Energy per Nucleon
PROBLEM: Iron-56 is an extremely stable nuclide. Compute the
binding energy per nucleon for 56Fe and compare it with that
for 12C (mass of 56Fe atom = 55.934939 u; mass of 1H atom
= 1.007825 u; mass of neutron = 1.008665 u).
PLAN: Iron-56 has 26 protons and 20 neutrons. We calculate the
mass difference, Δm, when the nucleus forms by subtracting
the given mass of one 56Fe atom from the sum of the masses
of 26 1H atoms and 30 neutrons. To find the binding energy
per nucleon, we multiply Δm by the equivalent in MeV and
divide by the number of nucleons.
SOLUTION:
Calculating the mass difference:
Δm = [(26 x mass 1H atom) + (30 x mass neutron)] – mass 56Fe atom
= [(26 x 1.007825 u) + (30 x 1.008665 u)] - 55.934939
= 0.52856 u
24-64
Sample Problem 24.6
Binding energy per nucleon =
0.52846 u x 931.5 MeV/u
56 nucleons
= 8.790 MeV/nucleon
An 56Fe nucleus would require more energy per nucleon to break up
into its nucleons than would 12C, so 56Fe is more stable than 12C.
24-65
Figure 24.16
The variation in binding energy per nucleon.
The greater the binding energy per nucleon, the more stable the nuclide.
24-66
Fission or Fusion
The binding energy per nucleon peaks at elements with
mass number A ≈ 60.
- Nuclides become more stable with increasing number up to around
60 nucleons, after which stability decreases.
There are two ways nuclides can increase their binding
energy per nucleon:
A heavier nucleus can split into lighter ones by
undergoing fission.
Lighter nuclei can combine to form a heavier nucleus in a
process called fusion.
24-67
Figure 24.17
24-68
Fission of 235U caused by neutron bombardment.
Nuclear Fission
Nuclear fission involves the splitting of large nuclei into
smaller nuclei, using neutron bombardment to start the
process.
Fission releases energy and generates more high-energy
neutrons, which cause further fission to occur.
The fission process becomes self-sustaining by a chain
reaction. The mass required to achieve this is called the
critical mass.
The energy from nuclear fission can be harnessed and
converted to other forms of energy.
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Figure 24.18
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A chain reaction involving fission of 235U.
Figure 24.19
An atomic bomb based on 235U.
An atomic bomb uses an uncontrolled chain reaction to produce
a powerful explosion.
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Figure 24.20
A light-water nuclear reactor.
The dome-shaped structure is the containment shell for the
nuclear reactor.
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Figure 24.20
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A light-water nuclear reactor.
Figure 24.21 The tokamak design for magnetic containment of
a fusion plasma.
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Chemical Connections
Figure B24.1
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Element synthesis I the life cycle of a star.