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NUCLEAR CHEM
23-1
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
Orbital electrons are involved as bonds
break and form; nuclear particles do
not take part.
Protons neutrons, and other particles
are involved; orbital electrons take part
much less often
Reactions are acoompanied by
relatively small changes in energy and
no measurable changes in mass.
Reactions are coompanied by relatively
large changes in energy and
measurable changes in mass.
Reactions are influenced by
temperature, concentrations catalysts,
and the compound in which the
element occurs.
Reaction rates depend on number of
nuclei but are not affected by
temperature, catalysts.
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23-3
23-4
THREE TYPES OF RADIOACTIVE EMISSIONS IN AN ELECTRIC FIELD
23-5
23-6
TYPES OF RADIOACTIVE DECAY: BALANCING NUCLEAR EQUATIONS
TotalA
TotalZ
TotalA
Reactants TotalZ
Products
Alpha decay - A decreases by 4 and Z decreases by 2. Every element
heavier than Pb undergoes a decay.
Beta decay - ejection of a b particle from the nucleus from the conversion
of a neutron into a proton and the expulsion of 0-1b. The product nuclide
will have the same A but will be one atomic number higher.
Positron decay - a positron (0-1b) is the antiparticle of an electron. A
proton in the nucleus is converted into a neutron with the expulsion of the
positron. A remains the same but the atomic number decreases.
Electron capture - a nuclear proton is converted into a neutron by the
capture of an electron. A remains the same but the atomic number
decreases.
Gamma emission - energy release; no change in Z or A.
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23-8
23-9
WRITING EQUATIONS FOR NUCLEAR REACTIONS
Write balanced equations for the following nuclear reactions:
(a) Naturally occurring thorium-232 undergoes a decay.
(b) Zirconium-86 undergoes electron capture.
Write a skeleton equation; balance the number of neutrons and
charges; solve for the unknown nuclide.
(a)
232
90Th
228
88X
+
A = 228 and Z = 88
(b) 8640Zr + 0-1e
A = 86 and Z = 39
23-10
4
2α
232
A
90Th
228
88Ra
ZX
86
40Zr
+ 0-1e
86
39Y
+
4
2α
23-11
23-12
23-13
23-14
A PLOT OF NUMBER OF NEUTRONS vs. NUMBER OF
PROTONS FOR THE STABLE NUCLIDES.
23-15
NUCLEAR STABILITY AND MODE OF DECAY

Very few stable nuclides exist with N/Z < 1

The N/Z ratio of stable nuclides gradually increases a Z increases.

All nuclides with Z > 83 are unstable

Elements with an even Z usually have a larger number of stable nuclides
than elements with an odd Z

Well over half the stable nuclides have both even N and even Z
23-16
PREDICTING THE MODE OF DECAY

Neutron-rich nuclides undergo b decay.

Neutron-poor nuclides undergo positron decay or electron capture.

Heavy nuclides undergo a decay.
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23-18
PREDICTING NUCLEAR STABILITY
Which of the following nuclides would you predict to be stable and which
radioactive? Explain.
(a) 1810Ne
(b) 3216S
(c) 23690Th
(d) 12356Ba
Stability will depend upon the N/Z ratio, the value of Z, the value
of stable N/Z nuclei, and whether N and Z are even or odd.
23-19
Predicting Nuclear Stability
Which of the following nuclides would you predict to be stable and which
radioactive? Explain.
(a) 1810Ne
(b) 3216S
(c) 23690Th
(d) 12356Ba
PLAN:
Stability will depend upon the N/Z ratio, the value of Z, the value
of stable N/Z nuclei, and whether N and Z are even or odd.
SOLUTION:
(a) Radioactive
(b) Stable
N/Z = 0.8; there are too few
neutrons to be stable.
N/Z = 1.0; Z < 20 and N and
Z are even.
(c) Radioactive
(d) Radioactive
Every nuclide with Z > 83 is
radioactive.
N/Z = 1.20; Figure 23.2A
shows stability when N/Z ≥
1.3.
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23-24
Predicting the Mode of Nuclear Decay
PROBLEM:
Use the atomic mass of the element to predict the mode(s) of
decay of the following radioactive nuclides:
(a)
PLAN:
23-25
12
5B
(b)
234
92U
(c)
81
33As
(d)
127
57La
Find the N/Z ratio and compare it to the band stability. Then predict
which of the modes of decay will give a ratio closer to the band.
Predicting the Mode of Nuclear Decay
PROBLEM:
Use the atomic mass of the element to predict the mode(s) of
decay of the following radioactive nuclides:
(a)
PLAN:
12
5B
(b)
234
92U
81
33As
(d)
127
57La
Find the N/Z ratio and compare it to the band stability. Then predict
which of the modes of decay will give a ratio closer to the band.
SOLUTION:
(a) N/Z = 1.4 which is high.
The nuclide will probably undergo bdecay altering Z to 6 and lowering the
ratio.
(c) N/Z = 1.24 which is in the band
of stability. It will probably undergo
b- decay or positron emission.
23-26
(c)
(b) The large number of
neutrons makes this a good
candidate for a decay.
(d) N/Z = 1.23 which is too low for
this area of the band. It can
increase Z by positron emission or
electron capture.
Figure 23.3
23-27
The 238U decay series.
Decay rate (A) = -
DN
Dt
SI unit of decay is the becquerel (Bq) = 1 d/s.
curie (Ci) = number of nuclei disintegrating each second in
1 g of radium-226 = 3.70 x 1010 d/s
Nuclear decay is a first-order rate process.
Large k means a short half-life and vice versa.
23-28
Figure 23.4
23-29
Decrease in the number of 14C nuclei over time.
23-30
Sample Problem 23.4
Finding the Number of Radioactive Nuclei
PROBLEM: Strontium-90 is a radioactive byproduct 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
1.2 x 1012 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). We are
given the A0 and can find At from the integrated form of the first-order
0.693
rate equation. 2
SOLUTION: t1/2 = ln
so
k=
= 0.024 yr -1
29 yr
k
N0
A0
(1.2 x1012 - 2.9 x 1011)
ln
= ln
= kt ln At = -kt + ln A0
Fraction =
Nt
At
(1.2 x 1012)
ln At = -(0.024 yr -1)(59 yr) + ln(1.2 x 1012 d/s) decayed
Fraction =
ln At = 26.4
At = 2.9 x 1011 d/s
0.76
decayed
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Sample Problem 23.5
Applying Radiocarbon Dating
PROBLEM: The charred bones of a sloth in a cave in Chile represent the earliest
evidence of human presence in 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 the 12C/14C ratio for living
organisms results in a specific activity of 15.3 d/min•g, how old are
the bones
(t1/2 constant
of 14C = 5730
PLAN: Calculate
the rate
usingyr)?
the given half-life. Then use the
first-order rate equation to find the age of the bones.
SOLUTION:
ln 2
0.693
k=
=
= 1.21 x 10-4 yr -1
t1/2
5730 yr
1
1
A0
15.3
t=
ln
=
ln
= 8.89 x 103 yr
-4
-1
k
1.21 x 10 yr
At
5.22
The bones are about 8900 years old.
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Figure 23.5
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Schematic of a linear accelerator.
Figure 23.6
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Schematic of a cyclotron accelerator.
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Figure 23.7
Penetrating power of
radioactive emissions.
Penetrating power is
inversely related to the mass
and charge of the emission.
23-36
Nuclear changes
cause chemical
changes in
surrounding matter
by excitation and
ionization.
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Figure 23.8
The use of radioisotopes to image the
thyroid gland.
Figure 23.9
PET and brain activity.
23-39
Figure 23.10
23-40
The increased shelf life of irradiated food.
The Interconversion of Mass and Energy
E = mc2
DE = Dmc2
DE
Dm = 2
c
The mass of the nucleus is less than
the combined masses of its nucleons.
The mass decrease that occurs when
nucleons are united into a nucleus is
called the mass defect.
The mass defect (Dm) can be used to
calculate the nuclear binding
energy in MeV (mega-electron volts).
1 amu = 931.5 x 106 eV = 931.5 MeV
23-41
Sample Problem 23.6 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 amu; mass of 1H atom =
1.007825 amu; mass of neutron = 1.008665 amu).
PLAN: Find the mass defect, Dm; multiply that by the MeV equivalent and
divide by the number of nucleons.
SOLUTION:
Mass Defect = [(26 x 1.007825 amu) + (30 x 1.008665 amu)] - 55.934939
Dm = 0.52846 amu
Binding energy =
12C
23-42
(0.52846 amu)(931.5 MeV/amu)
= 8.790 MeV/nucleon
56 nucleons
has a binding energy of 7.680 MeV/nucleon, so 56Fe is more stable.
Figure 23.11
23-43
The variation in binding energy per nucleon.
Figure 23.12
23-44
Induced fission of 235U.
Figure 23.13
23-45
A chain reaction of 235U.
Figure 23.14
23-46
A light-water nuclear reactor.
Figure 23.15
The tokamak design for
magnetic containment of a
fusion plasma.
23-47