23-1

COMPARISON OF CHEMICAL AND NUCLEAR REACTIONS
Chemical Reactions
One substance is converted into another but atoms never change identity
Orbital electrons are involved as bonds break and form; nuclear particles do not take part.
Reactions are acoompanied by relatively small changes in energy and no measurable changes in mass.
Nuclear Reactions
Atoms of one element typically are converted into atoms of another element
Protons neutrons, and other particles are involved; orbital electrons take part much less often
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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23-4

THREE TYPES OF RADIOACTIVE EMISSIONS IN AN ELECTRIC FIELD
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TYPES OF RADIOACTIVE DECAY: BALANCING NUCLEAR EQUATIONS
TotalA
TotalZ
 TotalA
TotalZ
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
-1 b
. The product nuclide will have the same A but will be one atomic number higher.
Positron decay - a positron ( 0
-1 b
) 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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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
90
Th 228
88
X + 4
2
α
A = 228 and Z = 88
(b) 86
40
Zr + 0
-1 e A
Z
X
232
90
Th 228
88
Ra + 4
2
α
A = 86 and Z = 39
86
40
Zr + 0
-1 e 86
39
Y
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A PLOT OF NUMBER OF NEUTRONS vs. NUMBER OF
PROTONS FOR THE STABLE NUCLIDES.
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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
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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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PREDICTING NUCLEAR STABILITY
Which of the following nuclides would you predict to be stable and which radioactive? Explain.
(a) 18
10
Ne (b) 32
16
S (c) 236
90
Th (d) 123
56
Ba
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.
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Predicting Nuclear Stability
Which of the following nuclides would you predict to be stable and which radioactive? Explain.
(a) 18
10
Ne (b) 32
16
S (c) 236
90
Th (d) 123
56
Ba
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
N / Z = 0.8; there are too few neutrons to be stable.
(c) Radioactive
Every nuclide with Z > 83 is radioactive.
(b) Stable
N / Z = 1.0; Z < 20 and N and
Z are even.
(d) Radioactive
N / Z = 1.20; Figure 23.2A shows stability when N / Z ≥
1.3.
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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) 12
5
B (b) 234
92
U (c) 81
33
As (d) 127
57
La
PLAN: 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.
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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) 12
5
B (b) 234
92
U (c) 81
33
As (d) 127
57
La
PLAN: 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 b decay 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.
(b) The large number of neutrons makes this a good candidate for a decay.
(d) N / Z increase
= 1.23 which is too low for this area of the band. It can
Z by positron emission or electron capture.
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Figure 23.3
The 238 U decay series.

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Decay rate (A) = -
D
N
D t
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 10 10 d/s
Nuclear decay is a first-order rate process.
Large k means a short half-life and vice versa.

Figure 23.4
Decrease in the number of 14 C nuclei over time.
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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 90 Sr is ingested by mammals, it is found in their milk and eventually in the bones of those drinking the milk. If a sample of 90 Sr has an activity of
1.2 x 10 12 d/s, what are the activity and the fraction of nuclei that have decayed after 59 yr (
PLAN: t of 90 Sr = 29 yr).
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 ln A t
N
N
0 given the A
0
0 and can find A t rate equation.
SOLUTION:
A
A t
1/2 ln = ln = t from the integrated form of the first-order
2
= ln so k = kt k ln A t
0.693
29 yr
= -k t + ln A
0 t
= -(0.024 yr -1 )(59 yr) + ln(1.2 x 10 12 d/s)
= 0.024 yr
Fraction decayed
=
-1
(1.2 x10 12 - 2.9 x 10
(1.2 x 10 12 )
11 ) ln A t
= 26.4
A t
= 2.9 x 10 11 d/s
Fraction decayed
= 0.76
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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 12 C/ 14 C ratio for living organisms results in a specific activity of 15.3 d/min•g, how old are
PLAN: the bones ( t
1/2 of 14 C = 5730 yr)?
first-order rate equation to find the age of the bones.
SOLUTION: ln 2 k = =
1 k t
1/2
0.693
5730 yr
A
A t
0 t = ln =
= 1.21 x 10
1
-4
1.21 x 10 -4 yr -1 yr ln
-1
15.3
5.22
= 8.89 x 10 3 yr
The bones are about 8900 years old.
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Figure 23.5
Schematic of a linear accelerator.
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Figure 23.6
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.
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Nuclear changes cause chemical changes in surrounding matter by excitation and ionization.

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Figure 23.9
PET and brain activity.
Figure 23.8
The use of radioisotopes to image the thyroid gland.

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Figure 23.10
The increased shelf life of irradiated food.

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The Interconversion of Mass and Energy
E = mc 2
D
E =
D mc 2
D
m =
D
E c 2
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 (
D m ) can be used to calculate the nuclear binding energy in MeV ( mega-electron volts ).
1 amu = 931.5 x 10 6 eV = 931.5 MeV

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 56 Fe and compare it with that for 12 C
(mass of 56 Fe atom = 55.934939 amu; mass of 1 H atom =
PLAN:
1.007825 amu; mass of neutron = 1.008665 amu).
Find the mass defect,
D m ; 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
D m = 0.52846 amu
Binding energy =
(0.52846 amu)(931.5 MeV/amu)
56 nucleons
= 8.790 MeV/nucleon
12 C has a binding energy of 7.680 MeV/nucleon, so 56 Fe is more stable.
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Figure 23.11
The variation in binding energy per nucleon.
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Figure 23.12
Induced fission of 235 U.
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Figure 23.13
A chain reaction of 235 U.
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Figure 23.14
A light-water nuclear reactor.
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Figure 23.15
The tokamak design for magnetic containment of a fusion plasma.