Transcript Nuclear Stability and Radioactive Decay

Chapter 19
The Nucleus:
A Chemist’s View
Section 19.1
Nuclear Stability and Radioactive Decay
Review
 Atomic Number (Z) – number of protons
 Mass Number (A) – sum of protons and neutrons
A
Z
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X
2
Section 19.1
Nuclear Stability and Radioactive Decay
Radioactive Decay
 Nucleus undergoes decomposition to form a
different nucleus.
 Nuclides with 84 or more protons are unstable.
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Section 19.1
Nuclear Stability and Radioactive Decay
The Zone of Stability
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Section 19.1
Nuclear Stability and Radioactive Decay
Types of Radioactive Decay
 Alpha production (α):
 Beta production (β):
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Section 19.1
Nuclear Stability and Radioactive Decay
Types of Radioactive Decay
 Gamma ray production (γ):
 Positron production:
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Section 19.1
Nuclear Stability and Radioactive Decay
Types of Radioactive Decay
 Electron capture:
Inner-orbital electron
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Section 19.1
Nuclear Stability and Radioactive Decay
Decay Series (Series of Alpha and Beta Decays)
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Balancing Nuclear Equations
1. Conserve mass number (A).
The sum of protons plus neutrons in the products must equal
the sum of protons plus neutrons in the reactants.
235
92 U
+ 10n
138
55 Cs
+
96
37 Rb
+ 2 10n
235 + 1 = 138 + 96 + 2x1
2. Conserve atomic number (Z) or nuclear charge.
The sum of nuclear charges in the products must equal the
sum of nuclear charges in the reactants.
235
92 U
+ 10n
138
55 Cs
+
96
37 Rb
92 + 0 = 55 + 37 + 2x0
+ 2 10n
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Section 19.1
Nuclear Stability and Radioactive Decay
CONCEPT CHECK!
Which of the following produces a  particle?
a)
68
31
b)
62
29
c)
212
87
d)
129
51
Ga +
0
1
Cu 
Fr 
Sb 
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e 
0
1
4
2
e+
He +
0
1
e+
68
30
62
28
Zn
electron capture
Ni
positron
208
85
At
alpha particle
Te
beta particle
129
52
10
Example 19.1
Balance the following nuclear equations (that is, identify the
product X):
(a) 212 Po
(b) 137 Cs
84
55

208
82 Pb
+X

137
56 Ba
+X
11
Example 19.1
Strategy
In balancing nuclear equations, note that the sum of atomic
numbers and that of mass numbers must match on both
sides of the equation.
Solution
(a) The mass number and atomic number are 212 and 84,
respectively, on the left-hand side and 208 and 82,
respectively, on the right-hand side. Thus, X must have a
mass number of 4 and an atomic number of 2, which
means that it is an α particle. The balanced equation is
212
84 Po

208
82 Pb
+
4
2
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Example 19.1
(b) In this case, the mass number is the same on both sides of
the equation, but the atomic number of the product is 1
more than that of the reactant. Thus, X must have a mass
number of 0 and an atomic number of -1, which means that
it is a β particle. The only way this change can come about
is to have a neutron in the Cs nucleus transformed into a
proton and an electron; that is, 01 n  11 p + -10 Β (note that
this process does not alter the mass number). Thus, the
balanced equation is
137
55 Cs

137
56 Ba
+
0
-1 Β
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Section 19.2
The Kinetics of Radioactive Decay
Rate of Decay
Rate = kN
 The rate of decay is proportional to the number of
nuclides. This represents a first-order process.
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Section 19.2
The Kinetics of Radioactive Decay
Half-Life
 Time required for the number of nuclides to reach half
the original value.
t1/ 2
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ln  2  0.693
=
=
k
k
15
Section 19.2
The Kinetics of Radioactive Decay
Nuclear Particles
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Section 19.2
The Kinetics of Radioactive Decay
Half-Life of Nuclear Decay
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Section 19.2
The Kinetics of Radioactive Decay
EXERCISE!
A first order reaction is 35% complete at the
end of 55 minutes. What is the value of k?
ln[A] = –kt + ln[A]o
 ln(0.65) = –k(55) + ln(1)
k = 7.8 × 10-3 min-1
Section 19.3
Nuclear Transformations
Nuclear Transformation
 The change of one element into another.
27
13
249
98
Al + He 
4
2
30
15
1
0
P+ n
263
Cf + 188 O  106
Sg + 4 01 n
 (Happens by bombarding the nucleus, here with alpha
particle or another element)
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Section 19.3
Nuclear Transformations
 Accelerators like cyclotron and linear accelerators
is used to bombard a nucleus with high velocity
positive ion to produce other element
 Neutrons are also used for this purpose
Section 19.4
Detection and Uses of Radioactivity
Measuring Radioactivity Levels
 Geiger counter
 Scintillation counter
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Section 19.4
Detection and Uses of Radioactivity
Geiger Counter
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Section 19.4
Detection and Uses of Radioactivity
Carbon–14 Dating
 Used to date wood and cloth artifacts.
 Based on carbon–14 to carbon–12 ratio.
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Section 19.4
Detection and Uses of Radioactivity
Radiotracers
 Radioactive nuclides that are introduced into organisms
in food or drugs and whose pathways can be traced by
monitoring their radioactivity.
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Section 19.4
Detection and Uses of Radioactivity
Radiotracers
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Section 19.5
Thermodynamic Stability of the Nucleus
Energy and Mass
 When a system gains or loses energy it also gains or
loses a quantity of mass.
E = mc2
Δm = mass defect
ΔE = change in energy
 If ΔE is negative (exothermic), mass is lost from
the system.
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Section 19.5
Thermodynamic Stability of the Nucleus
Binding Energy
 The energy required to decompose the nucleus into its
components.
 Iron-56 is the most stable nucleus and has a binding
energy of 8.79 MeV.
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Example 19.2
127
53 I
The atomic mass of
is 126.9004 amu. Calculate the
nuclear binding energy of this nucleus and the corresponding
nuclear binding energy per nucleon.
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Example 19.2
Strategy
To calculate the nuclear binding energy, we first determine the
difference between the mass of the nucleus and the mass of all
the protons and neutrons, which gives us the mass defect.
Next, we apply Equation (19.2) [ΔE = (Δm)c2].
Solution
There are 53 protons and 74 neutrons in the iodine nucleus.
1
The mass of 53 1H atom is
53 x 1.007825 amu = 53.41473 amu
and the mass of 74 neutrons is
74 x 1.008665 amu = 74.64121 amu
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Example 19.2
127
53 I
Therefore, the predicted mass for
is 53.41473 + 74.64121
= 128.05594 amu, and the mass defect is
Δm = 126.9004 amu - 128.05594 amu
= -1.1555 amu
The energy released is
ΔE = (Δm)c2
= (-1.1555 amu) (3.00 x 108 m/s)2
= -1.04 x 1017 amu · m2/s2
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Example 19.2
Let’s convert to a more familiar energy unit of joules. Recall that
1 J = 1 kg · m2/s2. Therefore, we need to convert amu to kg:
2
amu

m
1.00 g
1 kg
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E  -1.04×10
×
×
2
23
s
6.022×10 amu 1000 g
= -1.73×10-10
kg  m 2
-10
=
-1.73×10
J
2
s
Thus, the nuclear binding energy is 1.73 x 10-10 J . The nuclear
binding energy per nucleon is obtained as follows:
1.73×10-10 J
=
= 1.36 ×10-12 J / nucleon
127 nucleons
(1.60 x 10-13 J = 1 MeV)
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Section 19.5
Thermodynamic Stability of the Nucleus
Binding Energy per Nucleon vs. Mass Number
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Section 19.6
Nuclear Fission and Nuclear Fusion
Nuclear Fission and Fusion
 Fusion – Combining two light nuclei to form a heavier,
more stable nucleus.
 Fission – Splitting a heavy nucleus into two nuclei with
smaller mass numbers.
1
0
n+
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235
92
U  142
56 Ba +
91
36
Kr + 3 01 n
33
Section 19.6
Nuclear Fission and Nuclear Fusion
Nuclear Fission
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Section 19.6
Nuclear Fission and Nuclear Fusion
Fission Processes
 A self-sustaining fission process is called a chain
reaction.
Neutrons
Causing
Fission
Event
Event
subcritical
<1
critical
=1
supercritical
>1
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Result
reaction stops
sustained reaction
violent explosion
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Section 19.6
Nuclear Fission and Nuclear Fusion
Schematic of a Nuclear Reactor
Section 19.6
Nuclear Fission and Nuclear Fusion
Schematic Diagram
of a Reactor Core
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Section 19.6
Nuclear Fission and Nuclear Fusion
Nuclear Fusion
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Section 19.7
Effects of Radiation
Biological Effects of Radiation
Depend on:
1.
2.
3.
4.
Energy of the radiation
Penetrating ability of the radiation
Ionizing ability of the radiation
Chemical properties of the radiation source
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Section 19.7
Effects of Radiation
rem (roentgen equivalent for man)

The energy dose of the radiation and its effectiveness in causing
biologic damage must be taken into account.
Number of rems = (number of rads) × RBE
rads = radiation absorbed dose
RBE = relative effectiveness of the
radiation in causing biologic damage
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Section 19.7
Effects of Radiation
Effects of Short-Term Exposures to Radiation
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