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J. David Robertson
University of Missouri
Nuclear Chemistry
Chapter 23
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REACTIONS
CHEMICAL
NUCLEAR
RADIOACTIVE DECAY
NUCLEAR TRANSMUTATION
Unstable nuclei emits
particles and/or
electromagnetic
radiation spontaneously.
Bombardment of nuclei by
neutrons, protons or other
nuclei; occurrence are either
naturally or artificially.
23.1
Atomic number (Z) = number of protons in nucleus
Mass number (A) = number of protons + number of neutrons
= atomic number (Z) + number of neutrons
Mass Number
Atomic Number
proton neutron
1p
1H
1n
or
0
1
1
A
ZX
Element Symbol
electron
0b
0e
or
-1
-1
positron
0b
0e
or
+1
+1
a particle
4He
4a
or
2
2
A
1
1
0
0
4
Z
1
0
-1
+1
2
23.1
Radioactive Decay
A stable nucleus remains intact, but the majority of nuclei are
unstable.
An unstable nucleus exhibits radioactivity – it spontaneously
disintegrates or decays, by emitting radiation.
Each type of unstable nucleus has a characteristic rate of
radioactive decay – might range from a fraction of a second
to several billion years.
“when a nuclide of one
element decays, it changes
into a nuclide of a different
element”
23.2
When a nuclide decays, it forms a nuclide of lower energy,
and the excess energy is carried off by the emitted radiation.
The decaying nuclide is called ‘the parent’ and the product
nuclide is called ‘the daughter’.
Type of Radioactive Decay
Alpha decay: loss of an α particle
212
4
208
A decreases by 4
84
2
82
Z decreases by 2
Po He  Pb
Beta decay: ejection of a β particle
A does not change
40
40
0
19 K  20 Ca  1 b  
Z increases by 1
Positron decay: emission of a positron
Positron is the ‘antiparticle’ of the electron. Positron decay has
the opposite effect of β decay.
C  B b
11
6
11
5
0
1
A does not change
Z decreases by 1
23.2
Electron capture:
Occurs when the nucleus of an atom draws in an electron
from the lowest energy level. The symbol
is used to
distinguished it from a β particle.
A does not change
37
0
37
18 Ar  1 e17 Cl
Z decreases by 1
Gamma Emission
Loss of a -ray
(high-energy
radiation) that
almost always
accompanies the
loss of a nuclear
particle.
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
+ 2 10n
92 + 0 = 55 + 37 + (2x0)
23.1
QUESTION #1:
a) What radioactive isotope is produced in
the following bombardment of boron?
Answer:
Note: Identify the element using the atomic number
(proton number). Refer to the Periodic Table.
b) Write the nuclear equation for the beta
emitter Co–60 .
Answer:
Note: ‘Beta emitter’ means the reaction
emits/produces beta particles.
c) Complete the following nuclear reactions:
a)
b)
40Ar(a,
p)?
59Co(n, a)?
Answer:
Mn
Question #2: 212Po decays by alpha emission.
Write the balanced nuclear equation for the decay
of 212Po.
4
4
Alpha particle: 2 He or
2
a
212
84
Po He 
4
2
208
82
X
Checking the Periodic Table for the element
with atomic number (proton number) of 82
gives the answer as Pb.
Therefore:
212
84
Po He 
4
2
208
82
Pb
23.1
Question #3: In the following reactions,
identify X:
7
(a) 10
5 B ( X , a ) 3 Li
(b)
238
92
U (15n, X ) 253
99 Es
Answer:
5 B  X  a  Li
(a) 10
4
2
7
3
U  15 n  X  Es
(b) 238
92
1
0
253
99
X n
1
0
 X 7 b
0
1
Nuclear Stability & Mode of Decay
Neutron-proton ratio (n/p ratio):
Any element with more than one proton (i.e., anything
but hydrogen) will have repulsions between the protons
in the nucleus.
A strong nuclear force helps keep the nucleus from
flying apart.
*Nuclear force = the force between nucleons
i.e.: protons & neutrons
Neutrons play a key role stabilizing the nucleus.
Therefore, the ratio of neutrons to
protons (n/p) is an important factor in
determining the stability of nuclides.
For smaller nuclei (Z  20)
stable nuclei have a
neutron-to-proton ratio
close to 1:1.
As nuclei get larger, it
takes a greater number
of neutrons to stabilize
the nucleus.
The shaded region in the
figure shows what
nuclides would be stable,
the so-called belt of
stability.
Nuclei above this belt
have too many neutrons.
They tend to decay by
emitting beta particles.
Nuclei below the belt
have too many
protons. They tend
to become more
stable by positron
emission or electron
capture.
23.2
Nuclear Stability
Certain numbers of neutrons and protons are extra
stable:
• Nuclei with n or p = 2, 8, 20, 50, 82 and 126
(magic numbers);
• Extra stable numbers of electrons in noble
gases (e = 2, 10, 18, 36, 54 and 86).
Nuclei with even numbers of both
protons and neutrons are more
stable than those with odd
numbers of neutron and protons.
23.2
All isotopes of the elements with
atomic numbers higher than 83
are radioactive.
All isotopes of Tc (Technetium) and
Pm (Promethium) are radioactive.
23.2
Question #4: Which of the following
nuclides would you predict to be stable
and which is radioactive? Explain.
18
10
Ne
32
16
S
236
90
Th
123
56
Ba
Answer:
18
10
Ne
Radioactive. The n/p = 0.8. Despite having even
number of proton and neutron, the minimum ratio
for stability is 1.0. This nuclide has too few
neutron to be stable.
32
16
S
Stable. The n/p = 1.0 and it has even number of
proton and neutron.
236
90
Radioactive. Every nuclide with Z > 83 is radioactive.
123
56
Radioactive. The n/p = 1.20, therefore this
nuclide is not stable.
Th
Ba
Nuclear Binding Energy
Nuclear binding energy is the energy required to
disassemble a nucleus into free unbound neutrons
and protons.
Nuclear binding energy can be calculated
from the difference of mass of a nucleus,
and the sum of the masses of the number
of free neutrons and protons that make up
the nucleus.
This mass difference is called the mass defect or mass
deficiency, next, Einstein's formula: E = mc2 can be
used to compute the binding energy of any nucleus.
Nuclear binding energy per nucleon vs Mass number
nuclear binding energy
nucleon
nuclear stability
23.2
Calculation Steps of Binding Energy per Nucleon:
Step 1: Calculate mass defect.
Mass defect is the difference between the mass of
an atom and the sum of the masses of its protons
and neutrons. Unit is in amu.
Step 2: Convert amu to kg.
Step 3: Calculate nuclear binding energy, BE.
Use the formula: E = mc2 .
Step 4: Divide binding energy by number of nucleons
to get nuclear binding energy per nucleon.
Question #5: Iron-56 is an extremely stable
nuclide. Calculate the binding energy (BE) per
nucleon for iron-56.
Mass of 56Fe atom = 55.934939 a.m.u
Mass of proton = 1.007825 a.m.u
Mass of neutron = 1.008665 a.m.u
Answer:
56
25
Fe  25 p  31 n
1
1
1
0
Step 1: Calculate mass defect.
Mass defect, m = mass product – mass reactant
= (mass p + mass n) – mass 56Fe
mass p+n = (25 x 1.007825 amu) + (31 x 1.008665 amu)
m = 56.46424 amu – 55.934939 amu
= 0.529301 amu
23.2
Step 2: Convert amu to kg.
m = 0.529301 amu x
m = 8.789 x 10-28 kg
27
1.6605 10 kg
1.amu
Step 3: Calculate nuclear binding energy, BE.
Binding energy, BE:
E = mc2 = 8.789 x 10-28 kg x (3.00 x 108 m/s)2
E = 7.91 x 10-11 J
Step 4: Divide BE by number of nucleons.
Binding energy per nucleon:
E
7.911011 J
12

 1.4110 J / nucleon
# nucleon 56.nucleon
Question #6: What is the nuclear binding
energy per nucleon for 25Mg?
Mass of 25Mg = 24.985839 a.m.u
Mass of proton = 1.007825 a.m.u
Mass of neutron = 1.008665 a.m.u
Answer: 25
12
Mg  12 p  13 n
1
1
1
0
Mass defect, m = mass product – mass reactant
= (mass p + mass n) – mass 25Mg
mass p+n = (12 x 1.007825 amu) + (13 x 1.008665 amu)
m = 25.206545 amu – 24.985839 amu
= 0.220706 amu
23.2
m = 0.220706 amu x
m = 3.665 x 10-28 kg
27
1.6605 10 kg
1.amu
Binding energy, BE:
E = mc2 = 3.665 x 10-28 kg x (3.00 x 108 m/s)2
E = 3.30 x 10-11 J
Binding energy per nucleon:
E
3.30  10 11 J

 1.32  10 12 J / nucleon
# nucleon
25.nucleon
Kinetics of Radioactive Decay
For each duration (half-life), one half of the substance
decomposes.
For example: Ra-234 has a half-life of 3.6 days. If you start
with 50 grams of Ra-234:
After 3.6 days  25 grams
After 7.2 days  12.5 grams
After 10.8 days  6.25 grams
Half-Life
The time taken for ½ of a sample to
decompose.
The rate of a nuclear transformation depends only
on the “reactant” concentration.
All radioactive decays obey first order kinetics.
Kinetics of Radioactive Decay
N
daughter
N
rate = t
rate = lN
N
= lN
t
N = N0exp(-lt)
ln N = ln N0 - lt
N = the number of atoms at time t
N0 = the number of atoms at time t = 0
l is the decay constant
ln2
l =
t½
23.3
Kinetics of Radioactive Decay
ln[N] = ln[N]0 - lt
ln [N]
[N]
[N] = [N]0exp(-lt)
23.3
Question #7: The half-life of radium-226 is
1.60  103 years. How many hours will it
take for a 2.50 g sample to decay until
0.185 g of the isotope remains?
Answer:
ln 2
ln 2
4
1
l

 4.33 10 years
3
t1
1.60 10 years
2
ln Nt = ln N0 - lt
ln 0.185 = ln 2.50 – (4.33 x 10-4 yrs-1)t
t = 6013 years or 5.27  107 hours
Question #8: If 12% of a certain
radioisotope decays in 5.2 years, what is
the half life of this isotope?
Answer: ln Nt = ln N0 - lt
If N0 = 1.00 then Nt = 1.00 – 0.12 = 0.88
ln 0.88 = ln 1.00 – l(5.2)
l = 0.0246 years–1
ln 2
ln 2
t1 

 28.2 years

1
2
l
0.0246 years
Natural
Radioactivity
The figure shows the
Decay Series.
238U
Radioactive decay series is a
sequence of nuclear
reactions that ultimately
results in the formation of a
stable isotopes.
The beginning radioactive
isotope is called the parent
and the product is called the
daughter.
Natural Radioactivity : Radiocarbon Dating for
Determining the Age of Artifacts
Natural Radioactivity : Radiocarbon Dating for
Determining the Age of Artifacts
Radioactive C-14 is formed in the upper atmosphere by
nuclear reactions initiated by neutrons in cosmic
radiation
14N + 1 n ---> 14C + 1H
o
The C-14 is oxidized to CO2, which circulates through the
biosphere.
When a plant dies, the C-14 is not replenished.
But the C-14 continues to decay with t1/2 = 5730 years.
Activity of a sample can be used to date the sample.
Natural Radioactivity : Dating based on
Radioactive Decay
Dating using uranium-238 isotopes:
Some of the intermediate products in the uranium decay
series have very long half-lives, therefore they are
suitable for estimating the age of rocks in the earth and of
extraterrestrial objects.
Dating using potassium-40 isotopes:
The radioactive potassium-40 isotope
decays by several modes but the
important one is of that electron
capture. This particular mode can be
use to gauge the age of a specimen in
geochemistry.