Chapter
19
Radioactivity
and Nuclear
Chemistry
GOALS
Types of radioactivity
Identify radioactive nuclides
Nuclear equations
Binding energy; per nucleon; units
Kinetics of radioactive decay
2
Facts About the Atomic Nucleus
• Every atom of an element has the same number
of protons (+ve)
atomic number (Z)
• Atoms of the same elements can have different
numbers of neutrons (no charge)
• Isotopes: atoms of the same element, having
same atomic number, Z, but different mass
number, A (diff no. of neutrons).
• Isotopes are identified by their mass number (A)
mass number = number of protons + neutrons
3
Facts About the Atomic Nucleus
• mass number = number of protons + neutrons
neutrons = mass number – number of protons
• The nucleus of an isotope is called a nuclide
• Each nuclide is identified by a symbol
mass number
Element
atomicnumber

A
X
Z
4
Nuclide Symbols
• Boron-10 (105B) has 5 p and 5 n
• Boron-11 (115B) has 5 p and 6 n
11B
10B
• Oxygen-16 (168O) has 8 p and 8 n
• Oxygen-17 (178O) has 8 p and 9 n
• Oxygen-18 (188O) has 8 p and 10 n
The Discovery of Radioactivity
• Becquerel discovered that certain minerals were
constantly producing penetrating energy rays he called
uranic rays (1896)
Marie Curie discovered 2 new elements (Po, Ra) which also
emitted uranic rays.
• Curie changed term uranic rays to radioactivity
(present in elements other than uranium).
Some nuclei are unstable; they emit particles
and/or electromagnetic radiation spontaneously.
This is radioactivity.
6
Types of Radioactive Rays
• Rutherford discovered there were 3 types of
radioactivity;
• 2 additional types were later discovered.
• alpha (a) & beta (b) decay, gamma ray (g); then
positron emission, and electron capture.
Another type of radioactivity (nuclear
transmutation) results from the bombardment of
nuclei (heavy) by neutrons, protons or other nuclei
(lighter).
7
Penetrating Ability of Radioactive Rays
a
g
b
0.01 mm
1 mm
Pieces of lead
100 mm
8
Important Atomic Symbols
Particle
Symbol
Nuclear Symbol
proton
p+
neutron
n0
1
0
electron
e-
0
1
alpha
a
beta
b, b-
positron
b, b+
1
1
1
1
H p
4
2
n
e
α He
4
2
β
0
1
β
0
1
0
1
0
1
e
e
9
Nuclear Equations
• nuclear processes are described using nuclear
equations
• use the symbol of
the
nuclide
to
represent
the
238
234
4
nucleus
92
U
Th 
90
2
He
• atomic numbers and mass numbers are conserved
 use this to predict identity of daughter nuclide if parent and
emitted particle are known
p a r e n tn u c lid e
emitted particle: product
captured particle: reactant

238
92
U
234
90
Th 
4
2
He
d a u g h t e rn u c lid
es
10
Alpha Emission
• an a particle contains 2 protons and 2 4
α
He
2
neutrons
• most ionizing, but least penetrating
4
2
• loss of an alpha particle means
atomic number decreases by 2
mass number decreases by 4
238
92
U
234
90
Th 
4
2
He
222
Ra
88

4
He
2

218
Rn
86
11
Beta Emission
• An unstable nucleus emits an electron
0
• when an atom loses a b particle its 1
β
0
1
e
atomic number increases by 1
mass number remains the same
• in beta decay, a neutron changes into a proton
Th  e 
234
90
0
1
234
91
Pa
12
If californium-251 decays by successive α, α, β
emissions, what nucleus is produced?
a)
243
95
Am
d)
b)
247
92
243
93
Pa
c)
Np
e)
242
94
Pu
244
94
Pu
Gamma Emission
0
• gamma (g) rays are high energy photons of light 0
γ
• least ionizing, but most penetrating
• generally occurs after the nucleus undergoes some other
type of decay and the remaining particles rearrange
238
92
U  He 
4
2
Th  g
234
90
0
0
14
Positron Emission
• The positron has a charge of +1 and negligible
mass
anti-electron
• when an atom loses a positron from the
nucleus, its
0
1
mass number remains the same
atomic number decreases by 1
β
0
1
e
• A positron appears to result from a proton
changing into a neutron
22
11
Na  e 
0
1
1
1
p
22
10
Ne
0
1
1
0
e n
15
Electron Capture
0
1
• occurs when an inner orbital electron is
pulled into the nucleus
• no particle emission, but atom changes
e
same result as positron emission
• proton combines with the electron to make a
neutron
1
1
p
1
0
e n
0
1
mass number stays the same
atomic number decreases by one
92
44
Ru  e 
0
1
92
43
Tc
16
Summary of Decay Processes
Decay
a
Emission
4
2
He
At #
Mass #
N
Z
-2
-4
inc
b
b-
+1
0
dec
g
g-ray
0
0
-
b
b
-1
0
inc
e-capt
X-ray
-1
0
inc
(Table 19.1; pg 871)
17
Write the nuclear equation for positron emission from K-40
a) Write the nuclide symbols for both the starting
radionuclide and the particle
K  40 
40
19
positron 
K
0
1
e
b) Set up the equation (emitted particles are products;
captured particles are reactants)
40
19
K
0
1
e X
A
Z
c) Determine the mass number and atomic number of the
missing nuclide (mass and atomic numbers are
conserved)
40
19
K
0
1
e
40
18
X
18
Write the nuclear equation for positron emission from K-40
4) Determine the element from the atomic number
40
19
K
0
1
e
40
18
Ar
Q. In a decay series, U-238 emits 8 alpha particles and 6 beta
particles. What nuclide is formed?
238
92
U  6 01 e  8 42 He  ?
Mass dec by 32; charge = +6 & -16
238
92
U  6 01 e  8 42 He 
206
82
X
X  Pb
19
Write a nuclear equation for each of the following
alpha emission from U-238
beta emission from Ne-24
positron emission from N-13
electron capture by Be-7
20
Stability
of Nuclei
stable isotopes
fall in a very
narrow range
called the island
of stability.
-
What Causes Nuclei to Break Down?
• the particles in the nucleus are held together
by a very strong attractive force found in the
nucleus called the strong force
acts only over very short distances
• the neutrons play an important role in
stabilizing the nucleus, as they add to the
strong force, but do not repel each other like
the protons do
22
Neutron to Proton (N/Z) Ratio
• the ratio of neutrons : protons is an important
measure of the stability of the nucleus
• if the N/Z ratio is too high (neutron rich) –
neutrons are converted to protons via b decay
• if the N/Z ratio is too low (proton rich) –
protons are converted to neutrons via positron
emission or electron capture
or via a decay – though not as efficient
23
Valley (Island) of Stability (Plot of # Neutrons vs #
Protons)
for Z = 1  20 (H - Ca),
stable N/Z ≈ 1
for Z = 20  40,
stable N/Z approaches 1.25
for Z = 40  80,
stable N/Z approaches 1.5
low N/Z
Heavy nuclei: for Z > 83,
there are no stable nuclei
24
Determine the kind of radioactive decay that Mg-22 undergoes
• Mg-22
 Z = 12 (protons)
 N = 22 – 12 = 10 (neutrons)
• N/Z = 10/12 = 0.83
• from Z = 1  20, stable
nuclei have N/Z ≈ 1
• Mg-22 has low N/Z; it should
convert 11p into 10n, therefore
it will undergo positron
emission or electron capture
25
Determine the kind of radioactive decay that N-18 undergoes
• N-18
 Z = 7 (protons)
 N = 18 – 7 = 11 (neutrons)
• N/Z = 11/7 = 1.57
• from Z = 1  20, stable
nuclei have N/Z ≈ 1
26
Q. Which of the following will undergo beta decay?
16O, 20F, 13N
27
Magic Numbers
besides the N/Z ratio, the numbers of protons and neutrons effects
stability
most stable nuclei have even
numbers of protons and neutrons
only a few have odd numbers of protons and neutrons
if the total number of nucleons adds to a magic number,
the nucleus is more stable (compare # electrons in
noble gases)
most stable when N
or Z = 2 (He), 8 (O),
20 (Ca), 28 (Ni), 50
(Sn), 82 (Pb)
28
Binding Energy, Eb
-All atoms are a little lighter than they are really
supposed to be.
Missing mass: ∆m = mass defect.
-This missing mass is converted to energy,
and released when 1 mole of atoms is
formed from its subatomic particles
(protons + neutrons + electrons).
-Energy holds the nucleus together.
Calculating Binding Energy, Eb
Eb is the energy required to separate the nucleus
of an atom into protons, neutrons, electrons.
For stability, Eb > electrostatic repulsive forces
between protons.
In deuterium, 21H
1H 
2
1 p
1
+
1 n
0
Eb = 2.15  108 kJ/mol 21H
Eb per mol nucleon = Eb/2 nucleons
= 1.08  108 kJ/mol nucleons
Also, calc Eb per nucleon (6.022  1023 nucleons)
Calculating Binding Energy, Eb
For deuterium, 21H: 21H  11p + 10n
Actual mass of 21H = 2.01410 g/mol (given or PT)
Mass of proton = 1.007825 g/mol
Mass of neutron = 1.008665 g/mol
Theoretical mass = 2.016490 g/mol
Mass defect (‘missing mass’) = 2.016490 – 2.01410
= 0.00239 g/mol
Calculate Binding Energy, Eb
Mass defect
= 0.00239 g/mol
= (0.002391000) kg/mol
= 2.39  10-6 kg/mol
From Einstein’s equation:
Eb = (∆m)c2 = 2.39  10-6 kg  (3.00 × 108 m/s)2
= 2.15 ×1011 kgm2/s2
(but 1 kgm2/s2 = 1 J)
= 2.15  1011 J/mol 1000 J = 2.15  108 kJ/mol
Two nucleons for deuterium, 21H:

1 p
1
+
1 n
0
Eb /mol nucleon = 1.08  108 kJ/mol nucleons
Calculating Binding Energy, Eb
For I-127, 12753I: 53p + 74n
(i.e. 127 nucleons)
Actual mass of 12753I = 126.9045 g/mol (given or PT)
53 protons = 531.007825 g/mol = 53.41473 g/mol
74 neutrons = 74  1.008665 g/mol = 74.64121 g/mol
Theoretical mass defect = 128.05594 g/mol
Mass defect = (128.05594 -126.9045) g/mol
= 1.1514 g/mol
= 1.1514  10-3 kg/mol
Calculate Binding Energy, Eb
Eb = 1.1514  10-3 kg/mol (3.00 × 108 m/s)2
= 1.04 ×1014 kgm2/s2
(but 1 kgm2/s2 = 1 J)
= 1.04 ×1014 J/mol
Eb /mol nucleon = 1.04 ×1014 J/ (127 nucleons)
= 8.19 ×1011 J
Eb /nucleon = 8.19×1011 J  (6.022 ×1023)
= 1.36 ×10-12 J
Also, can express Eb in MeV:
1 MeV = 1.602 ×10-13 J
Eb /nucleon = ? MeV
Plot of Eb vs Mass
-the greater the binding energy
per nucleon, the more stable the
nucleus is
35
Nuclear Fission
The
splitting of a heavy unstable nucleus of an
atom into two or more fragments; Pu, U & Th!
-induced reaction to produce energy!
235
92
U  n
1
0
142
56
Ba 
91
36
Kr  n
1
30
Energy released  16,800,000,000 kJ/mol
(235 g Uranium)
Nuclear Fusion
Light nuclei fuse to generate heavier
nuclei (more stable)
Free of long-lived radioactive waste.
6 H  2 He  2 p  2 n
2
1
4
2
1
1
1
0
More difficult to achieve. Nuclei must
travel at v. large KE’s at each other.
More destructive than fission bombs
(WWII)!
Kinetics of Radioactive Decay
Rate = kN It is a first order process
N = number of radioactive nuclei
t1/2
0.693

k
Nt
ratet
ln
 kt  ln
N0
rate0
time
no.of half - lives 
t1/2
ln Nt  kt  ln N0
the shorter the half-life, the more nuclei decay
every second (sample is hot!), the higher the rate38
The half life of Pu-236 is 2.86 years. Starting with a 1.35 mg sample
of Pu-236, calculate the mass that will remain after 5.00 years
Given:
mass Pu-236 = 1.35 mg, t = 5.00 yr, t1/2 = 2.86 yr
Find:
mass, mg
Concept Plan: t
k + m0 , t
mt
1/2
Relationships:
mt
0.693
t 
1
2
ln
k
m0
 kt
0.693
t 12 
k
0.693 0.693
k

 0.2423 yr-1
t 12
2.86 yr
39
Starting with a 1.35 mg sample of Pu-236, calculate the mass
that will remain after 5.00 years
t1/2
k
+
m0, t
mt
k  0.2423 yr
-1
Nt
ln
 kt
N0
Solve:
N t  N0e
 kt


)
 1.35mg e
)
 0.2423 yr - 1 5.00 yr )
Nt  ?
Check:
units are correct, the magnitude makes sense since it is less
than ½ the original mass for longer than 1 half-life
40
An ancient skull gives 4.50 dis/min∙gC. If a living organism gives
15.3 dis/min∙gC, how old is the skull? 14C-t1/2 = 5730 yr
dis = disintegrations
Given: ratet = 4.50 dis/min∙gC, ratet = 15.3 dis/min∙gC
Find: time, yr
Concept Plan: t
k + rate0, ratet
t
1/2
Relationships:
0.693
ratet
t 
ln
 kt
k
rate0
1
Solve:
2
0.693
t 
k
0.693
0.693
k

 1.209  10  4 yr -1
t
5730 yr
1
2
1
2
41
An ancient skull gives 4.50 dis/min∙gC. If a living organism gives
15.3 dis/min∙gC, how old is the skull? 14C-t1/2 = 5730 yr
Solve:
ratet
ln
 kt
rate0
rat et
ln
rat e0

t
k
4.50dis min gC
ln
15.3dis min gC
t
?
-4
-1
1.209 10 yr
Check:
units are correct, the magnitude makes sense since it is less
than 2 half-lives
42
An artifact containing carbon taken from the tomb of a king of
ancient Egypt gave 8.1 dpm/gC. How old is the artifact? Carbon
from a living organism gives 15.3 dis/min∙gC; 14C-t1/2 = 5730 yr.
dis = disintegrations
0.693
k
t1/2
ratet
0.693t
ln
rate0
t1/2
5730 15.3
ln
t
0.693 8.1
ratet
ln
 kt
rate0
t1/ 2
rate0
ln
t
0.693 ratet
5730
 0.63599  t
0.693
43
Artificial Nuclear Reactions
• bombardment of one nucleus with
another (2H, 4He, 10B, 12C) causing new
atoms to be made
 can also bombard with neutrons; protons
• reaction done in a particle accelerator
 linear
 cyclotron
Tc-97 is made by bombarding Mo-96
with deuterium, releasing a neutron
96
42
Joliot-Curies
Mo  H  Tc  n
2
1
97
43
1
0
44
Artificial Nuclear Reactions
Reactions using neutrons are called n,g reactions because a g
ray is usually emitted.
Radioisotopes used in medicine are often made by n,g reactions.
An example of a n,g reaction is production of
radioactive 31P for use in studies of P uptake in
the body.
31
15P +
0n 
1
32
15P
+ g
Transuranium Elements
Elements beyond 92 (transuranium) made
starting with an n, g reaction
238
92U +
1
239 U + g
n

0
92
239 Np +
U

92
93
239
239
239 Pu +
Np

93
94
-1b
0
0
-1b
Q.
56Fe
when bombarded with deuterium, produces 54Mn
and one other particle. Write a balanced equation
for the reaction & identify the other particle.
56
26He
+
2
1H
 5425Mn + ?
47
Medical Uses of Radioisotopes
Nuclide
Iodine-131
Iron-59
Molybdenum-99
Phosphorus-32
Strontium-87
Technetium-99
Half-life
8.1 days
45.1 days
67 hours
14.3 days
2.8 hours
6 hours
Organ/System
thyroid
red blood cells
metabolism
eyes, liver
bones
heart, bones, liver,
lungs
48
Nonmedical Uses of Radioactive Isotopes
• smoke detectors
 Am-241
 smoke blocks ionized air, breaks circuit
• insect control
 sterilize males
• food preservation
• radioactive tracers
 follow progress of a “tagged” atom in a
reaction
49
Nonmedical Uses of Radioactive Isotopes
• authenticating art object
 many older pigments and ceramics were made from minerals
with small amounts of radioisotopes
• crime scene investigation
• measure thickness or condition of industrial
materials
 corrosion
 track flow through process
 gauges in high temp processes
 weld defects in pipelines
 road thickness
50