Transcript Chapter 26

Unit 3: Nuclear Chemistry
1
Unit Objectives
 Describe the makeup of the nucleus
 Describe the relationships between neutron-proton
ratio and nuclear stability
 Discuss what is meant by the band of stability
 Calculate mass deficiency and nuclear binding
energy
 Describe the common types of radiation emitted
when nuclei undergo radioactivity decay
 Write and balance equations that describe nuclear
reactions
2
Unit Objectives
 Predict the different kinds of nuclear reactions
undergone by nuclei, depending on their positions
relative to the band of stability
 Describe methods for detecting radiation
 Understand half-lives of radioactive elements
 Carry out calculations associated with radioactive
decay
 Interpret decay series
3
Unit Objectives
 Discuss some uses of radionuclides, including the
use of radioactive elements for dating objects
Describe some nuclear reactions that are induced
by bombardment of nuclei with particles
Discuss nuclear fission and some of its
applications including nuclear reactors
Discuss nuclear fusion and some prospects for and
barriers to its use for the production of energy
So What’s Nuclear Chemistry?
Nuclear chemistry is concerned with the
behavior of protons and neutrons in the
atomic nucleus
Radioactive tracers
Carbon dating
Radiation therapy
Beginning of Nuclear Science
• Acceptance of
Dalton’s theory
convinced
scientists that
one element
could not be
converted in
another
(transmutation)
Before there was Chemistry there was Alchemistry
Beginning of Nuclear Science
In 1896, Henri Becqurel
accidentally discovered
radioactivity in Uranium (U)
salts.
In 1898, Marie and Pierre Curie
discovered two new radioactive
elements in U mine residue.
 Po (polonium) and Ra (radium)
In 1898, Ernest Rutherford
discovered that radioactivity has
two distinct forms.
  and  radiation
Marie Curie (1867-1934)
7
Comparison Of Chemical
and Nuclear Reactions
• Nuclear Reactions
• Chemical Reactions
1. Particles within the
1. Usually only the outer
nucleus, such as
most electrons
protons and neutrons,
participate in reactions.
are involved in
reactions.
2. No new elements can
2. Elements may be
be produced, only new
converted from one
chemical compounds.
element to another.
8
Comparison Of Chemical
and Nuclear Reactions
• Nuclear Reactions
3 Release or absorb
immense amounts of
energy, typically 1000
times more.
4 Rates of reaction are
not influenced by
external factors.
• Chemical Reactions
3 Release or absorb much
smaller amounts of
energy.
4 Rates of reaction
depend on factors such
as concentration,
pressure, temperature,
and catalysts.
9
Fundamental Particles of Matter
So… what do you remember about these
particles?
10
The Nucleus
 The nucleus consists of protons and neutrons in a very
small volume.
 Since nearly all the mass of an atom resides in the
nucleus  nucleus very DENSE (2 x 1014 g/cm3)
 So how can positively charged protons be packed
so closely together without causing most nuclei to
spontaneously decompose ?
 There are over 100 short-lived subatomic particles as
products of nuclear reactions (quarks)
 They help overcome proton-proton repulsion
11
Neutron-Proton Ratio
and Nuclear Stability
 Recall…
 Nuclides denotes different nuclei

Nuclide symbol for an element AZE
• What is the nuclide symbol for an element with 79p, 118n
and 76e?

Isotopes are nuclei that have the same number of
protons but different neutron numbers.
 Isotopes are the same element.
12
Neutron-Proton Ratio
and Nuclear Stability
 Experimentally, it can be shown that nuclei have
a preference for even numbers of protons and
neutrons
Abundance of naturally occurring nuclides
Proton Number Neutron Number Number of Nuclides
Even
Even
157
Even
Odd
52
Odd
Even
50
Odd
Odd
4
13
Neutron-Proton Ratio
and Nuclear Stability
Special stability is associated with certain proton
and neutron numbers (or sum of the two)
Magic numbers are:
2 8 20 28 50 82 126
14
Neutron-Proton Ratio
and Nuclear Stability
• Example nuclides with magic numbers of
nucleons includes:
4
2
He 2
2 protons, 2 neutrons
16
8
O8
8 protons, 8 neutrons
40
20
Ca
20 protons, 20 neutrons
120
50
Sn 70
208
82
50 protons, 70 neutron
Pb126 82 protons, 126 neutrons
15
Plot of # neutrons
versus atomic #
 stable nuclei (green dots)
are located in the band of
stability
 all other nuclei are unstable
and radioactive (white, blue
and pink)
 To note:
 For low atomic numbers (
20), the most stable nuclei
have equal numbers of
protons & neutrons (N =Z)
 Above Z=20, the most
stable nuclei have more
neutrons than protons
Nuclear Stability and
Binding Energy
• Experimentally shown that masses of atoms other
than hydrogen are always LESS than sum of their
particles
– Referred to as mass deficiency (Δm)


m  sum of masses of all p , n and e-  actualmass of atom
17
Nuclear Stability and
Binding Energy
Example 26-1: Calculate the mass deficiency for
39K. The actual mass of 39K is 39.32197 amu per
atom.
39
K has 19 protons, 20 neutrons and 19 electrons
1 proton has a mass of 1.0073 amu
1 neutron has a mass of 1.0087 amu
1 electron has a mass of 0.0005458 amu
18
Nuclear Stability and
Binding Energy
Example 26-1: Calculate the mass deficiency for
39K. The actual mass of 39K is 39.32197 amu
per atom.
The sum of the masses of the protons, neutrons, and electrons is
 19  1.0073 amu   20 1.0087 amu   19  0.0005458 amu 
 19.1387 amu  20.1740 amu  0.0104 amu
 39.32307 amu
Therefore,
m  39.32307 amu - 39.32197 amu
m  0.00110 amu
in one atom
19
Nuclear Stability and
Binding Energy
You try it!
Questions: 14 (a) , 16 (a) & (b)
20
Nuclear Stability and
Binding Energy
Some what happened to the “lost mass”???
Einstein’s Theory of Relativity: Matter & energy are equivalent (E = mc2)
 Therefore the mass lost was converted to energy
( To this date the reverse has not yet been achieved
on a large scale, i.e. converting energy into matter)
The mass defect is the mass of the nuclear
particles that has been used to bind the nucleus in
the nuclear binding energy or strong nuclear
force.
21
Nuclear Stability and
Binding Energy
Due to the Einstein relationship, we can
calculate the nuclear binding energy for a
nucleus.
E  mc
2
E  m c
or
2
Binding Energy m c
2
22
Nuclear Stability and
Binding Energy
Example 26-2: Calculate the nuclear binding
energy of 39K in J/mol of K atoms. 1 J = 1 kg m2/s2.
Step 1: Find mass deficiency for 1 mol of atoms 
Δm = 0.00100 amu * 6.022 x 1023 atoms = 6.022 x 1020 amu / mol
atom
1 mol
 Step 2: Convert amu to grams  This is the SAME value as
amu/ atom:
Δm = 0.00100 amu * 6.022 x 1023 atoms *
atom
1 mol
1g
= 0.0010 g
6.022 x 1023 amu
mol
23
Nuclear Stability and
Binding Energy
Example 26-2: Calculate the nuclear binding
energy of 39K in J/mol of K atoms. 1 J = 1 kg m2/s2.
 Step 3: Convert grams to kilograms 
0.0011 g
mol
x
1 kg
= 1.1 x 10-6 kg /mol
1000 g
24
Nuclear Stability and
Binding Energy
Example 26-2: Calculate the nuclear binding
energy of 39K in J/mol of K atoms. 1 J = 1 kg m2/s2.
Step 4: Substituting for m in Energy equation 

3.00 10 
9.00 10 
E = mc  1.10 10
2

 1.10 10 6 kg mol
 9.90 10
 6 kg
8 m
mol
16 m 2
2
s
s2
10 kg m 2
s 2 mol
 Step 5: Converting kg m2 / s2 mol to J 
 9.90 1010 J mol
This is a huge amount of energy. Enough to heat 60,000 to 70, 000 tons of water from
0
25
-100 0C. It’s also the amount of energy needed to separate 1 mole of 39K nuclei into its
Nuclear Stability and
Binding Energy
 You try it!
 Questions 14 (b), 16 (c) (d) & (e)
 Remember:
 1 J = 1 kg m2/s2
 1g = 6.022 x 1023 amu
1 proton has a mass of 1.0073 amu
1 neutron has a mass of 1.0087 amu
1 electron has a mass of 0.0005458 amu
Radioactive Decay
Nuclei whose neutron-to-proton ratio lies outside the
band of stability experience spontaneous radioactive
decay.
 Emit one or more particles, electromagnetic rays or
both.
27
 The major types of
natural radioactivity:





Alpha emission
Beta emission
Gamma emission
Positron emission
Electron capture
 The type emitted
depends on where
the nucleus is
relative to the band
of stability
28
Radioactive Decay
or He ion (He 2+)
Nucleon = number of particles in the nucleus; protons and neutrons
Radioactive decay
Chemical equations show equal numbers of
each kind of atom and equal number of charges
on either side of the equation
– Law of conservation of mass
In nuclear reactions a proton can be converted
to a neutron and vice versa
– So, the total number of nucleons remain the same
Equations for Nuclear Reactions
 Two conservation principles hold for ALL nuclear
reaction equations.
1. The sum of the mass numbers of the reactants equals
the sum of the mass numbers of the products.
2. The sum of the atomic numbers of the reactants
equals the sum of the atomic numbers of the
products.
31
Equations for Nuclear Reactions
• For the general reaction:
M1
Z1
Q R  Y
M2
Z2
M3
Z3
• The two conservation principles demand
M1 =
M 2 + M3
and
Z1
=
–
–
Z2
+ Z3
Where the M's are mass numbers,
And the Z's are atomic numbers.
32
Neutron Rich Nuclei (Above the Band of Stability)
 These nuclei have too high a
ratio of neutrons to protons.
 Decays must lower this ratio
and include:
 neutron emission
 beta emission
 A beta particle is an
electron ejected from the
nucleus when a neutron is
converted to a proton;
1
1
0
0 n1 p  1

33
Neutron Rich Nuclei (Above the
Band of Stability)
• Beta emission simultaneously decreases the number
of neutrons (by one) and increases the number of
protons (by one).
– Efficiently changes the neutron to proton ratio.
• Examples of beta emission:
14
6
C N+ 
226
88
14
7
0
-1
Ra  Ac + 
226
89
0
-1
NOTE: the SUM of mass
numbers are the same on
either side of equation.
Likewise the atomic
numbers
Note: Because there is a
change in atomic number,
the identity of the atom
changes: C  N
34
Neutron Rich Nuclei (Above the
Band of Stability)
Neutron emission does not change the atomic number,
but it decreases the number of neutrons.
– The product isotope is less massive by the mass of 1 neutron.
Examples of neutron emission
17
7
N N+ n
137
53
16
7
1
0
Because the atomic numbers
do not change  identity of
the atom remains the same
I I+ n
136
53
1
0
35
Neutron Poor Nuclei (Below the Band of Stability)

These nuclides have too
low a ratio of neutrons
to protons.
 Nuclear radioactive
decays must raise this
ratio
 The possible decays
include:
1. electron capture
2. positron emission
36
Neutron Poor Nuclei (Below the
Band of Stability)
Electron capture involves the capture of an electron in
the lowest energy level in the atom by the nucleus.
– conversion of a proton to a neutron
1
1
p e n
37
18
0
-1
1
0
Ar  e Cl
0
-1
37
17
37
Neutron Poor Nuclei (Below the Band of
Stability)
A positron has the mass of an electron but has a
positive charge.
– The symbol is 0+1e.
Positron emission is associated with the conversion of
a proton into a neutron.
1
1
p
0
1
e n
1
0
39
19
K e 
15
8
O e  N
0
1
0
1
39
18
15
7
Ar
38
Nuclei with Atomic Number Greater
than 83
Alpha emission occurs for some nuclides, especially
heavier ones.
Alpha () particles are helium nuclei, 42He,
containing two protons and two neutrons.
 Alpha emission increases the neutron-to-proton ratio.
204
82
Pb Hg  He
200
80
4
2
39
Nuclei with Atomic Number Greater
than 83
All nuclides having atomic numbers greater than 83 are
beyond the belt of stability and are radioactive.
 Many of these isotopes decay by emitting alpha particles.
238
92
U Th  He
234
90
4
2
40
Nuclei with Atomic Number Greater
than 83
The transuranium elements (Z>92) also decay by
nuclear fission in which the heavy nuclide splits into
nuclides of intermediate mass and neutrons.
252
98
Cf  Ba 
142
56
106
42
Mo  4 n
1
0
41
Detection of Radiation
 Present radiation detection schemes depend on the fact
that particles and radiations emitted by radioactive
decay are energetic and some carry charges.
1. Photographic Detection


Radioactivity affects photographic plates or film as does
ordinary light.
Medical and dental x-ray photographs are made using this
technique.
42
Detection of Radiation
2. Fluorescence Detection


Fluorescent substances absorb energy from high energy
rays and then emit visible light.
A scintillation counter is an instrument using this principle.
43
Detection of Radiation
3. Cloud Chambers contain air saturated with a vapor.



Radioactive decay particles emitted ionize the air molecules
in the chamber.
The vapor condenses on these ions.
Then the ion tracks are photographed.
44
Detection of Radiation
Diagram of a Simple Cloud Chamber
45
Detection of Radiation
 A Cloud Chamber Photo from a Large Detector.
46
Detection of Radiation
4. Gas Ionization Counters


The ions produced by ionizing radiation are passed between
high voltage electrodes causing a current to flow between
the electrodes and the current is amplified.
This is the basis of operation of gas ionization counters such
as the Geiger-Mueller counter.
47
Detection of Radiation
Schematic of Geiger Counter
48
Detection of Radiation
Picture of a Geiger Counter
49
Rates of Decay and Half-Life
Radionuclides have different stabilities and decay at
different rates ( seconds to millions yrs)
The rates of all radioactive decays are independent of
temperature and obey first order kinetics.
 Rate proportional to the conc. of only one substance
Rate of decay  kA  or
 A0 
ln 
  k t or
 A
 N0 
ln 
kt
 N
Where:
A0 and A can either be molar conc. or
masses of reactant
Where:
A0 = original mass or conc
A = remaining mass or conc
where N = number of disintegrations
per unit time
Rates of Decay and Half-Life
In nuclear chemistry, decay is expressed in terms of
half-life, t½
 The amount of time required for HALF of the original sample
to react
 The half-life, t1/2, is related to the rate constant by the simple
relationship:
t ½ = 0.693
k
Figure 26-6. The decay of a 16 ug sample of 9038Sr, a radioactive nuclide with
half life of 29.1 yrs
Rates of Decay and Half-Life
• Example 26-3: How much 60Co remains 15.0 years
after it is initially made? 60Co has a half-life of 5.27
years.
A = Assume initial amt of sample is 1
0
t = 15.0 yrs
t½ = 5.27 yrs
k= find using half life
A= ?
 A0 
ln 
kt
 A 
 A0 
kt
ln 
A 


 A0  k t

e
 A 
Take antilogs on both
sides to get rid of Ln
0.693
k
0.693 0.693
k

t 12
5.27 y
t 12 
k  0.132 y -1
53
Rates of Decay and Half-Life
A0  A ek t

0.132 y 15.0 y 
1.0  A * e
-1
A  1.0

0.132 y 15.0 y 
e
Rearranging to make A the subject of
the formula
-1
A  1.0 / 7.243
A  0.138 or 13.8% remains
54
Rates of Decay and Half-Life
Cobalt treatments used in medicine to arrest
certain types of cancer rely on the ability of gamma
rays to destroy cancerous cells. Coblat-60 decays
with the emission of beta particles and gamma
rays, with a half-life of 5.27 years
 How much of a 3.42-ug sample of cobalt-60 remains
after 30.0 years.
55
Plan
 Determine specific value of rate constant, k
t ½ = 0.693
k
 Use value of k in to determine ratio of A0 to A after 30 yrs
 A0 
ln 
kt
 A 
56
Decay Series
 Some nuclides are so far
away from the belt of
stability, that it takes a
series nuclear
disintegrations to attain
nuclear stability.
 For any step,
 the decaying nuclide is
called the parent
 and the product nuclide,
the daughter
The 238U decay series. Alpha emissions
shown in red. Beta emissions shown in blue
57
Decay Series
•
•
“branchings” are
possible at various
points but they
result in the same
final product
Less prevalent
decays are shown
in blue
Table 26-4 Emissions and half lives of the
238 U,
Natural Radioactive Decay Series of58
235 U and 232 Th (Thorium)
Uses of Radionuclides
Radioactive Dating
• Radiocarbon dating can be used to estimate the
ages of items of organic origin.
• 14C is produced continuously in the upper
atmosphere by the bombardment of 14N by
cosmic-ray neutrons:
14
N  n  C p
1
0
14
6
1
1
59
Uses of Radionuclides
 14C atoms react with O2 to form 14CO2
 The 14CO2 then is incorporated into plant life by
photosynthesis.
 After the organism dies the 14C content
decreases via radioactive decay
 The 14C half-life is 5730 years.
14
6
C N 
14
7
0
-1
Remains of a man found
frozen in a glacier in the
Italian Alps for about 4000 yrs
 The activity per gram of Carbon is a measure of
the length of time elapsed since death
60
Uses of Radionuclides
 Example 26-4: Estimate the age of an object whose
14C activity is only 55% that of living wood.
1. Determine the rate constant for 14C.
0.693 0.693
t 12 


ak
k
0.693 0.693
 4 1
k

 1.2110 y
t 12
5730 y
61
Uses of Radionuclides
2. Determine the age of the object.
 A0 
ln 
kt
 A 
 100% 
4
1
ln 
  1.21 10 y t
 55% 


ln 1.82   1.21 10 y  t
0.598  1.2110 y  t
4
4
1
1
0.598
t
1.2110  4 y 1
t  4940 y
62
Recall :
 A0 
ln 
kt
 A 
Can be used for radioactive decay calculations
where A0 and A can either be expressed as mass
or concentration
The equations can also be expressed as:
 N0 
ln 
kt
 N
where N = number of disintegrations per unit time
63
Uses of Radionuclides
Carbon dating only useful for objects less than 50,000
years.
The potassium-argon and uranium-lead methods are
used for dating older objects.
Potassium-argon method relies on the electron capture
decay of 40K to 40Ar
40
19
K  Ar  e
40
18
0
1
t 12  1.3  10 y
9
64
Uses of Radionuclides
The uranium-lead method relies on the alpha decay of
238U to 234Th.
238
92
U
Th  He
234
90
4
2
t 12  4.5  10 y
9
65
Uses of Radionuclides
 Example: Uranium lead dating
A sample of uranium ore is found to contain 4.64 mg of
238U and 1.22 mg of 206Pb. Estimate the age of the
ore. The half-life of 238 U is 4.51 x 109 years.
Plan:
 Find original mass of 238U
The original mass of 238U is = mass of 238U
remaining + mass 206Pb.
 Find rate constant, k
 A0 
 Solve for t using  ln 
kt
 A 
Uses of Radionuclides
Medical Uses
Agricultural Uses
Industrial Uses
Research Applications
Read up on these and come prepared
to discuss or write out answers at next
class. Be sure to cite examples.
Artificial Transmutations
of Elements
TRANSMUTATUON – converting one element into
another
Bombardment of a nuclide with a nuclear particle can
make an unstable compound nucleus that decays to a
new nuclide by emission of a different particle.
The rules for balancing equations for nuclear
reactions which were presented in the section on
radioactivity still hold.
68
Artificial Transmutations
of Elements
This can be achieved by:
1.
Bombardment with positive ions
 This requires particle accelerators
Example reactions are:
96
42
Mo  H Tc  n
2
1
97
43
1
0
69
Artificial Transmutations
of Elements
96
42
Mo  H Tc  n
209
83
2
1
97
43
1
0
Bi  He 3 n 
210
85
Th  H  3 n 
?
4
2
230
90
1
1
1
0
1
0
At
70
Artificial Transmutations
of Elements
96
42
Mo  H Tc  n
209
83
2
1
97
43
1
0
Bi  He 3 n 
210
85
Th  H  3 n 
228
91
4
2
230
90
1
1
1
0
1
0
At
Pa
71
Artificial Transmutations
of Elements
2. Neutron Bombardment
Because neutrons have no charge, there is no
coulomb repulsion to their nuclear penetration, so
they do not have to be accelerated.
Nuclear reactors are often used as neutron sources.
72
Artificial Transmutations
of Elements
Transmutation reactions are often indicated in an
abbreviated form, with the bombarding particle and
emitted subsidiary particles shown in parentheses
between the parent and daughter nuclides
200
80
6
3
Hg  n 
1
0
Hg   n,   reaction
201
80
0
0
Li  n  H  He n,   reaction
1
0
3
1
4
2
73
Nuclear Fission
 Some nuclides with atomic numbers greater than 80 are
able to undergo fission.
 These nuclei split into nuclei of intermediate masses and emit one
or more neutrons.
 Some fissions are spontaneous while others require
activation by neutron bombardment.
 Enormous amounts of energy are released in these fissions.
 Some of the numerous possible fission paths for 235U (after
bombardment by a neutron) are:
74
Nuclear Fission
A self-propagating
nuclear chain reaction.
A stray neutron
induces a single
fission, liberating more
neutrons. Each of then
induces another
fission, each of which
is accompanied by the
release of 2 or more
neutrons. The chain
continues branching,
very quickly resulting in
an explosive rate of
fission
Lots of energy is
released!
75
Nuclear Fission
Fission is energetically favorable for elements with Z
greater than 80
 The product nuclides are more stable (near the high part of the
nuclear binding energy curve).
76
Nuclear Fission
Recall that nuclear binging energy is energy needed by nucleus to break it into
its subatomic particles. Graph shows that atoms of intermediate mass number
have the highest binding energies per nucleon, therefore they are most stable.
Thus, fission is energetically favaourable for heavier atoms because it produces
atoms of intermediate mass and greater binding energies
77
Nuclear Fusion
Fusion, the merging of light nuclei to make
heavier nuclei, is favorable for very light
atoms.
 Extremely high energies or temperatures are
necessary to initiate fusion reactions.
The energy source for stars is fusion.
 The fusion reaction in main sequence stars is:
2
1
H  H  He  n  energy
3
1
4
2
1
0
78
Nuclear Fusion
79