CHE 116: General Chemistry
1
CHAPTER TWENTY ONE
Copyright © Tyna L. Heise 2002
All Rights Reserved
Chem. 116 Prof. T.L. Heise
Nuclear Chemistry
2
Nuclear Reactions: changes in matter that occur
in the nucleus of an atom
- spontaneous changes of nuclei, which emit
radiation, are said to be radioactive
Chem. 116 Prof. T.L. Heise
Radioactivity
3
Nucleus - made up of two subatomic particles
PROTON
NEUTRON
Both molecules are
called nucleons
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
4
All atoms of a given element have the same
number of protons, known as atomic number
All atoms of a given element can have
different numbers of neutrons, and
therefore different mass numbers
- mass number is the number of nucleons
in nucleus
- same atomic number, different mass
number is an ISOTOPE
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
5
Different isotopes have different abundancies
in nature.
Different nuclei also have different stabilities:
- nuclear properties of an atom depend
on the number of protons and neutron
- nuclei that are radioactive are called
radionuclides
- atoms containing these nuclei are called
radioisotopes
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
6
The vast majority of nuclei found in nature
are stable and remain intact indefinately
Radionuclides - unstable and spontaneously
emit particles and electromagnetic energy.
- emission of radiation is one way an
unstable nuclide can become a stable
nuclide with less energy
- when a nuclide spontaneously
decomposes, it is called radioactive decay
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
7
Alpha decay (a)- emission of the nucleus of a
helium atom : 4He2
238U
234Th + 4He
--->
92
90
2
** all mass numbers and atomic numbers
are similarly balanced in all nuclear
equations
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
8
Sample exercise: What element undergoes
alpha decay to form lead-208?
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
9
Sample exercise: What element undergoes
alpha decay to form lead-208?
X ---> 208Pb82 + 4He2
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
10
Sample exercise: What element undergoes
alpha decay to form lead-208?
X ---> 208Pb82 + 4He2
atomic numbers add up to 212
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
11
Sample exercise: What element undergoes
alpha decay to form lead-208?
X ---> 208Pb82 + 4He2
atomic numbers add up to 212
mass numbers add up to 84
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
12
Sample exercise: What element undergoes
alpha decay to form lead-208?
212X
208Pb + 4He
--->
84
82
2
look up atomic number 84 to identify
symbol
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
13
Sample exercise: What element undergoes
alpha decay to form lead-208?
212Po
Chap. 21.1
208Pb + 4He
--->
84
82
2
Chem. 116 Prof. T.L. Heise
Radioactivity
Beta decay (b)- emission of the nucleus of a 14
high speed electron : 0e-1
131I ---> 131Xe + 0e
53
54
-1
** beta emission is equivalent to the
conversion of a neutron to a proton, thereby
increasing the atomic number by 1
1n --> 1p + 0e
0
1
-1
the electron only comes into existence during
nuclear reaction, it was NOT there all along
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
Gamma radiation (g)- emission of the nucleus
of a high energy photons : 0g0
** not shown when writing nuclear
equations
Chap. 21.1
15
Chem. 116 Prof. T.L. Heise
Radioactivity
16
nope
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
Positron emission - emission of the nucleus of a17
high speed positive electron : 0e+1
11C ---> 11B + 0e
6
5
+1
** positron emission is equivalent to the
conversion of a proton to a neutron, thereby
decreasing the atomic number by 1
1p --> 1n + 0e
1
0
+1
the positron only comes into existence during
nuclear reaction, it was NOT there all along
.
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
Electron capture - capture by the nucleus of a 18
high speed electron : 0e-1
81Rb + 0e --> 81Kr
37
-1
36
** electron capture is equivalent to the
conversion of a proton to a neutron, thereby
decreasing the atomic number by 1
1p + 0e --> 1n
1
-1
0
.
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
along
19
.
Chap. 21.1
Chem. 116 Prof. T.L. Heise
Radioactivity
Write a balanced nuclear equation for the
reaction in which oxygen-15 undergoes
positron emission.
Chap. 21.1
20
Chem. 116 Prof. T.L. Heise
Radioactivity
Write a balanced nuclear equation for the
reaction in which oxygen-15 undergoes
positron emission.
15O
Chap. 21.1
21
0e + X
-->
8
+1
Chem. 116 Prof. T.L. Heise
Radioactivity
Write a balanced nuclear equation for the
reaction in which oxygen-15 undergoes
positron emission.
15O
Chap. 21.1
22
0e + 15X
-->
8
+1
7
Chem. 116 Prof. T.L. Heise
Radioactivity
Write a balanced nuclear equation for the
reaction in which oxygen-15 undergoes
positron emission.
15O
Chap. 21.1
23
0e + 15N
-->
8
+1
7
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
24
The stability of a particular nucleus depends on
a variety of factors, and no single rule allows
us to predict whether a particular nucleus is
radioactive and how it might decay, however
empirical observations can be made
- neutron to proton ratio is most important
Chap. 21.2
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
25
neutron to proton ratio
- the more protons packed into the nucleus, the
more neutrons needed to bind the nucleus
together
stable nuclei with low atomic numbers have
approximately equal numbers of neutrons and
protons
Chap. 21.2
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
26
neutron to proton ratio
- the more protons packed into the nucleus, the
more neutrons needed to bind the nucleus
together
nuclei with higher atomic numbers, the number
of neutrons exceeds the number of protons
because the number of neutrons necessary to
create a stable nucleus increases more rapidly
than the number of protons
Chap. 21.2
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
The belt of stability ends at 83 27
- above the belt can lower their
ratio by emitting a beta
- below the belt can increase their
ratio by either positron
emission or electron capture
- nuclei with atomic numbers above
84 tend to undergo alpha
emission
Chap. 21.2
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
Sample exercise: Predict the mode of decay of
(a) plutonium-239
Chap. 21.2
28
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
Sample exercise: Predict the mode of decay of
(a) plutonium-239
29
atomic number of 94, alpha emission
Chap. 21.2
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
Sample exercise: Predict the mode of decay of
(a) indium-120
Chap. 21.2
30
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
Sample exercise: Predict the mode of decay of
(a) indium-120
31
atomic number of 49, neutrons are 71, above the
belt of stability; beta emission
Chap. 21.2
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
Keep in mind that the previous slides describe
32
guidelines to follow, and not all nuclei abide by the
guidelines given.
Certain nuclei can not gain stability by a single
emission. Elements like this have a series of
emissions called a disintegration series.
Chap. 21.2
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
Uranium-238 is an
excellent example of
a nuclei which has a
disintegration series
Chap. 21.2
33
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
Two other observations have proven useful in the
34
determination of stable nuclei
Nuclei with 2, 8, 20, 28, 50, or 82 protons OR 2,
8, 20, 28, 50 or 82 neutrons are generally more
stable. These numbers have been called the
magic numbers
Nuclei with even numbers of both protons and
neutrons are generally more stable than those
with odd numbers of nucleons
Chap. 21.2
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
Sample exercise: Which of the following nuclei would35
you expect to exhibit a special stability:
118Sn
Chap. 21.2
210At , 208Pb
,
50
85
82
Chem. 116 Prof. T.L. Heise
Patterns of Nuclear stability
Sample exercise: Which of the following nuclei would36
you expect to exhibit a special stability:
118Sn
Chap. 21.2
50
208Pb
82
Chem. 116 Prof. T.L. Heise
Nuclear Transmutations
Another way a nucleus can change identity is to be 37
struck by a neutron or by another nucleus. Nuclear
reactions that have been induced this way are called
Nuclear (Artificial) Transmutations
Nuclear Transmutations are listed in the following
order:
target nucleus + bombarding particle -->
ejected particle + product nucleus
14N
Chap. 21.3
4He --> 1H + 17O
+
7
2
1
8
14N (a,
7
p) 17O8
Chem. 116 Prof. T.L. Heise
Nuclear Transmutations
Charged particles must be moving very fast in order to
38
overcome the electrostatic repulsion between them
and the target nucleus.
- the higher the nuclear charge on either the
projectile or the target, the faster the particle
must be going
- Strong magnetic and electric fields are used to
accelerate the particles.
Chap. 21.3
Chem. 116 Prof. T.L. Heise
Nuclear Transmutations
Particle Accelerators
Chap. 21.3
39
Chem. 116 Prof. T.L. Heise
Nuclear Transmutations
Particle Accelerators
Chap. 21.3
40
Chem. 116 Prof. T.L. Heise
Nuclear Transmutations
Most synthetic isotopes in quantity in medicine and
scientific research are made using neutrons as
projectiles
- neutrons are neutral so there is no nuclear
repulsion to overcome
- no need to be accelerated
Chap. 21.3
41
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Different nuclei undergo radioactive decay at different
42
rates.
Radioactive decay is a first order kinetic process
- characteristic half life
- independent of initial concentration
- unaffected by external forces such as
temperature, pressure, or state of chemical
combination
- radioactive atoms cannot be rendered harmless
by a chemical reaction or by any other
practical treatment
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: Carbon-11, used in medical imaging,43
has a half life of 20.4 min. The carbon-11 nuclides
are formed and then incorporated into a desired
compound. The resulting sample is injected into the
patient, and the image is obtained. The entire process
takes five half lives. What percentage of original
carbon remains at this time?
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: Carbon-11, used in medical imaging,44
has a half life of 20.4 min. The carbon-11 nuclides
are formed and then incorporated into a desired
compound. The resulting sample is injected into the
patient, and the image is obtained. The entire process
takes five half lives. What percentage of original
carbon remains at this time?
100
50
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: Carbon-11, used in medical imaging,45
has a half life of 20.4 min. The carbon-11 nuclides
are formed and then incorporated into a desired
compound. The resulting sample is injected into the
patient, and the image is obtained. The entire process
takes five half lives. What percentage of original
carbon remains at this time?
100
50
Chap. 21.4
25
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: Carbon-11, used in medical imaging,46
has a half life of 20.4 min. The carbon-11 nuclides
are formed and then incorporated into a desired
compound. The resulting sample is injected into the
patient, and the image is obtained. The entire process
takes five half lives. What percentage of original
carbon remains at this time?
100
50
Chap. 21.4
25
12.5
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: Carbon-11, used in medical imaging,47
has a half life of 20.4 min. The carbon-11 nuclides
are formed and then incorporated into a desired
compound. The resulting sample is injected into the
patient, and the image is obtained. The entire process
takes five half lives. What percentage of original
carbon remains at this time?
100
Chap. 21.4
50
25
12.5
6.25
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: Carbon-11, used in medical imaging,48
has a half life of 20.4 min. The carbon-11 nuclides
are formed and then incorporated into a desired
compound. The resulting sample is injected into the
patient, and the image is obtained. The entire process
takes five half lives. What percentage of original
carbon remains at this time?
100
Chap. 21.4
50
25
12.5
6.25
3.125
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Due to the constancy of half lives, they can be used as49a
molecular clock to determine the ages of different
objects
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Shroud of Turin - face
Chap. 21.4
50
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Shroud of Turin - hands
Chap. 21.4
51
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Calculation based on Half-lives
 Rate = kN
 the first order rate constant is called a decay
constant
 The rate at which a sample decays is called its
activity, units are disintegrations/sec
 ln(Nt/No) = -kt
 k = 0.693/t1/2
Chap. 21.4
52
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A wooden object from an
53
archeological site is subjected to radiocarbon dating.
The activity of the sample due to carbon-14 is
measured to be 11.6 disintegration per second. The
activity of a carbon sample of equal mass from fresh
wood is 15.2 disintegrations per second. The half-life
of 14C is 5,715 yr. What is the age of the archeological
sample?
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A wooden object from an
54
archeological site is subjected to radiocarbon dating.
The activity of a carbon sample of equal mass from
fresh wood is 15.2 disintegrations per second. The
half-life of 14C is 5,715 yr. What is the age of the
archeological sample?
k = 0.693/t1/2
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A wooden object from an
55
archeological site is subjected to radiocarbon dating.
The activity of a carbon sample of equal mass from
fresh wood is 15.2 disintegrations per second. The
half-life of 14C is 5,715 yr. What is the age of the
archeological sample?
k = 0.693/t1/2
k = 0.693/5,715 yr
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A wooden object from an
56
archeological site is subjected to radiocarbon dating.
The activity of a carbon sample of equal mass from
fresh wood is 15.2 disintegrations per second. The
half-life of 14C is 5,715 yr. What is the age of the
archeological sample?
k = 0.693/t1/2
k = 0.693/5,715 yr
k = 1.21 x 10-4 yr-1
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A wooden object from an
57
archeological site is subjected to radiocarbon dating.
The activity of a carbon sample of equal mass from
fresh wood is 15.2 disintegrations per second. The
half-life of 14C is 5,715 yr. What is the age of the
archeological sample?
k = 0.693/t1/2
t = (-1/k)ln(Nt/No)
k = 0.693/5,715 yr
k = 1.21 x 10-4 yr-1
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A wooden object from an
58
archeological site is subjected to radiocarbon dating.
The activity of a carbon sample of equal mass from
fresh wood is 15.2 disintegrations per second. The
half-life of 14C is 5,715 yr. What is the age of the
archeological sample?
k = 0.693/t1/2
t = (-1/k)ln(Nt/No)
k = 0.693/5,715 yr
t = (-1/1.21x10-4)ln(11.6/15.2)
k = 1.21 x 10-4 yr-1
t = (-8264)(-0.2702)
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A wooden object from an
59
archeological site is subjected to radiocarbon dating.
The activity of a carbon sample of equal mass from
fresh wood is 15.2 disintegrations per second. The
half-life of 14C is 5,715 yr. What is the age of the
archeological sample?
k = 0.693/t1/2
t = (-1/k)ln(Nt/No)
k = 0.693/5,715 yr
t = (-1/1.21x10-4)ln(11.6/15.2)
k = 1.21 x 10-4 yr-1
t = (-8264)(-0.2702)
t = 2233 yr
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A sample to be used for medical 60
imaging is labeled with 18F, which has a half-life of
110 minutes. What percentage of the original activity
in the sample remains after 300 minutes?
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A sample to be used for medical 61
imaging is labeled with 18F, which has a half-life of
110 minutes. What percentage of the original activity
in the sample remains after 300 minutes?
k = 0.693/t1/2
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A sample to be used for medical 62
imaging is labeled with 18F, which has a half-life of
110 minutes. What percentage of the original activity
in the sample remains after 300 minutes?
k = 0.693/t1/2
k = 0.693/110 min.
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A sample to be used for medical 63
imaging is labeled with 18F, which has a half-life of
110 minutes. What percentage of the original activity
in the sample remains after 300 minutes?
k = 0.693/t1/2
k = 0.693/110 min.
k = 0.0063 min.-1
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A sample to be used for medical 64
imaging is labeled with 18F, which has a half-life of
110 minutes. What percentage of the original activity
in the sample remains after 300 minutes?
k = 0.693/t1/2
ln(Nt/No) = -kt
k = 0.693/110min
k = 0.0063 min-1
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A sample to be used for medical 65
imaging is labeled with 18F, which has a half-life of
110 minutes. What percentage of the original activity
in the sample remains after 300 minutes?
k = 0.693/t1/2
ln(Nt/No) = -kt
k = 0.693/ 110 min
ln(x/100g) = -0.0063(300)
k = 0.0063 min-1
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A sample to be used for medical 66
imaging is labeled with 18F, which has a half-life of
110 minutes. What percentage of the original activity
in the sample remains after 300 minutes?
k = 0.693/t1/2
ln(Nt/No) = -kt
k = 0.693/ 110 min
ln(x/100g) = -0.0063(300)
k = 0.0063 min-1
x/100 g = e-1.89
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A sample to be used for medical 67
imaging is labeled with 18F, which has a half-life of
110 minutes. What percentage of the original activity
in the sample remains after 300 minutes?
k = 0.693/t1/2
ln(Nt/No) = -kt
k = 0.693/110 min
ln(x/100g) = -0.0063(300)
k = 0.0063 min-1
x/100 g = e-1.89
x/100 g = 0.151
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Rates of Radioactive Decay
Sample exercise: A sample to be used for medical 68
imaging is labeled with 18F, which has a half-life of
110 minutes. What percentage of the original activity
in the sample remains after 300 minutes?
k = 0.693/t1/2
ln(Nt/No) = -kt
k = 0.693/ 110 min
ln(x/100g) = -0.0063(300)
k = 0.0063 min-1
x/100 g = e-1.89
x/100 g = 0.151
x = 15.1 g or 15.1%
Chap. 21.4
Chem. 116 Prof. T.L. Heise
Detection of Radiation
A variety of methods have been designed to detect 69
emissions from radioactive substances.
 Photographic film and plates, the greater the
exposure, the darker the area exposed
 Geiger counters, uses the conduction of electricity
by ions and electrons produced by radioactive
substances
 Phosphors glow when as electrons excited by
radiation fall back down to ground state
 Scintillation counter detects tiny flashes of light
from phosphors
Chap. 21.5
Chem. 116 Prof. T.L. Heise
Detection of Radiation
 Geiger counters
Chap. 21.5
70
Chem. 116 Prof. T.L. Heise
Detection of Radiation
Radiotracers: a radioactive element that can be traced
71
so easily they are used to follow the pathway a
chemical reaction takes
- ability to do this comes from the fact that all
isotopes of an element have essentially identical
chemical properties
- the chemicals pathway is revealed by the
radioactivity of the radioisotope
Chap. 21.5
Chem. 116 Prof. T.L. Heise
Energy Changes
The energies involved in nuclear reactions must be
considered using Einstein’s famous equation
72
E = mc2
This equation states that the mass and energy of an
object are proportional, if a system loses mass, it loses
energy and vice versa.
The proportionality constant c2 is so large, even small
changes in mass cause large changes in energy
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
The mass changes and the associated energy changes 73
in
nuclear reactions are much greater than those in
chemical reactions.
- the mass change in the decay of 1 mole of
Uranium-238 is 50,000 times greater than
that for the combustion of one mole of
methane.
238U
Chap. 21.6
234Th + 4He
-->
92
90
2
Chem. 116 Prof. T.L. Heise
Energy Changes
238U
234Th + 4He
-->
92
90
2
74
mass of
nuclei: 238.0003
233.9942 + 4.0015
(amu)
238.0003 = 237.9957
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
238U
234Th + 4He
-->
92
90
2
75
mass of
nuclei: 238.0003
233.9942 + 4.0015
(amu)
238.0003 = 237.9957
0.0046 amu are LOST, so proportional
energy is LOST
**Lost energy is exothermic
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
238U
234Th + 4He
-->
92
90
2
76
mass of
nuclei: 238.0003
233.9942 + 4.0015
(amu)
238.0003 = 237.9957
0.0046 amu
If 1 mole of U-238 is considered, amu turns into grams
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
238U
234Th + 4He
-->
92
90
2
77
mass of
nuclei: 238.0003
233.9942 + 4.0015
(g)
238.0003 = 237.9957
0.0046 g
E = mc2
E = 0.0000046 kg(3.00x108m/s)2
E = 4.14x1011 kg m2/s2
E = 4.14x1011 J
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
Sample exercise: Positron emission form 11C,
78
11C --> 11B + 0e
6
5
1
occurs with release of 2.87x1011 J per mole of 11C. What
is the mass change per mole of 11C in this nuclear
reaction?
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
Sample exercise: Positron emission form 11C,
79
11C --> 11B + 0e
6
5
1
occurs with release of 2.87x1011 J per mole of 11C. What
is the mass change per mole of 11C in this nuclear
reaction?
E = mc2
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
Sample exercise: Positron emission form 11C,
80
11C --> 11B + 0e
6
5
1
occurs with release of 2.87x1011 J per mole of 11C. What
is the mass change per mole of 11C in this nuclear
reaction?
E = mc2
2.87x1011 J = m(3.00x108m/s)2
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
Sample exercise: Positron emission form 11C,
81
11C --> 11B + 0e
6
5
1
occurs with release of 2.87x1011 J per mole of 11C. What
is the mass change per mole of 11C in this nuclear
reaction?
E = mc2
2.87x1011 J = m(3.00x108m/s)2
2.87x1011 J = m
(3.00x108m/s)2
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
Sample exercise: Positron emission form 11C,
82
11C --> 11B + 0e
6
5
1
occurs with release of 2.87x1011 J per mole of 11C. What
is the mass change per mole of 11C in this nuclear
reaction?
E = mc2
2.87x1011 J = m(3.00x108m/s)2
2.87x1011 J = m
(3.00x108m/s)2
3.18x 10-6 kg = m
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
Sample exercise: Positron emission form 11C,
83
11C --> 11B + 0e
6
5
1
occurs with release of 2.87x1011 J per mole of 11C. What
is the mass change per mole of 11C in this nuclear
reaction?
E = mc2
2.87x1011 J = m(3.00x108m/s)2
2.87x1011 J = m
(3.00x108m/s)2
3.19x 10-6 kg = m
0.00319 g = m
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
Scientists discovered in the 1930’s that the masses of 84
nuclei are always less than the masses of the
individual nucleons of which they are composed.
 The mass difference between a nucleus and its
constituent nucleons is called the mass defect
 The origin of the mass defect is readily understood
if we consider that energy is used to break into the
nucleons
 The larger the binding energy, the more stable the
nucleus
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Energy Changes
85
nuclei of intermediate mass numbers are more tightly
bound than those with smaller or larger mass
numbers
- a larger atom will break up into two intermediates
- 2 or more smaller atoms will fuse into an
intermediate
Chap. 21.6
Chem. 116 Prof. T.L. Heise
Nuclear Fission
86
Chap. 21.7
Chem. 116 Prof. T.L. Heise
Nuclear Fission
2.4 neutrons produced by every fission of uranium-235.
87
Number of fissions and energy released quickly
escalates exponentially is unchecked
In order for a fission chain reaction to occur a
minimum mass of material must be present
(critical mass) - with minimum present only one
neutron is effective in producing another fission
Chap. 21.7
Chem. 116 Prof. T.L. Heise
Nuclear Fission
88
2.4
Chap. 21.7
Chem. 116 Prof. T.L. Heise
Nuclear Fission
89
To trigger the fission
reaction, two subcritical
masses are slammed
together using chemical
explosives.
The two combined
masses are supercritical
which rapidly leads to an
uncontrolled nuclear
explosion
Chap. 21.7
Chem. 116 Prof. T.L. Heise
Nuclear Fission
Nuclear Reactors:
90
Uranium is enriched to about 3% U-235 and
then used to form UO2 pellets that are encased
in zirconium or stainless steel tubes
Rods composed of materials such as cadmium
or boron control the fission process by
absorbing neutrons
Moderators slow down neutrons so they can be
captured more readily by the fuel
Chap. 21.7
Chem. 116 Prof. T.L. Heise
Nuclear Fission
Nuclear Reactors:
91
A cooling liquid is circulated through the core to
carry off heat generated by the nuclear fission.
Cooling liquid and moderator could be one and the
same substance
Steam is used to drive a turbine connected to an
electrical generator, however steam must be
condensed so additional cooling liquid is
required, generally acquired from lake or river
Chap. 21.7
Chem. 116 Prof. T.L. Heise
Nuclear Fission
Nuclear Reactors:
92
Reactor is surrounded by a concrete shell to
shield personnel and nearby residents from
radiation
Reactor must be stopped periodically so that the
fuel can be replaced or reprocessed
Spent fuel rods are being kept in storage at
reactor sites
20 half-lives are required for their radioactivity
to reach levels acceptable for biological exposure
(600 years)
Chap. 21.7
Chem. 116 Prof. T.L. Heise
Nuclear Fission
93
Chap. 21.7
Chem. 116 Prof. T.L. Heise
Nuclear Fusion
Fusion
94
appealing as an energy source because of
availability of light isotopes and because fusion
products are generally not radioactive
not presently used to generate energy because
high energies are needed to overcome the
repulsion between nuclei
reaction requires temps of about 40,000,000 K
these temps have only been achieved using a
hydrogen bomb
Chap. 21.8
Chem. 116 Prof. T.L. Heise
Nuclear Fusion
Fusion
95
also a problem with confining the reaction - no
known structural material can withstand such
temps
possibilities? Tokamak
Lasers
Chap. 21.8
Chem. 116 Prof. T.L. Heise
Nuclear Fusion
Tokamak
Chap. 21.8
96
Chem. 116 Prof. T.L. Heise
Biological Effects
We are continually bombarded with radiation!
97
When matter absorbs radiation, the energy of
radiation can cause either excitation or
ionization of the matter
- ionizing radiation is more harmful
When living tissue is irradiated, most of the
energy is absorbed by the 70% water by mass of
living tissue
Chap. 21.9
Chem. 116 Prof. T.L. Heise
Biological Effects
Ionizing radiation
98
electrons are removed from water forming
highly reactive H2O+ ions
H2O+ + H2O --> H3O+ + OH
the unstable and highly reactive OH molecule is
an example of a free radical due to the unpaired
electron, •OH
in tissue, free radicals attack a host of
surrounding biomolecules to produce more free
radicals
Chap. 21.9
Chem. 116 Prof. T.L. Heise
Biological Effects
Damage depends on
activity and energy of the radiation
length of exposure
whether source is inside or outside the body
Tissue that shows most damage
reproduce at rapid rates
bone marrow
blood forming tissue
lymph nodes
Chap. 21.9
99
Chem. 116 Prof. T.L. Heise
Biological Effects
Extended Exposure to Low Doses
100
cancer
damage to growth regulation mechanism in cell,
inducing cells to reproduce in an uncontrolled
manner
Chap. 21.9
Chem. 116 Prof. T.L. Heise
Biological Effects
Units used to measure radiation
101
becquerel (Bq) = 1 nuclear disintegration per
second
curie (Ci) = 3.7 x 1010 disintegrations per
second
gray (Gy) = 1 J absorbed per kilogram of tissue
rad (radiation absorbed dose) = 1 x 10-2 J per
kilogram of tissue
to correct for differences in strengths of varying
radiation, a multiplication factor is used
Chap. 21.9
Chem. 116 Prof. T.L. Heise
Biological Effects
Radon
102
Rn-222 is a product of nuclear disintegration of
U-238
being a noble gas, radon is extremely unreactive
and easily escapes the ground
radon has a short half life and emits alpha
particles
222Rn --> 218Po + 4He
86
84
2
 polonium is also an alpha emitter
Chap. 21.9
Chem. 116 Prof. T.L. Heise