chapter 4 nuclear chemistry: the heart of matter daniel fraser university of toledo, toledo oh...
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Chapter 4 Nuclear Chemistry: The Heart of
Matter
Daniel Fraser
University of Toledo, Toledo OH
©2004 Prentice Hall
Chemistry for Changing Times 10th editionHill/Kolb
Chapter 4 2
Types of Radiation
• Ionizing radiation – knocks electrons out of atoms or groups of atoms– Produces charged species – ions– Charged species that cause damage
Chapter 4 3
Differences Between Chemical and Nuclear Reactions
Chapter 4 4
Half-Life
• Period for one-half of the original elements to undergo radioactive decay
• Characteristic for each isotope
• Fraction remaining = n = number of half-lives
n2
1
Chapter 4 5
Practice Problems
Chapter 4 6
Radioisotopic Dating
• Use certain isotopes to estimate the age of various items
• 235U half-life = 4.5 billion years– Determine age of rock
• 3H half-life = 12.3 years– Used to date aged wines
Chapter 4 7
Chapter 4 8
Carbon-14 Dating
• 99.9% 12C
• Produce 14C in upper atmosphere
• Half-life of 5730 years
• ~50,000 y maximum age for dating
H Cn N 11
146
10
147
Chapter 4 9
You obtain a new sample of cobalt-60, half-life 5.25 years, with a mass of 400 mg. How much cobalt-60 remains after 15.75 years (three half-lives)?
Example 4.2 Half-Lives
SolutionThe fraction remaining after three half-lives is
1
2n
1
23
1
2 x 2 x 2
1
8===
The amount of cobalt-60 remaining is ( ) (400 mg) = 50 mg.1
8
You have 1.224 mg of freshly prepared gold-189, half-life 30 min. How much of the gold-189 sample remains after five half-lives?
Exercise 4.2A
What percentage of the original radioactivity remains after five half-lives?
Exercise 4.2B
Chapter 4 10
You obtain a 20.0-mg sample of mercury-190, half-life 20 min. How much of the mercury-190 sample remains after 2 hr?
Example 4.3
There are 120 min in 2 hr. There are ( ) = 6 half-lives in 2 hr. The fraction remainingafter six half-lives is
The amount of mercury-190 remaining is ( ) (20.0 mg) = 0.313 mg.
Solution
1
2n
1
26
1
2 x 2 x 2 x 2 x 2 x 2
1
64===
1
64
12020
A sample of 16.0 mg of nickel-57, half-life 36.0 hr, is produced in a nuclear reactor. How much of the nickel-57 sample remains after 7.5 days?
Exercise 4.3A
Tc-99 decays to Ru-99 with a half-life of 210,000 years. Starting with 1.0 mg of Tc-99, how long will it take for 0.75 mg of Ru-99 to form?
Exercise 4.3B
Chapter 4 11
A piece of fossilized wood has carbon-14 activity one-eighth that of new wood. How old is the artifact? The half-life of carbon-14 is 5730 years.
Example 4.4
SolutionThe carbon-14 has gone through three half-lives:
It is therefore about 3 x 5730 = 17,190 years old.
1
8==
1
2
1
2x
1
2x( )1
2
3
How old is a piece of cloth that has carbon-14 activity that of new cloth fibers? The half-life of carbon-14 is 5730 years.
Exercise 4.41
16
Chapter 4 12
Shroud of Turin
• Alleged burial shroud of Jesus Christ– Contains faint human likeness– First documented in Middle Ages
• Carbon-14 dating done in 1988– Three separate labs– Shroud ~800 years old– Unlikely to be burial shroud
Chapter 4 13
Uses of Radioisotopes• Tracers
– Easy to detect– Different isotopes have similar chemical and
physical properties– Physical, chemical, or biological processes
• Agriculture– Induce heritable genetic alterations – mutations– Preservative
– Destroys microorganisms with little change to taste or appearance of the food
Chapter 4 14
Nuclear Medicine
• Used for two purposes
• Therapeutic – treat or cure disease using radiation
• Diagnostic – obtain information about patient’s health
Chapter 4 15
Radiation Therapy
• Radiation most lethal to dividing cells
• Makes some forms of cancer susceptible
• Try to destroy cancer cells before too much damage to healthy cells– Direct radiation at cancer cells– Gives rise to side effects
Chapter 4 16
Diagnostic Uses
• Many different isotopes used– See Table 4.6
• Can measure specific things– Iodine-131 to locate tumors in thyroid– Selenium-75 to look at pancreas– Gadolinium-153 to determine bone
mineralization
Chapter 4 17
Imaging
• Positron emission tomography (PET)
• Uses an isotope that emits a positron
• Observe amount of radiation released
e B C 01
115
116
Chapter 4 18
Chapter 4 19
Penetrating Power of Radiation• The more mass the particle has, the
less penetrating it is
• The faster the particle is, the more penetrating it is
Chapter 4 20
Prevent Radiation Damage• To minimize
damage – Stay a distance
from radioactive sources
– Use shielding; need more with more penetrating forms of radiation
Chapter 4 21
Energy from Nucleus
• E = mc2
• Lose mass, gain energy– For chemical
reactions, mass changes are not measurable
– For nuclear reactions, mass changes may be measurable
Chapter 4 22
Binding Energy• Holds protons and neutrons together in
the nucleus
• The higher the binding energy, the more stable the element
Chapter 4 23
Nuclear Fission• “Splitting the atom”
• Break a large nucleus into smaller nuclei
Chapter 4 24
Nuclear Chain Reaction• Neutrons from one
fission event split further atoms
• Only certain isotopes, fissile isotopes, undergo nuclear chain reactions
Chapter 4 25
Manhattan Project
• How to sustain the nuclear reaction?
• How to enrich uranium to >90% 235U?– Only 0.7% natural abundance
• How to make 239Pu (another fissile isotope)?
• How to make a nuclear fission bomb?
Chapter 4 26
Radioactive Fallout
• Nuclear bomb detonated; radioactive materials may rain down miles away and days later– Some may be unreacted U or Pu– Radioactive isotopes produced during the
explosion
Chapter 4 27
Nuclear Power Plants
• Provide ~20% U.S. electricity– France >70%
• Slow controlled release of energy
• Need 2.5–3.5% 235U
• Problem with disposal of radioactive waste
Chapter 4 28
Nuclear Fusion
• Reaction takes smaller nuclei and builds larger ones– Also called thermonuclear reactions
e2 He H4 01
42
11
• Releases tremendous amounts of energy–1 g of H would release same as 20 tons of coal
Chapter 4 29
End of Chapter 4
Chapter 4 30
Radioisotopes
• Radioactive decay
– Many isotopes are unstable
• Radioisotopes– Nuclei that undergo radioactive decay– May produce one or more types of
radiation
Chapter 4 31
Natural Radioactivity
• Background radiation– What occurs from
natural sources– >80% of
radioactivity exposure
Chapter 4 32
Nuclear Equations
• Elements may change in nuclear reactions
• Total mass and sum of atomic numbers must be the same
• MUST specify isotope
Po He Rn 21884
42
22286
Chapter 4 33
Alpha Decay• Nucleus loses particle
– Mass decreases by 4 and atomic number decreases by 2
He42
Chapter 4 34
Beta Decay• Nucleus loses particle
– No change in mass but atomic number increases
e01
Chapter 4 35
Positron Emission• Loses a positron
– Equal mass but opposite charge of an electron
– Decrease in atomic number and no change in mass
+
Chapter 4 36
Electron Capture• Nucleus
absorbs an electron and then releases an X-ray
• Mass number stays the same and atomic number decreases
Chapter 4 37
Gamma Radiation
• Release of high-energy photon
• Typically occurs after another radioactive decay
• No change in mass number or atomic number
Chapter 4 38
Artificial Transmutation
• Transmutation changes one element into another– Middle Ages: change lead to gold
• In 1919 Rutherford established protons as fundamental particles– Basic building blocks of nuclei
H O He N 11
178
42
147
Chapter 4 39
Example 4.1 Balancing Nuclear EquationsWrite balanced nuclear equations for each of the following processes. In each case, indicate what new element is formed.a. Plutonium-239 emits an alpha particle when it decays.b. Protactinium-234 undergoes beta decay.c. Carbon-11 emits a positron when it decays.d. Carbon-11 undergoes electron capture.
Solutiona. We start by writing the symbol for plutonium-239 and a partial equation showing that one of the products is an alpha particle (helium nucleus):
23994
Pu42
He + ?
Mass and charge are conserved. The new element must have a mass of 239 – 4 = 235 and a charge of 94 – 2 = 92. The nuclear charge (atomic number) of 92 identifies the element as uranium (U):
23994
Pu42
He +235
92U
Chapter 4 40
Example 4.1 Balancing Nuclear Equations (cont.)
b. Write the symbol for protactinium-234 and a partial equation showing that one of the products is a beta particle (electron):
23491
Pa0
–1e + ?
The new element still has a mass number of 234. It must have a nuclear charge of 92 in order for the total charge to be the same on each side of the equation. The nuclear charge identifies the new atom as another isotope of uranium (U):
23491
Pr0
–1e +
23492
U
c. Write the symbol for carbon-11 and a partial equation showing that one of the products is a positron:
116
C0
+1e + ?
To balance the equation, a particle with a mass number of 11 and an atomic number of 5 (B) is required:
116
C0
+1e +
115
B
Chapter 4 41
As mentioned in the text, positron emission and electron capture result in identicalchanges in atomic number, and therefore the identical elements are formed! Also, asparts (c) and (d) illustrate, C-11 (and certain other nuclei) can undergo several differenttypes of radioactive decay processes.
Example 4.1 Balancing Nuclear Equations (cont.)
Write balanced nuclear equations for each of the following processes. In each case, indicate what new element is formed.
Exercise 4.1
a. Radium-226 decays by alpha emission.b. Sodium-24 undergoes beta decay.c. Gold-188 decays by positron emission.d. Argon-37 undergoes electron capture.
116
C0
–1e
d. We write the symbol for carbon-11 and a partial equation showing it capturing an electron:
+ ?
116
C0
–1e
115
B
To balance the equation, the product must have a mass number of 11 and an atomic number of 5 (B):
+
Chapter 4 42
Example 4.5When potassium-39 is bombarded with neutrons, chlorine-36 is produced. What other particle is emitted?
Cl + ?3617
10
n3919
K +
SolutionWrite a balanced nuclear equation. To balance the equation, we need four mass unitsand two charge units (that is, a particle with a nucleon number of 4 and an atomicnumber of 2). That’s an alpha particle.
Cl +3617
10
n3919
K + He42
Technetium-97 is produced by bombarding molybdenum-96 with a deuteron (hydrogen-2 nucleus). What other particle is emitted?
Exercise 4.5
Tc + ?9743
21
H9642
Mo +
Chapter 4 43
Example 4.6One of the isotopes used for PET scans is oxygen-15, a positron emitter. What new element is formed when oxygen-15 decays?
Phosphorus-30 is a positron-emitting radioisotope suitable for use in PET scans. What new element is formed when phosphorus-30 decays?
Exercise 4.6
SolutionFirst write the nuclear equation
0+1
e + ?15
8O
The nucleon number does not change, but the atomic number becomes 8 – 1, or 7; and sothe new product is nitrogen-15:
0+1
e +15
8O
157
N
Chapter 4 44
Chapter 4 45
Chapter 4 46
Chapter 4 47
Chapter 4 48
Chapter 4 49
Chapter 4 50
Chapter 4 51
Chapter 4 52
Chapter 4 53
Chapter 4 54
Chapter 4 55
Chapter 4 56
Chapter 4 57
Chapter 4 58