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Chapter 21: Nuclear Chemistry
Chemistry: The Molecular Nature of Matter, 6E
Jespersen/Brady/Hyslop
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
2
How are atoms formed? Big Bang—Intense heat ~109 K
Cooled quickly to 106 K—T of stars e–, p, n formed and joined into nuclei—atoms
Mostly H and He (as in our sun) Rest of elements formed by nuclear
reactions Fusion—two nuclei come together to
form another heavier nucleus Fission—one heavier nucleus splits into
lighter nuclei
Various other types of reactions+
+
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Nuclear Shorthand Nucleons
Subatomic particles found in the nucleus Protons (p) Neutrons (n)
Nuclide Specific nucleus with given atomic number
(Z ) Atomic Number (Z )
Number of protons in nucleus Determines chemical properties of nuclide Z = p
Mass Number (A)—mass of nuclide A = n + p
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Shorthand for Writing Nuclides Where X = atomic symbol
e.g. In the neutral atom: e– = p = Z Isotopes
Nuclides with same Z (same number of p), but different A (different n)
Hydrogen Deuterium Tritium
1 p 1 p + 1 n 1 p + 2 n
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Radioactivity Radioactive isotopes
Isotopes with unstable atomic nuclei Emit high energy streams of particles or
electromagnetic radiation Radionuclides
Another name for radioactive isotopes Undergo nuclear reactions
Uses Dating of rocks and ancient artifacts Diagnosis and treatment of disease Source of energy
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Mass Not Always Constant Mass of particle not constant under all
circumstances It depends on velocity of particle relative to
observer As approaches speed of light, mass increases
When v goes to zero Particle has no velocity relative to observer v/c 0 Denominator 1 and m = mo
2)/(1 cv
mm
m = mass of particle
v = velocity of particlem = rest massc = speed of light
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Why don’t we observe mass change?
In lab and ordinary life, velocity of particle is small
Only see mass vary with speed as velocity approaches speed of light, c As v c, (v/c) 0 and m ∞
In lab, m = mo within experimental error Difference in mass too small to measure
directly Scientists began to see relationship
between mass and total energy Analogous to potential and kinetic energies
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Law of Conservation of Mass and Energy
Mass and energy can neither be created nor destroyed, but can be converted from one to the other.
Sum of all energy in universe and all mass (expressed in energy equivalents) in universe is constant
Einstein Equation E = (mo)c
2
Where c = 2.9979 × 108 m/s
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Mass Defect Rest mass of nuclide is always less than sum of
masses of all individual nucleons (neutrons and protons) in that same nuclide Mass is lost upon binding of neutrons and protons
into nucleus When nucleons come together, loss of mass
translates into release of enormous amount of energy by Einstein's relationEnergy released = – Nuclear Binding
Energy Nuclear Binding Energy
Amount of energy must put in to break apart nucleus
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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What is Mass Loss?
nucleonsnucleus mmm
For given isotope of given Z and A
or
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 1 Binding Energy CalculationWhat is the binding energy of 7Li3+ nucleus?
Step 1. Determine mass loss or mass defectA. Determine mass of nucleus
mass of 7Li3+ = m (7Li isotope) – 3me
= 7.016003 u – 3(0.0005485 u) = 7.0143573 u
B. Determine mass of nucleonsmass of nucleons = 3 mp + 4 mn
= 3(1.007276470 u) + 4(1.008664904 u) = 7.056489026 u
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 1 Binding Energy Calculation (cont.)C. m = mnucleus – mnucleons
= 7.0143573 u – 7.056489026 u= –0.0421317 u= mass lost by nucleons when they form
nucleusStep 2. Determine energy liberated by this
change in massE = (mo)c2
E = – 6.287817 × 10–12 J/atom
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 1 Binding Energy Calculation (cont.)
E = –6.287817 × 1012 J/atom × 6.0221367 x 1023
atoms/moleE = –3.78655 × 1012 J/mole = –3.78655 × 109 kJ/mole
Compare this to: 104 – 105 J/mol (102 – 103 kJ/mol) for
chemical reactions Nuclear ~ 1 – 10 million times larger
than chemical reactions!!
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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MeV (Energy Unit)
Nuclear scientists find it convenient to use a different Energy unit: MeV (per atom)
Electron volt (eV) Energy required to move e across energy
potential of 1 V 1 eV = 1.602 × 10–19 J M(mega) = 1 × 106 So 1 MeV = 1 × 106 eV 1 MeV = 1.602 × 10–13 J
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 1 Binding Energy Calculation in MeV
For Ex. 1. Converting E to MeV gives
Often wish to express binding energy per nucleon so we can compare to other nuclei For Li3+ with 3 1p and 4 n this would be
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 2 Calculate E Released
The overall reaction in the sun responsible for the energy it radiates is
How much energy is released by this
reaction in kJ/mole of He?
m (1H) = 1.00782 um (4He) = 4.00260 um (0+) = 0.00054858 u
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Ex. 2 Calculate E Released (cont.) m = mproducts – mreactants
m = m (4He) + 2m (e +) – 4m (1H)
m = 4.00260 u + 2(0.00054858 u) – 4(1.00782 u)
m = –0.02758 u [We will convert u to kg, kg m2/s2 to J, and atoms to moles
in the following calculation]
E = –2.479 × 1012 J/mol = –8.268 × 109 kJ/mol
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Determine the binding energy, in kJ/mol and MeV/atom, for an isotope that has a mass
defect of –0.025861 u.A. –2.3243 × 109 kJ/mol; 24.092
MeV/atomB. –3.8595 × 10–12 kJ/mol; 24.092
MeV/atomC. –7.7529 kJ/mol; 8.03620 × 10–8
MeV/atomD. –2.3243 × 109 kJ/mol; 4.1508 × 10–2
MeV/atom18
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn! - Solution
19
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Divide binding energy EB by mass number, EB/A
Get binding energy per nucleon
Binding Energies per Nucleon
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Implications of Curve Most EB /A in range of 6 – 9 MeV (per
nucleon) Large binding energy EB /A means stable
nucleus Maximum at A = 56
56Fe largest known EB /A Most thermodynamically stable Nuclear mass number (A) and overall charge
are conserved in nuclear reactions Lighter elements undergo fusion to form
more stable nuclei
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Implications of CurveFusion
Researchers are currently working to get fusion to occur in lab
Heavier elements undergo fission to form more stable elements
Fission
Reactions currently used in bombs and power plants (238U and 239Pu)
As stars burn out, they form elements in center of periodic table around 56Fe
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Radioactivity Spontaneous emission of high energy
particles from unstable nuclei Spontaneous emission of fundamental
particle or light Nuclei falls apart without any external stimuli
Discovered by Becquerel (1896) Extensively studied by Marie Curie and
her husband Pierre (1898 early 1920's) Initially worked with Becquerel
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Fun Facts Marie and Pierre Curie discovered
polonium and radium Nobel Prize in Physics 1903
For discovery of Radioactivity Becquerel, Marie and Pierre Curie—all three
shared Nobel Prize in Chemistry 1911
For discovery of Radium and its properties Marie Curie only
Marie Curie - first person to receive two Nobel Prizes and in different fields
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Discovery of Radioactivity
Initially able to observe three types of decay
Labeled them , , rays (after first three letters of Greek alphabet)
If they pass through an electric field, very different behavior
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Discovery of Radioactivity rays attracted to
negative pole so its positively charged
rays attracted to positive pole so its negatively charged
rays not attracted to either so its not charged
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Nuclear Equations Used to symbolize decay of nucleuse.g. 238U 234Th + parent daughter Produce new nuclei so need separate
rules to balanceBalancing Nuclear Equationsa. Sum of mass numbers (A, top) must be
same on each side of arrowb. Sum of atomic numbers (Z, bottom) must
be same on each side of arrow
He4292 90
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Types of Spontaneous Emission 1. Alpha () Emission
He nucleus2 n + 2 p A = 4 and Z = 2
Daughter nuclei has:A decreases by 4 A = – 4
Z decreases by 2 Z = – 2Very common mode of decay if Z > 83 (large radioactive nuclides)Most massive particle
e.g.
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Balancing Nuclear Equations
29
1. The sum of the mass numbers (A; the superscripts) on each side of the arrow must be the same
2. The sum of the atomic numbers (Z; the subscripts; nuclear charge) on each side of the arrow must be the same
e.g.
A: 234 = 230 + 4Z: 92 = 90 + 2
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Emission of electrons Mass number A = 0 and charge Z = –1 But How? No electrons in nucleus! If nucleus neutron rich — nuclide is too
heavy
2. Beta (– or e–) Emission
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
2. Beta (– or e–) Emission
31
Charge conserved, but not mass m E
Ejected e– has very high KE + emits Antineutrino variable energy particle Accounts for extra E generated
e.g.
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
3. Gamma () Emission Emission of high energy photons Often accompanies or emission Occurs when daughter nucleus of some
process is left in excited state Use * to denote excited state
Nuclei have energy levels analogous to those of e– in atoms
Spacing of nuclear E levels much larger light emitted as -rayse.g.
32
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
4. Positron (+ or e+) Emission
Emission of e+
Positive electron Where does + come from?
If nucleus is neutron poor) Nuclide too light
Balanced for charge, but NOT for mass
33
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
4. Positron (+ or e+) Emission Product side has much greater mass! Reaction costs energy Emission of neutrino
Variable energy particle Equivalent of antineutrino but in realm of
antimatter e+ emission only occurs if daughter
nucleus is MUCH more stable than parent
34
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
4. Positron (+or e+) EmissionWhat happens to e+? Collides with electron to give matter
anti-matter annihilation and two high energy -ray photons m E
Annihilation radiation photons Each with E = 511 keV
What is antimatter? Particle that has counterpart among
ordinary matter, but of opposite charge
High energy light, massless Detect by characteristic peak in
-ray spectrum
35
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
5. Electron Capture (EC) e– in 1s orbital
Lowest Energy e– Small probability that e– is
near nucleus e– actually passes through
nucleus occasionally If it does:
Net effect same as e+ emission
36
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Types of Spontaneous Emission
6. Neutron Emission = ( ) Fairly rare Occurs in neutron rich nuclides Does not lead to isotope of different
element
7. Proton Emission = ( ) Very rare
n10
p11
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Types of Spontaneous Emission
8. Spontaneous Fission No stable nuclei with Z > 83 Several of largest nuclei simply fall
apart into smaller fragments Not just one outcome, usually several
different—see distribution
Fm256100 SbIn 131
5112549
IAg 13653
12047
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Summary—Common Processes1. Alpha () Emission
Very common if Z > 83
2. Beta () Emission e–
Common for neutron-rich nuclides—below belt of stability
3. Positron (+) Emission e+
Common for neutron-poor nuclides—above belt of stability
4. Electron Capture (EC) Occurs in neutron-poor nuclides, especially if Z > 40
5. Gamma () Emission Occurs in metastable nuclei (in nuclear excited
state)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Learning Check Complete the following table which refers
to possible nuclear reactions of a nuclide:
Emission Z = p
n e A New Element?
–
+
EC
–2 –2 0 –4 yes
+1+1–1 0 yes
+1 –1 0 yes–1
0 0 0 0 no
+1+1–1 0 yes
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Learning CheckBalance each of the following equations
a.
b.
c.
d.
e.
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!What is the missing species, , in the
following nuclear reaction?
A.B.C.D.
42
23994Pu
24798Cf
23895Am
23894Pu
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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What Holds Nucleus Together? Consider nucleus
Neutrons and protons in close proximity Strong proton-proton repulsions Neutrons spread protons apart Neutron to proton ratio increases as Z
increases
Strong Forces Force of attraction between nucleons Holds nuclei together Overcomes electrostatic repulsions
between protons Binds protons and neutrons into nucleus
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Table of Nuclides Chart where Rows = different atomic
numberColumns = different number of neutrons
Symbol entered if element is known Stable nuclei
Natural abundance entered below symbol Shaded area Trend of stable nuclei = Belt of Stability Z ≈ number of neutrons (for elements 1 to 20)
Unstable nuclei Give type(s) of radioactive decay
(spontaneous) Outer edges, most of atoms
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Table of Nuclides
Atomic number (Z = number of protons)
Num
ber
of
neu
trons
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Table of Nuclides
Shaded area = stable nuclei Trend of stable nuclei = diagonal line
= Belt of Stability Z ≈ number of neutrons (for
elements 1 to 20) Note: only a small corner of table is
shown. (The complete is in Handbook of Chemistry and Physics)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Belt of Stability Each isotope is a dot Up to Z = 20
Ratio n /Z = 1 As Z increases, n >
Z and By Z = 82, n/Z ~1.5
n = number of neutrons
Z = number of protons
n
Z = p
1n:1p
Stable nuclide, natural Unstable nuclide, natural Unstable nuclide, synthetic
Band of Stability
1n:1p
1.1n:1p
1.2n:1p
1.3n:1p
1.4n:1p
1.5n:1p
e– emitters
e+ emitters
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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How To Predict if Nuclei are Stable1) Atomic Mass = weighted average of masses of
naturally occurring isotopes, i.e. most stable ones2) Compare atomic mass of element to A (atomic mass
number) of given isotope and see if it is more or less Atomic Mass > A too light to be stable Atomic Mass < A too heavy to be stable
Ex. Atomic Mass
Conclusion
180Os 135I Final note: All nuclei with Z > 83 are radioactive
190.2
126.9
Too light, neutron poor
Too heavy, neutron rich
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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More Patterns of Stability If we look at stable and unstable nuclei, other
patterns emerge 283 stable nuclides (out of several thousand
known nuclides) If we look at which have even and odd numbers of
protons (Z) and neutrons (n); patterns emerge
Z n # stable nuclides
even
even
165
even
odd 56
odd even
53
odd odd 5
2H, 6Li, 10B, 14N, 138La
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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More Patterns of Stability Clearly NOT random: even must imply
greater stability Not too surprising
Same is true of electrons in molecules Most molecules have an even number of
electrons, as electrons pair up in orbitals Odd electron molecules, radicals, are
very unstable, i.e. very reactive!!
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Magic Numbers Look at binding energies, see certain
numbers of protons and neutrons result in special stability
Called Magic Numbers 1n and 1p in separate shells Magic numbers (for both 1n and 1p)
are 2, 8, 20, 28, 50, 82, 126…
For e– pattern of stability is: 2, 10, 18, 36, 54, 86…(Noble gases)
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Magic Numbers Special stability of noble gases due
to closed shells of occupied orbitals Structure of nucleus can also be
understood in terms of shell structure With filled shells of neutrons and
protons having added stability At some point adding more neutrons
to higher energy neutron shells decreases stability of nuclei with too high a neutron to proton ratio
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Isotopes above the band of stability are
morelikely to:A. emit alpha particlesB. emit gamma raysC. capture electronsD. emit beta particles
53
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Radioactive Nuclei Found in Nature Non-naturally occurring elements (man-made
unstable) are denoted by having atomic mass in parentheses
All nuclei with Z > 83 are radioactive Yet some elements with Z between 83 and 92
occur naturally Atomic weight is NOT in parentheses
How can this be? There are three heavy nuclei, which have very
long half-lives Long enough to have survived for billions of years
Each parent of natural decay chain
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Decay Chains238U half-life (½) = 4.5 billion years emitter Daughter 234Th is also radioactive
– emitter Half-life much shorter
Long sequence of emissions, and – Recall that emission changes A by 4, while
– emission A = 0 Result: every member of chain has
A = (4n + 2) where n = some simple integer
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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238Uranium Decay Chain238U 234Th
5109 y
25 d
, 234Pa
7 hr
, 234U
5.7105 y
230Th,
8104 y226Ra
, 2103 y
222Rn
4 d218Po
3 m214Pb
,
27 m214Bi
20 m
, 214Po
1.610–4 s
210Pb
22 y,
210Bi
5 d210Po
138 d206Pb
92 90 91 92
90888684
82 83 84 82
82 84 83
A stable isotope
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Decay Chains Final stable member of sequence is 206Pb Some intermediate nuclides have
reasonably short half-lives Still found in nature because they are
constantly being replenished by decay of nuclei further up chain
Uranium-containing minerals (pitchblende is most famous) contain many radioactive elements
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!When the reaction, ,
occurs, the particle emitted is:A. an alpha particleB. a beta particleC. an electronD. a gamma ray
59
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Transmutation Change of one isotope for another Caused by
1. Radioactive decay 2. Bombardment of nuclei with high energy
particles particles from natural emitters Neutrons from atomic reactors Protons made by stripping electrons for hydrogen
Protons and particles can be accelerated in electrical field to give higher E
Mass and energy of bombarding particle enter target nucleus to form compound nucleus
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Non-Spontaneous Nuclear Processes
Fusion Occurs in stars—right now How elements formed
Induced Fission Bombard heavy nuclei with neutron
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Compound Nucleus Designated with * High energy due to velocity of incoming
particle Energy quickly redistributed among nucleons,
but usually unstable To get rid of excess energy, nucleus ejects
something Neutron ▪ Proton Electron ▪ Gamma radiation
Decay leaves new nucleus different from original
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Example: Transmutation
p11
178
189
147
42 O F N He
Compound nucleus
New nucleus
Target nucleus
Bombard-ing particle
High energy particle
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Transmutation Can synthesize given nucleus in many
ways:
Once formed, compound nucleus has no memory of how it was made
Only knows how much energy it has
Al Na He 2713
2311
42
Al Mg 2713
2612
11
p
Al Mg H 2713
2512
21
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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TransmutationDecay pathway depends on how much energy 0
02713 Al
p112612 Mg
n102613 Al
pn 11
10
2512 Mg
He Na 42
2311
Al 2713
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Transmutation Used to synthesize new isotopes
> 900 total Most not on band of stability
All elements above 93 (neptunium) are man- made Includes actinides above 93 + 104 – 112 +
114 Heavier elements made by colliding two
larger nuclei Also known as fusion n1
0269110
270110
20882
6228 Ds Ds Pb Ni
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!What would be the element produced
from the fusion of with ? The species
would be in a high energy state and in time
would undergo decay to other species.A. NoB. LrC. UD. Hs
67
5927Co 192
76Os
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Measuring Radioactive Decay Atomic radiation = ionizing radiation
Creates ions by knocking off electrons Geiger Counter
Consists of a tube with a mica window, low pressure argon fill gas and two high voltage electrodes
Detects and radiation with enough E to penetrate mica window
Inside tube, gas at low pressure is ionized when radiation enters
Ions allow current to flow between electrodes Amount of current relates to amount of radiation
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Measuring Radioactive Decay Scintillation Counter
Surface covered with chemical Emits tiny flash of light when hit by radiation Emission magnified electronically and counted
Film Dosimeters Piece of photographic film Darkens when exposed to radiation How dark depends on how much radiation
exposure over time Too much exposure, person using must be
reassigned to other work
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Activity Number of disintegrations per second Used to characterize radioactive material A = kN
k = first order decay constant in terms of number of nuclei rate than concentration
N = number of radioactive nuclides
Law of radioactive decay Radioactive decay is first order kinetics
process
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Units of Activity SI unit
Bequerel (Bq) 1 disintegration per second (dps) 1 liter of air has ~ 0.04 Bq due to 14C in
CO2
Older unit Curie (Ci)
3.7 × 1010 dps = 3.7 × 1010 Bq Activity in 1.0 g 226Ra = 1 Ci
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Half-Life Time it takes for number of nuclides, Nt ,
present at time, t, to fall to half of its value.
Half-lives are used to characterize nuclides If you know half-life:
Can use to compute k Can also calculate A of known mass of
radioisotope
kkt
693.02ln2
1
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
73
Ex. 3 Activity of Sr-90What is the activity of 1.0 g of strontium-90? The half-life = 28.1 yearsStep 1. Convert t½ to seconds
Step 2. Convert t½ to k
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
74
Ex. 3 Activity of Sr-90 (cont.)Step 3. Convert mass of 90Sr to number
of atoms (N)
Step 4. Calculate Activity = kN
A = 5.23 1012 atoms Sr/s 1 disintegration/atomA = 5.23 1012 dps or 5.23 1012 Bq
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
75
Ex. 4 Mass of 3H in Sample3H, tritium, is a emitter with a half-lifet½ = 12.26 yrs. MW = 3.016 g/mol. How many grams of 3H are in a 0.5 mCi sample?Step 1. Convert half-life to seconds as Ci is in disintegrations per second (dps)
Step 2. Convert t½ to k
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
76
Ex. 4 Mass of 3H in Sample (cont.)Step 3. Convert Ci to dps
Step 4. Calculate g 3H to get this activity
Step 5. Convert atoms to g
= 5.2 × 10–8 g
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
77
Exposure Units Not all materials equally absorb radiation,
thus activity doesn’t describe effect of exposure 1 gray (Gy) = 1 J absorbed energy/kg material
SI unit of absorbed radiation 1 rad = absorption of 10−2 J/ kilogram of tissue
Older unit 1 Gy = 100 rad
These units don’t take into account type of radiation
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Exposure UnitsSieverts (Sv)
SI unit of dose equivalent, H Depends on amount and type of radiation
as well as type of tissue absorbing it H=DQN
H = dose in Sv D = dose in Gy Q = radiation properties N = other factors
Rem = older unit 1 Rem = 10–2 Sv Still used in medicine
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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Exposure to Radiation Typically X ray = 0.007 rem or 7 mrem 0.3 rem/week is maximum safe exposure
set by US government 25 rem (0.25 Sv): Causes noticeable changes
in human blood 100 rem (1 Sv):
Radiation sickness starts to develop 200 rem (2 Sv):
Severe radiation sickness 400 rem (4 Sv):
50% die in 60 days Level of exposure or workers at Chernobyl when
steam explosion tore apart reactor 600 rem (6 Sv): lethal dose to any human
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Workers cleaning up the Fukushima reactors were exposed to as much as 400 mSv units of radiation per hour. How many rems of exposure does this correspond to?A. 4000 remB. 400 remC. 40 remD. 4 rem
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100 mrem 1 rem400 mSv 40 rem
mSv 1000 mrem
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Why is Radiation Harmful? Not heat energy
Ability of ionizing radiation to form unstable ions or neutral species with odd (unpaired) electrons
Free radicals Chemically very reactive Can set off other reactions Do great damage in cell
H O H+ +H+ O H
H O Hradiation
H O H+
+ 0e-1
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Which Types are Most Harmful? High energy gamma () radiation and X
rays Massless High velocity Penetrate everything but very dense
materials, such as lead
Which type is least harmful? Alpha () particles
Most massive Quickly slow after leaving nucleus Don’t penetrate skin
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Background Radiation Presence of natural radionuclides means we
can’t escape exposure to some background radiation Cosmic rays (from sun) hit earth
Turn 14N 13C 13C emits – particles Incorporated into food chain from CO2 via
photosynthesis Radiation from soil and building stone
From radionuclides native to Earth’s crust Top 40 cm of soil hold 1 g radium ( emmiter) /sq
kilometer 40K emit – particles
Total average exposure 360 mrem/year 82% natural radiation 18% man made
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Radiation Intensity Intensity of radiation varies with
distance from the source Farther from emitter, lower intensity of
exposure Relationship is governed by Inverse
Square Law, where: I is intensity and d is distance from source
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Ex. 5 Inverse Square LawIf the activity of a sample is 10 units at 5 meters from the source, what is it at 10 m?
What distance is needed to reduce 1 unit at 1 yd to the 0.05 units?
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!How far away from a radioactive source producing 40 rem/hr at a distance of 10 m would you need to be to reduce your exposure to 0.4 rem/hr?
A. 32 mB. 100 mC. 200 mD. 1000 m
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2
2
40 rem 10 m100 m
0.4 remd
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Radioactive Decay—Kinetics Spontaneous decay of any nuclide
follows first order kinetics May be complicated by decay of daughter
nuclide For now consider single step decay
processes Rate of reaction for first order process
A products In nuclear reaction, consider rate based on
number of nuclei N present
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Radioactive Decay—Kinetics The integrated form is:
ln N – ln No = – kt N = number of nuclei present
at time t No = number of nuclei present
at t = 0
Plot ln N (y axis) versus t (x axis)
Yields straight line—indicative of first order kinetics
Plot of N vs. time gives an exponential decay.
ln N
t
t
Nkt
oeNN
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Ex. 6 Activity Calculations131I is used as a metabolic tracer in hospitals. It has a half-life, t½ = 8.07 days. How long before the activity falls to 1% of the initial value?
ktoeNN
t = 53.6 days
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!How many hours will it take a radioisotope
with a half-life of 10.0 hours to drop to 12.5% of its original activity?A. 30.0 hrsB. 20.0 hrsC. 40.0 hrsD. 63.2 hrs
12.5% of original activity is 3 half-lives or 30.0 hrs.
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Radioisotope Dating How old is an object?
Fields — Geology, Archeology, and Anthropology
Nature provides us with natural clocks or stopwatches
A) Radiocarbon Dating (Willard Libby—Nobel Prize in 1960)
Cosmic rays (from space) enter atmosphere
Some react with N in atmosphere forming radioisotope 14C
– emitter with t½ = 5730 yr
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14C Dating 14C becomes incorporated into
atmospheric CO2 in very small quantities 14C/12C ratio in air is slightly greater than
Earth’s crust because of ongoing enrichment Living organisms breath, eat, etc…
14C/12C equilibrate with atmosphere Radioactive 14C is uniformly distributed
around globe Tested experimentally Checked vs. counting tree rings, etc. For precise work, use correction based on
alternate methods
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HOW? Freshly cut wood samples have ~15.3 cpm per gram of total carbon
cpm = counts per minute Ao = 15.3 cpm/g total C
Assumption: Ao was always 15.3 cpm, i.e. cosmic radiation is constant
When organism dies it stops eating, breathing, etc… 14C starts to decrease
14C Dating
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Wooden implement in Egyptian tomb (~3000 BC) Have about half activity of fresh sample ~5000 years have elapsed
Method is applicable for objects Few hundred to ~20,000 years
Beyond this Activity of sample is very low Experimental uncertainties too big
This method used for dating1.Charcoal in cave paintings2.Linen wraps on Dead Sea scrolls
14C Dating
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Ex. 7 C-14 DatingGeologists examine shells found in cliffs. Shells are CaCO3 and are made by living organisms. The activity of the shells is found to be 6.24 cpm/g total C. How old is the cliff formation?
21
2lnln
tt
ktAA
o
A = 6.24 cpm/g total CAo = 15.3 cpm/g total Ct½ = 5730 yr
Can use N/No and A/Ao interchangeably as A = kN
Since ratio, k cancels
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Ex. 7 C-14 Dating (cont.)
Rearranging and solving for t
t = 7414 yr
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B) Other Isotopes Provide Natural Clocks
Minerals (moon rocks) dated using isotopes with much longer half-lives
t½ = 1.27 × 109 yr
Compare ratios in rock
t½ = 4.5 × 109 yr
Rock with no other source of Pb can be dated using ratios
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Ex. 8 Dating with UA sample of lava is found to contain 0.232 g of 206Pb and 1.605 g 238U. Since lead is volatile at the temperature of molten lava, all the 206Pb now present came from the decay of 238U, calculate the time since the solidification of this rock.Step 1. Mass of 238U that decayed =
= 0.268 g 238U decayed
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Ex. 8 Dating with U (cont.) Step 2. Mass of 238U in lava initially (t =
0) No = 1.605 g + 0.268 g = 1.873 g
t = 1.0 × 109 yr
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!A wooden bowl fragment found at an old camp site thought to be approximately 11,000 years old was submitted for carbon-14 analysis. The sample was found to have 4.67 cpm/g total C. What is the
actual age of the sample?
A. 4260 yrsB. 3347 yrsC. 9810 yrsD. 2523 yrs
t = (5730 yrs × ln(4.67/15.3))/(ln 2) = 9810 yrs
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Fission Induce by
bombarding unstable nucleus with a slow neutron
Nuclear chain reaction Neutrons generated
keep going
With small mass of 235U reaction continues, but easily controlled Some neutrons are
lost to environment
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Fission “Critical mass”
Too much 235U in one place
Too many neutrons absorbed
Too few lost Uncontrollable
fission Leads to explosion
Use control rods to absorb excess neutrons and keep reaction from going critical
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Nuclear Reactor No chance of nuclear explosion
Critical mass requires pure 235U Reactor rods 2 – 4% 235U rest non-fissionable 238U
Core meltdown possible If heat of fission not carried away by cooling
water Or
Explosion possible High heat of fission splits H2O into H and O,
which recombine very exothermically and cause a chemical explosion
What happened at Chernobyl
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Nuclear Reactors Could it happen at U.S. reactors?
Extremely unlikely Chernobyl only single containment system U.S. has all double containment systems U.S. extra backup systems - both computer
and mechanical that would prevent
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Nuclear Reactor
Use heat from nuclear reaction to heat steam turbine
Use to generate electricity
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Which of the following fission reactions is Balanced?A.
B.
C.
D.
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Nuclear Fusion
Occurs when light nuclei join to form heavier nucleus
On a mass basis, fusion yields more than five times as much energy as fission
Source of the energy released in the explosion of a H-bomb The energy needed to trigger the fusion is
provided by the explosion of a fission bomb Source of energy in stars
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Thermonuclear Fusion Uses high temperatures to overcome
electrostatic repulsions between nuclei T required are >100 million °C Atoms want to fuse stripped of electrons
High initial energy cost Plasma
Electrically neutral, gaseous mixture of nuclei and electrons
Make plasma very dense (>200 g/cm–3) Brings nuclei within 2 fm = 2 × 10–15 m Pressures = several billion atm
Not there yet, major problem Containment of high temperature and pressures Magnetic field current approach
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