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Nuclear ChemistryNuclear ChemistryChapter 20Chapter 20
Glenn T. Seaborg1912-1999.*Transuraniumelements.
Pierre and Marie Curie.1859-1906,* 1867-1934.**Discovered radium; defined “radioactivity.”
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Basics: RadioactivityBasics: RadioactivityNuclear EquationsNuclear Equations• Nucleons: particles in the nucleus:
p+: proton n: neutron.
• Mass number A: the number of p+ + n.• Atomic number Z: the number of p+.
• Symbol: AzX; e.g. 14
6C is “carbon-14”
• Isotopes: have the same number of p+ and different numbers of n (and therefore, different mass)
• In nuclear equations, the total number of nucleons is conserved:
23892U 234
90Th + 42He
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RadioactivityRadioactivity There are three types of radiation which we consider:
-Radiation is the loss of 42He from the nucleus,
-Radiation is the loss of an electron from the nucleus
(electrons represented as either 0-1e or 0
-1β)
-Radiation is the loss of high-energy photon from the nucleus.
Example of α emission: HeThU 42
23490
23892
eXeI 01
13154
13153 Example of β emission:
Example of γ emission: Tc99m43
0hνTc 9943
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Radioactivity - Radioactivity - Separating the types of radiationSeparating the types of radiation
(-)
(++)
Charge 2+ 1- 0Mass(g) 6.64x10-24 9.11x10-28 0Rel. mass 7,300 1 0Rel. penetration 1 100 10,000
42He nucleus electron high energy photons
α β γ _
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RadioactivityRadioactivityComplete the following nuclear reactions:
____pnS 11
10
3216
_____eBe 01
74
____n2XenU 10
13554
10
23592
____HeHH 32
21
21
____nHMo 10
21
9842
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Patterns of Nuclear StabilityPatterns of Nuclear StabilityNeutron-to-Proton RatioNeutron-to-Proton Ratio
Neutron/proton ratio increases as atoms become larger
Above 83Bi, all nuclei are unstable and belt of stability ends.
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PbPo
BiPbPoBi
PbPoRnRa
ThUPaThU
20682
α21084
β
21083
β21082
α21484
β21483
β
21482
α21884
α22286
α22688
α
23090
α23492
β23491
β23490
α23892
238U Series
Stable
Radioactive Series
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Nuclear TransmutationsNuclear TransmutationsUsing Charged Particles - cyclotronUsing Charged Particles - cyclotron
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Rates of Radioactive DecayRates of Radioactive DecayCalculations Based on Half-LifeCalculations Based on Half-Life• Radioactive decay is a first order process:
Rate = kN• In radioactive decay the constant, k, is called the
decay constant, and N is number of nuclei.• The rate of decay is called activity (disintegrations per
unit time).
• If N0 is the initial number of nuclei and Nt is the number of nuclei at time t, then
ktN
N
t
o ln with half-life t1/2=0.693/k
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Rates of Radioactive DecayRates of Radioactive Decay
5.0
2.5
1.25
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Rates of Radioactive DecayRates of Radioactive DecayDatingDating• Carbon-14 is a radioactive isotope of carbon and is
used to determine the ages of organic compounds.• We assume the ratio of 12C to 14C has been constant
over time.• For us to detect 14C, the object must be less than
50,000 years old.• The half-life of 14C is 5,730 years.• Its abundance is <1% (the most common isotope of
carbon is C-12)
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Rates of Radioactive DecayRates of Radioactive Decay
14C is created in upper atmosphere by bombardment of nitrogen with cosmic neutrons:
pCnN 11
146
10
147
14C itself decays to stable 14N by β emission:
eNC 01
147
146
with a half-life of t1/2= 5715 yr
The amount of 14C in the environment is constant.
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How does 14C dating work?•When an organism dies, it no longer takes in carbon compounds but its 14C continues to decay.
•5715 years (or one half-life) after the death of the organism (or 1 half-life), the relative amount of 14C is half that found in living matter.•11,430 years after the death of the organism (two half-lives), the relative amount of 14C is ¼ that found in living matter.
First order rate law: ktC][
C][ln
t14
014
and t1/2 =.693/k
Since [14C] is proportional to radiation emitted (in counts/min or cpm), the law become:
ktt)cpm(curren
l)cpm(initialn
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Rates of Radioactive Decay-Rates of Radioactive Decay-1414C DatingC Dating
21.37 Artifact has 14C activity of 24.9 counts/m, compared to current count of 32.5 counts/m for a standard. What is the age of the artifact? Given: t1/2 = 5715 year.
k=.693/t1/2 = .693/5715 yr = 1.21x10-4 yr-1
Use first order expression ktN
Nln
t
o
)tyr(1.21x109.24
32.5ln 1-4
2200 yr = t
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4040K – K – 4040Ar DatingAr Dating
40K is a strange beast – with a half life of 1.3 x 109 yr, it has two simultaneous modes of decay.88.8% decays by electron-emission to give Ca-40:
Ca4020K40
19 e0
1
11.2% decays by electron-capture (of one of theorbital electrons) to give Ar-40:
AreK 4018
01
4019
40K constitutes 0.01% of the natural abundance ofpotassium in the earth’s crust.
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Problem: a mineral is found with a 40K/40Ar mass ratio of 3/1.How old is the mineral?
Answer: For the purposes of calculation, let’s assume a femtogramtotal mass of 40K and 40Ar. The respective masses of K-40 andAr-40 will be 0.75 fg K-40 and 0.25 fg Ar-40. Since the mode of decay giving Ar-40 is 11.2%, the mass of K-40 which decayed is0.25/0.112 = 2.23 fg. Hence, the original mass of K-40 was 0.75 + 2.23 = 2.98 fg.
ktN
Nln
t
o
(5.33 x 10-10 yr-1) t0.75
2.98ln
t = 2.58 x 109 yr
k = 0.693/t1/2
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Half-livesHalf-lives
Isotope Half-life Type of decayU-238 4.5 x 109 yr AlphaU-235 7.0 x 108 yr AlphaTh-232 1.4 x 1010 yr AlphaK-40 1.3 x 109 yr Beta-capture or -emissionC-14 5715 yr BetaRn-222 3.825 days AlphaTc-99 210,000 yr Beta
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Detection of RadioactivityDetection of Radioactivity
• Matter is ionized by radiation.• Geiger counter determines the amount of ionization by
detecting an electric current.• A thin window is penetrated by the radiation and causes the
ionization of Ar gas.• The ionized gas carried a charge and so current is produced.• The current pulse generated when the radiation enters is
amplified and counted.
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Energy Changes in Nuclear ReactionsEnergy Changes in Nuclear Reactions
• Einstein showed that mass and energy are proportional:
E = mc2
• If a system loses mass it loses energy (exothermic).• If a system gains mass it gains energy (endothermic).• Since c2 is a large number (8.99 1016 m2/s2) small
changes in mass cause large changes in energy.• Mass and energy changed in nuclear reactions are
much greater than in chemical reactions.
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Energy Changes in Nuclear ReactionsEnergy Changes in Nuclear Reactions Consider reaction: 238
92U 23490Th + 4
2He – Molar masses: 238.0003 g 233.9942 g + 4.0015 g.
– The change in mass during reaction is
233.9942 g + 4.0015 g - 238.0003 g = -0.0046 g. =m
– The process is exothermic because the system has lost mass.
– To calculate the energy change per mole of 23892U:
J 101.4s
m-kg101.4
g 1000
kg 1g 0046.0m/s 1000.3
)(
112
211
28
22
cmmcE
This is a very large number!!
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Nuclear FissionNuclear Fission• Splitting of heavy nuclei is exothermic for large mass
numbers.• Consider a neutron bombarding a 235U nucleus:
Un 23592
10
n2ZrTe 10
9740
13752
n3KrBa 10
9136
14256
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Nuclear FusionNuclear Fusion• Light nuclei can fuse to form heavier nuclei.• Most reactions in the Sun are fusion.
Fusion processes occurring in sun:
eHeHHe
HeHH
eHHH
01
42
11
32
32
11
21
01
21
11
11
Sun currently is about 75% H and 25% He.
Another 15b yrs or so to go before sun “burns” up all its H.
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Nuclear FusionNuclear Fusion• Fusion of tritium and deuterium requires about
40,000,000K:2
1H + 31H 4
2He + 10n
• These temperatures can be achieved in a nuclear bomb or a tokamak.
• A tokamak is a magnetic bottle: strong magnetic fields contained a high temperature plasma so the plasma does not come into contact with the walls. (No known material can survive the temperatures for fusion.)
• To date, about 3,000,000 K has been achieved in a tokamak.
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