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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.