Chapter 24 : Nuclear Reactions and Their Applications

Chapter 24 : Nuclear Reactionsand Their Applications

24.1 Radioactive Decay and Nuclear Stability

24.2 The Kinetics of Radioactive Decay

24.3 Nuclear Transmutation: Induced Changes in Nuclei

24.4 The Effects of Nuclear Radiation on Matter

24.5 Applications of Radioisotopes

24.6 The Interconversion of Mass and Energy

24.7 Applications of Fission and Fusion

Comparison of Chemical andNuclear Reactions

Chemical Reactions Nuclear Reactions1. One substance is converted to 1. Atoms of one element typically another, but atoms never change change into atoms of another. identity.2. Orbital electrons are involved as 2. Protons, neutrons, and other bonds break and form; nuclear particles are involved; orbital particles do not take part. electrons rarely take part.3. Reactions are accompanied by 3.Reactions are accompanied by relatively small changes in energy relatively large changes in energy and no measurable changes in mass. and measurable changes in mass.4. Reaction rates are influenced by 4. Reaction rates are affected by temperature, concentration, number of nuclei, but not by catalysts, and the compound in temperature, catalysts, or the which an element occurs. compound in which an element

occurs.Table 24.1

AZX A - mass number = p+ + n0

Z - number of protons(atomic no.) = p+

Number of neutrons(N) in a nucleus

N = A - Z238

92U N = 238 - 92 = 146 of n0

23892U has 92e- 92 p+

235U - atomic bomb

Subatomic elementary particles

Electron 0-1 e

Proton 11 p

Neutron 10n

Nuclide - nuclear species with specificnumbers of two types of nucleonsNucleons - made up ofprotons and neutrons

Properties of Fundamental Particles

• Particle Symbol Charge Mass

• (x10-19 Coulombs) (x 10-27 kg)

• Proton p +1.60218 1.672623

• Neutron n 0 1.674929

• Electron e -1.60218 0.0005486

NUCLEAR STABILITYModes of Radioactive Decay

• Alpha decay–heavy isotopes: 42He2+ or

• Beta decay–neutron rich isotopes: e- or

• Positron emission–proton rich isotopes:

• Electron capture–proton rich isotopes: x-rays• Gamma-ray emission( )–Decay of nuclear

excited states

• Spontaneous fission–very heavy isotopes

Fig. 24.1

(p. 1046)

Alpha Decay–Heavy Elements• 238U 234Th + α + e

t1/2 = 4.48 x 10 9 years

• 210Po 206Pb + α + e

t1/2 = 138 days• 256Rf 252No + α + e

t1/2 = 7 ms

• 241Am 237Np + α + e

t1/2 = 433 days

Beta Decay–Electron Emission

• n p+ + β− + Energy

• 90Sr 90Y + β− + Energy

t1/2= 30 years• 14C 14N + β− + Energy

t1/2= 5730 years• 247Am 247Cm + β− + Energy

t1/2= 22 min• 131I 131Xe + β− + Energy

t1/2 = 8 days

Electron Capture–Positron Emission

p+ + e- n + Energy = Electron capture

p+ n + e+ + Energy = Positron emission

51Cr + e- 51V + Energy t1/2 = 28 days

7Be 7Li + β+ + Energy t1/2 = 53 days

177Pt + e- 177Ir + Energy t1/2 = 11 s

144Gd 144Eu + β+ + Energy t1/2 = 4.5 min

Fig. 24.2

Number of Stable Nuclides for Elements 48 through 54

Atomic Number ofElement Number (Z) Nuclides

Cd 48 8

In 49 2

Sn 50 10

Sb 51 2

Te 52 8

I 53 1

Xe 54 9

Table 24.3 (p. 1049)

Distribution of StableNuclides

• Protons Neutrons Stable Nuclides %

• Even Even 157 58.8

• Even Odd 53 19.9

• Odd Even 50 18.7

• Odd Odd 7 2.6

267 100.0%Total =

(c.f. Table 24.4, p. 1050)

The 238U DecaySeries

Fig. 24.3

CHEM 1332

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Room 117 SR1

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Predicting the Mode of DecayUnstable nuclide generally decays in a modethat shifts its N/Z ratio toward the band ofstability.1. Neutron-rich nuclides undergo ß decay, whichconverts a neutron into a proton, thus reducingthe value of N/Z.

2. Proton-rich nuclides undergo positron decayor electron capture, thus increasing the value ofN/Z.3. Heavy Nuclides Z > 83 undergo alpha decay,thus reducing the Z and N values by two unitsper emission.

24.26 24395Am 42He + 239

93Np

23993Np 0-1ß + 239

94Pu

23994Pu 42He + 235

92U

23592U 42He + 231

90Th

Kinetics of Radioactive DecayThe decay rate or activity(A) = - N/ t

N = Change in number of nuclei

t = change in timeSI unit of radioactivity is the becquerel(Bq), definedas one disintegration per second(d/s): 1Bq = 1 d/s

1 curie(Ci) equals the number of nucleidisintegrating each second in a 1g ofradium- 226. 1 Ci = 3.70 X 1010 d/s

For a large collection of radioactive nuclei,thenumber decaying per unit time is proportionalto the number present:

Decay rate (A) α N or A = kN

k = decay constant

The larger the value of k, the higher is thedecay rate. A = -∆N/∆t = kN

The activity depends only on N raised to thefirst order so radioactivity decay is a first-order process. We consider the number ofnuclei than their concentration.

Half-life of Radioactive Decay t1/2

The half-life (t1/2) of a nuclide is the time ittakes for half the nuclei present to decay.

146C 14

7N + 0-1ß

The number of nuclei remaining is halved aftereach half-life.

Natural Decay Series of Existing Isotopes

40K 40Ar t1/2 = 1.29 x 109 years

232 Th 208 Pb t1/2 = 1.4 x 1010 years

235U 207 Pb t1/2 = 7 x 108 years

238U 206 Pb t1/2 = 4.5 x 109 years

Natural Decay Series for Uranium-238

238U 234 Th 234Pa

234U 230 Th 226Ra 222Rn 218Po 214Pb 218At 214Bi 210 Tl

214Po 210Pb 206Hg

= α decay 210Bi 206Tl

= β− decay 210 Po 206Pb

238U: 8 decays and 6 decays leaves you with 206Pb

Natural Decay Series for Uranium-235235U 231 Th

231Pa 227Ac 223Fr 219At 215Bi

227 Th 223Ra 219Ra 215Po 211Pb

215At 211Bi 207 Tl

211Po 207Pb

= decay

= decay

235U: 8 decays and 4 decays leaves you with 207Pb

Natural Decay Series for Thorium-232232 Th 228Ra

228Ac

228 Th 224Ra 220Rn 216Po 212Pb

212Bi 208Tl

212Po 208Pb

= decay

= decay

232 Th: 7 decays and 4 decays leaves you with 208Pb

Fig. 24.A

Fig. 24.4

Decrease in Number of 14CNuclei Over Time

ln Nt / No = - kt

ln No / Nt = kt No is the number of nucleiat t = 0 and Nt is the number of nuclei remainingat any time t.

To calculate the half-time ; set Nt = 1/2 N0

ln No / (1/2No) = k t 1/2 ; t 1/2 = ln2 / k

The half-life is not dependent on the number ofnuclei and is inversely related to the decayconstant. Large k short t 1/2.

Decay Constants (k) and Half-lives (t1/2)of Beryllium Isotopes

Nuclide k t1/2

74Be 1.30 x 10-2/day 53.3 day

84Be 1.0 x 1016/s 6.7 x 10-17s

94Be Stable

104Be 4.3 x 10-7/yr 1.6 x 106 yr

114 Be 5.02 x 10-2/s 13.8 s

Table 24.5 (p. 1054)

Radioisotopic Dating

147N + 10n 14

6C + 11p

12C / 14C ratio is constant in living organisms.

Dead organism 12C / 14C ratio increases.

14C decays 146C 14

7N + 0-1ß

Radiocarbon Dating for Determiningthe Age of Artifacts

Fig. 24.5

Nuclear Transmutation

147N + 42He 11H + 17

8O

Notation 14N (α , p) 17O

2713Al + 42He 10n + 30

15P

Notation 27Al(α , n) 30P

Fig. 24.6A

The Cyclotron Accelerator

Fig. 24.7

Formation of Some Transuranium Nuclides

Reaction Half-life of Product

23994Pu + 42He 240

95Am + 11H + 2 10n 50.9 h

23994Pu + 42He 242

96Cm + 10n 163 days

24496Cm + 42He 245

98Bk + 11H + 2 10n 4.94 days

23892U + 12

6C 24698Cf + 4 10n 36 h

25399Es + 42He 256

101Md + 10n 76 min

25298Cf + 10

5B 256103Lr + 6 10n 28 s

Table 24.6 (p. 1059)

Fig. 24.8

Units of Radiation Dose

rad = Radiation-absorbed dose The quantity of energy absorbed per kilogram of tissue: 1 rad = 1 x 10-2 J/kg

rem = Roentgen equivalent for man The unit of radiation dose for a human:

1 rem = 1 rad x RBE

RBE = 10 for α RBE = 1 for x-rays, γ-rays, and β’s

Examples of Typical Radiation Doses from Natural and Artificial Sources–I

Source of Radiation Average Adult Exposure

Cosmic radiation 30-50 mrem/yrRadiation from the ground From clay soil and rocks ~25-170 mrem/yr In wooden houses 10-20 mrem/yr In brick houses 60-70 mrem/yr In light concrete houses 60-160 mrem/yrRadiation from the air (mainly radon) Outdoors, average value 20 mrem/yr In wooden houses 70 mrem/yr In brick houses 130 mrem/yr In light concrete houses 260 mrem/yrInternal radiation from minerals in tap water and daily intake of food ~40 mrem/yr ( 40K, 14C, Ra)

Natural

Table 24.7 (p. 1062)

Examples of Typical Radiation Doses fromNatural and Artificial Sources–II

Source of Radiation Average Adult Exposure

Diagnostic x-ray methods Lung (local) 0.04-0.2 rad/film Kidney (local) 1.5-3.0 rad/film Dental (dose to the skin) ≤ 1 rad/filmTherapeutic radiation treatment locally ≤ 10,000 radOther sources Jet flight (4 hr) ~1 mrem Nuclear tests < 4 mrem/yr Nuclear power industry < 1 mrem/yr

Total Average Value 100-200 mrem/yr

Artificial

Table 24.7 (p. 1062)

Acute Effects of a Single Dose ofWhole-Body Irradiation–I

Dose Lethal Dose(rem) Effect Population (%) No. of Days

5-20 Possible late effect; possible — — chromosomal aberrations20-100 Temporary reduction in — — white blood cells50+ Temporary sterility in men — — (100+ rem = 1 yr duration)100-200 “Mild radiation sickness”: vomiting, diarrhea, tiredness in a few hours Reduction in infection resistance Possible bone growth retardation in children

Table 24.8 (p. 1063)

Acute Effects of a Single Dose of Whole-Body Irradiation–II

Dose Lethal Dose(rem) Effect Population (%) No. of Days

300+ Permanent sterility in women — —

500 “Serious radiation sickness”: 50-70 30 marrow/intestine destruction

400-1000 Acute illness, early deaths 60-95 30

3000+ Acute illness, death in hours 100 2 to days

Table 24.8 (p. 1063)

(p. 1064)

24.6 Inter conversion of Mass and Energy

Nuclear Fission: A heavy nucleus splits into two

lighter nuclei.

Nuclear Fusion: Two lighter nuclei combine toform a heavier one.

The Mass Defect (∆m) - The mass decrease thatoccurs when nucleons combine to form a nucleus

The size of the mass change is nearly 10 million timesthat of bond breakage.

NUCLEAR ENERGY• EINSTEIN’S

EQUATION FOR THECONVERSION OFMASS INTO ENERGY

• E = mc2

• m = mass (kg)

• c = Speed of light

• c = 2.998 x 108 m/s

Nuclear Binding Energy

Energy required to break up 1 mol ofnuclei into their individual nucleons.

Nucleus + nuclear binding energy nucleons

For 12C ∆E = ∆mc2

= (9.894 x 10-5kg/mol)(2.9979 x108m/s)2

= 8.8921 x 1012 J/mol

Fig. 24.12

Units Used for Nuclear Energy Calculations

Electron volt (ev) The energy an electron acquires when it moves through a potential difference of one volt:

1 ev = 1.602 x 10-19J

Binding energies are commonly expressed in units of megaelectron volts (Mev)

1 Mev = 106 ev = 1.602 x 10 -13J

A particularly useful factor converts a given mass defect in atomic mass units to its energy equivalent in electron volts: 1 amu = 931.5 x 106 ev = 931.5 Mev

Binding Energy per Nucleon of Deuterium

Deuterium has a mass of 2.01410178 amu.

Hydrogen atom = 1 x 1.007825 amu = 1.007825 amu Neutrons = 1 x 1.008665 amu = 1.008665 amu 2.016490 amu

Mass difference = theoretical mass - actual mass = 2.016490 amu - 2.01410178 amu = 0.002388 amu

Calculating the binding energy per nucleon:

Binding energy 0.002388 amu x 931.5 Mev/amu Nucleon 2 nucleons

= 1.1123 Mev/nucleon

=

Calculation of the Binding Energy per Nucleon for Iron-56

The mass of iron-56 is 55.934939 amu; it contains 26 protons and 30 neutronsTheoretical mass of Fe-56 : Hydrogen atom mass = 26 x 1.007825 amu = 26.203450 amu Neutron mass = 30 x 1.008665 amu = 30.259950 amu 56.463400 amu

Mass defect = theoretical mass - actual mass: 56.463400 amu - 55.934939 amu = 0.528461 amu



=


N = A - Z

Calculation of the Binding Energy per Nucleon for Uranium-238

The actual mass of uranium-238 = 238.050785 amu; it has 92 protons and 146 neutrons

Theoretical mass of uranium-238: Hydrogen atom mass = 92 x 1.007825 amu = 92.719900 amu Neutron mass = 146 x 1.008665 amu = 147.265090 amu 239.98499 amu

Mass defect = theoretical mass - actual mass: 239.98499 amu - 238.050785 amu = 1.934205 amu



=


N = A - Z

Mass and Energy in Nuclear Decay–I

Consider the alpha decay of 212Po t1/2 = 0.3 µs

212Po 208Pb + α + Energy

211.988842 g/mol 207.976627 g/mol + 4.00151 g/mol

Products = 207.976627 + 4.00151 = 211.97814 g/mol

Mass = Po - Pb + α = 211.988842 -211.97814 0.01070 g/mol

E = mc2 = (1.070 x 10-5 kg/mol)(3.00 x 108m/s)2

= 9.63 x 1011 J/mol

9.63 x 1011 J/mol6.022 x 1023 atoms/mol

= 1.60 x 10-12J/atom

Mass and Energy in Nuclear Decay–II

The energy for the decay of 212Po is 1.60 x 10-12J/atom

1.60 x 10-12J/atom1.602 x 10-19 J/ev

= 1.00 x 107 ev/atom

10.0 x 106 ev 1.0 x 10-6 Mev atom ev

x = 10.0 Mev/atom

The decay energy of the alpha particle from 212Po is 8.8 Mev.

Fig. 24.13

Fig. 24.14 Critical Mass: Mass needed to achieve achain reaction.

Fig. 24.15B

Neutron Induced Fission– Bombs and Reactors

There are three isotopes with sufficiently long half-lives andsignificant fission cross-sections that are known to undergoneutron induced fission, and are useful in fission reactors, andnuclear weapons. Of these, only one exists on earth (235U whichexists at an abundance of 0.72% of natural uranium) and that is theisotope that we use in nuclear reactors for fuel and some weapons. The three isotopes are:

233U t1/2 = 1.59 x 105 years sigma fission = 531 barns

235U t1/2 = 7.04 x 108 years sigma fission = 585 barns

239Pu t1/2 = 2.44 x 105 years sigma fission = 750 barns

Breeding Nuclear FuelThere are two relatively common heavy isotopes that will not undergo neutron induced fission, that can be used to make other isotopes that do undergo neutron induced fission, and can be used as nuclear fuel in a nuclear reactor.

Natural thorium is 232 T which is common in rocks.

232 Th + 10n 233 Th + Energy t1/2 = 22.3 min 233 Th 233Pa + β− + Energy t1/2 = 27.0 days 233Pa 233U + β− + Energy t1/2 = 1.59 x 105 years

Natural uranium is 238U which is common in rocks as well.

238U + 10n 239U + Energy t1/2 = 23.5 min 239U 239Np + β− + Energy t1/2 = 2.36 days 239Np 239Pu + β− + Energy t1/2 = 24400 years

Hydrogen Burning in Stars and Nuclear Weapons

1H + 1H 2H + β+ + 1.4 Mev

1H + 2H 3He + 5.5 Mev

2H + 2H 3He + 1n + 3.3 Mev

2H + 2H 3H + 1H + 4.0 mev

2H + 3H 4He + 1n + 17.6 Mev Easiest! Highest2H + 3He 4He + 1H + 18.3 Mev cross- section!1H + 7Li 4He + 4He + 17.3 Mev

Helium Burning Reactions in Stars

12C + 4He 16O

16O + 4He 20Ne

20Ne + 4He 24Mg

24Mg + 4He 28Si

28Si + 4He 32S

32S + 4He 36Ar

36Ar + 4He 40Ca

ElementSynthesisin theLife Cycleof a Star

Fig. 24.C (p. 1077)

The Tokamak Design for Magnetic Containment of a Fusion Plasma

Fig. 24.16

24.42

k = 0.693/t 1/2 = 0.693/1.6 x 103 yr = - 4.331x10 -4 yr-1

ln Nt / No = -kt

t = ln (0.185g/2.50g) / -4.33 x 10-4 yr-1 = 6.01 x105 yr

24.44

k = 0.693/5730yr = 1.21 x 10-4 yr-1

t = ln (Nt/No)/ -k = ln(0.735/1.000)/ -1.21x10-4 yr-1

t = 2.54 x 10 3 yr

Chapter 24 : Nuclear Reactions and Their Applications

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