Brdy 6Ed Ch21 NuclearReactions

8/9/2019 Brdy 6Ed Ch21 NuclearReactions

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Chapter 21:Nuclear Chemistry

Chemistry: The Molecular Natureof Matter, 6E

Jespersen/Brady/Hyslop


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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 2

How are atoms formed?

Big BangIntense heat ~109 K Cooled quickly to 106 KT of stars

e, p+, no formed and joined into nucleiatoms

Mostly H and He (as in our sun)

Rest of elements formed by nuclear reactions

Fusiontwo nuclei come together to formanother heavier nucleus

Fissionone heavier nucleus splits intolighter nuclei

Various other types of reactions

+

+


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Nuclear Shorthand

Nucleons Subatomic particles found in the nucleus

Protons (p+)

Neutrons (no)

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 = no + p+


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Shorthand for Writing Nuclides

WhereX= atomic symbol

Ex.

In the neutral atom: e = p+ = Z Isotopes

Nuclides with same Z (same number of p+), but

differentA (different no)

Th23090

Cd11348

Hydrogen Deuterium Tritium

1 p+ 1 p+ + 1 no 1 p+ + 2 no

H11

H21

H31

XAZ


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Radioactivity

Radioactive isotopes

Isotopes with unstable atomic nuclei

Emit high energy streams of particles orelectromagnetic 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


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Mass Not Always Constant

Mass of particle not constant under allcircumstances

It depends on velocity of particle relative toobserver

As approaches speed of light, mass decreases

When vgoes to zero

Particle has no velocity relative to observer

v/c 0

Denominator 1

and m = mo

2)/(1 cvmm

= o m = mass of particlev= velocity of particle

m= rest mass

c= speed of light


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Why dont we observe mass change?

In lab and ordinary life, velocity of particle issmall

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

2)/(1 cvmm

= o


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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 tothe other.

Sum of all energy in universe and all mass

(expressed in energy equivalents) in universeis constant

Einstein Equation E = (mo)c

2

Where c= 2.9979 x 108 m/s


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Mass Defect

Rest mass of nuclide is always less than sum ofmasses of all individual nucleons (neutrons andprotons) in that same nuclide

Mass is lost upon binding of neutrons and protons intonucleus

When nucleons come together, loss of mass translates

into release of enormous amount of energy byEinstein's relation

Energy released = Nuclear Binding Energy

Nuclear Binding Energy Amount of energy must put in to break apart

nucleus


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What is Mass Loss?

nucleonsnucleus mmm =

pneisotope

mZmNmZmm +=

For given isotope of given Z and A

or


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Ex.1. Binding Energy Calculation

What 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) 3 me

= 7.016003 u 3(0.0005485 u)

= 7.0143573 u

B. Determine mass of nucleons

mass of nucleons = 3 mp + 4 mn= 3(1.007276470 u) + 4(1.008664904 u)

= 7.056489026 u

E43 373 ++ ++ Linp o


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Ex.1. (cont.) Binding Energy

CalculationC.m = mnucleus mnucleons= 7.0143573 u 7.056489026 u

= 0.0421317 u

= mass lost by nucleons when they form nucleus

Step 2. Determine energy liberated by this

change in massE = (mo)c

2

E= 6.287817 x 1012 J/atom

( )

( )22

28

27

/

1/10997925.2*

/106605402.10421217.0

smkg

Jsm

ukguE

=


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Ex. 1. (continued)

E= 6.287817 x 1012 J/atom *6.0221367 x 1023 atoms/mole

E= 3.78655 x 1012 J/mole= 3.78655 x 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!!


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Ex. 1. but in MeV

For Ex. 1. Converting E to MeV gives

Often wish to express binding energy pernucleon so we can compare to other nuclei

For Li3+ with 3 1p and 4 0n this would be

MeV/atom24539

J/MeV106021771

J/atom102878176

13

12

.

.

.E

B

=

=

MeVMeVEB 61.5

7

245.39

nucleon

)Li( 3=


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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E

Your Turn!

Determine the binding energy, in kJ/mol andMeV/atom, for an isotope that has a mass defect of

0.025861 u.

A. -2.3243 x 109 kJ/mol; 24.092 MeV/atom

B. -3.8595 x 10-12

kJ/mol; 24.092 MeV/atomC. -7.7529 kJ/mol; 8.03620x 10-8 MeV/atom

D. -2.3243 x 109 kJ/mol; 4.1508 x 10-2 MeV/atom

18


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Your Turn! - Solution

19

( ) ( )

( )

27

28

2 2

12

12 23

12 9

1

0.025861 1.6605402 10 kg /

1 J

* 2.997925 10 m/s kgm /s

3.8595 x 10 J/atom

J atoms3.8595 x 10 6.0221367 x 10atom mol

J kJ2.3243 x 10 or -2.3243 x 10

mol mol

3.8595 x 10

E u u

=

=

=

2-13

J 1 MeV MeV 24.092atom atom1.602 x 10 J

=


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Implications of Curve

Most EB/Ain range of 6 9 MeV (per nucleon) Large binding energy EB/Ameans stable

nucleus

Maximum at A = 56 56Fe largest known EB/A

Most Thermodynamically stableTD sink Nuclear mass number (A) and overall chargeare conservedin nuclear reactions

Lighter elements undergo fusionto form morestable nuclei


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Implications of Curve

Ex. Fusion

Researchers are currently working to get fusion tooccur in lab

Heavier elements undergo fissionto form morestable elements

Ex. Fission

Reactions currently used in bombs and powerplants (238U and 239Pu) As stars burn out, they form elements in center of

Periodic Table around 56Fe

nKrBanU 10

91

36

142

56

1

0

235

92 3

MeVnHeHH . 61710

4

2

3

1

2

1


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Radioactivity

Spontaneous emission of high energyparticles from unstable nuclei

Spontaneous emission of fundamental particle orlight

Nuclei falls apart withoutany external stimuli

Discovered by Becquerel (1896) Extensively studied by Marie Curie and her

husband Pierre (1898 early 1920's) Initially worked with Becquerel


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Fun Facts

Marie and Pierre Curie discovered Poloniumand Radium

Nobel Prize in Physics 1903 For discovery of Radioactivity

Becquerel, Marie and Pierre Curieall three shared

Nobel Prize in Chemistry 1911 For discovery of Radium and its properties

Marie Curie only

Marie Curie - first person to receive two NobelPrizes and in different fields


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Discovery of Radioactivity

Initially able toobserve 3 types of

decay Labeled them , ,

rays (after 1st threeletters of Greekalphabet)

If they pass throughan electric field,very different

behavior


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Discovery of Radioactivity

rays attracted to pole positively charged

rays attracted to +pole

negatively charged rays not attracted to

either

not charged


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Nuclear Equations

Used to symbolize decay of nucleusEx. 238U 234Th +

parent daughter Produce new nuclei so need separate rules

to balance

Balancing Nuclear Equationsa. Sum of mass numbers (A, top) must be same

on each side of arrow

b. Sum of atomic numbers (Z, bottom) must besame on each side of arrow

He4292 90


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Types of Spontaneous Emission

1. Alpha () Emission = He nucleus

= 2 n+ 2p

A = 4 and Z = 2

Daughter nuclei has:

A by 4 A= 4Z by 2 Z= 2 Very common mode of decay

if Z > 83 (large radioactive nuclides) Most massive particle

Ex.ThU

4

2

230

90

234

92


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Balancing Nuclear Equations

29

1. The sum of the mass numbers (A;superscripts) on each side of the arrow must

be the same2. The sum of the atomic numbers (Z;subscripts; nuclear charge) on each side of

the arrow must be the same Ex.

A: 234 = 230 + 4

Z: 92 = 90 + 2

ThU 42230

90234

92

HeRnRa 42222

86226

88


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2. Beta (

or e) Emission

~epn 01

1

1

1

0

30

Emission of e

Mass number A = 0 and charge Z = 1

But How? NO e's in nucleus!

If nucleus (nrich)nuclide too heavy


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Beta ( or e) Emission

~epn 011110

31

Charge conserved, but not mass mE Ejectede has very high KE + emits

Antineutrino variable energy particle

Accounts for extra EgeneratedEx.

~

~2359301

23592 NpU

~BiPb 01214

83214

82 e


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3. Gamma () Emission

Emission of high energy photons Often accompanies or emission

Occurs when daughter nucleus of someprocess is left in excited state

Use * to denote excited state

Nuclei have energy levels analogous to thoseof e in atoms

Spacing of nuclear E levels much larger

light emitted as -raysEx.

32

137

56

137

56 BaBa

i ( ) i i


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4. Positron (

+ or e+) Emission

Emission of e+

Positive electron

Where does + come from? If nucleus (npoor)

Nuclide too light

Balanced for charge,but NOT for mass

33

enp ++ +0

1

1

0

1

1

4 P i ( + +) E i i


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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

antimattere+emission only occurs if daughternucleus

is MUCH more stable thanparent

34

381801

3819 ++ + AreK

4 P it ( + +) E i i


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4. Positron (

+or e+) Emission

What happens to e+? Collides with electron to give matter

anti-matter annihilation and two highenergy -ray photons mE

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 spectrum35

20101 + + ee

5 El t C t (EC)


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5. Electron Capture (EC)

e

in 1sorbital Lowest Energy e

Small probability that e is

near nucleus

e actually passes throughnucleus occasionally

If it does:

Net effect same as e+emission

36

nep 1001

11 +

++ + raysXPtAu 19578captureelectron0119579 e

T pes of Spontaneo s Emission


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6. Neutron Emission = ( )

Fairly rare

Occurs in nrich nuclides Does not lead to isotope of different element

7. Proton Emission = ( )

Very rare

nII10

13653

13753 +

n10

p11

pCuZn 115629

5730 +



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8. Spontaneous Fission NO stable nuclei with Z > 83

Several of largest nuclei simply fall apartinto smaller fragments

Not just one outcome, usually several

differentsee distribution

Fm256

100 SbIn 131

51

125

49 +

IAg 136

53

120

47 +

Summary Common Processes


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SummaryCommon Processes

1. Alpha () Emission Very common if Z > 83

2. Beta () Emission e

Common for nrich nuclidesbelow belt of stability

3. Positron (+) Emission e+

Common for npoor nuclidesabove belt of stability4. Electron Capture (EC)

Occurs in npoor nuclides, especially if Z > 40

5. Gamma () Emission Occurs in metastable nuclei (in nuclear excited state)

He

4

2

Learning Check


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Learning Check

Complete the following table which refers topossible nuclear reactions of a nuclide:

EmissionZ =p n e A NewElement?

+

EC

2 2 0 4 yes

+1+11 0 yes

+1 1 0 yes1

0 0 0 0 no

+1+11 0 yes

Learning Check


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Learning Check

Balance each of the following equationsa.

b.c.

d.e.

f.

____UPu 23592239

94 He42

____XeI 1315413153

01e

____SiP 27142715

01e

____n4ZnUCa10

7030

23892

4020

EPb

204

82

____n2ZnnU 107230

10

23392

ESm16062

____PdKrSr 1164684368838 Ni5628

Your Turn!


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Your Turn!

What is the missing species, ,in the followingnuclear reaction?

A.

B.

C.

D.

42

4 242 1

2 96 0

X + He Cm + nnm

Xn

m

239

94Pu

247

98Cf

238

95Am

238

94Pu

What Holds Nucleus Together?


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What Holds Nucleus Together?

Consider nucleus no + p+ in close proximity

Strong p+ + p+ repulsions

no spread p+'s apart

n/p ratio as Z

Strong Forces Force of attraction between nucleons

Holds nuclei together

Overcomes electrostatic repulsions betweenprotons

Binds protons and neutrons into nucleus


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Table of Nuclides


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Table of Nuclides

Atomic number (Z = # of protons)

#o

fneutron

s(N=#

on)

Table of Nuclides


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Table of Nuclides

Note:only a small corner of table isshown. (complete is in CRC)

Shaded area = Stable Nuclei Trend of stable nuclei = diagonalline =

Belt of Stability

~ Z = N (for 1 to 20)

Belt of Stability


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Belt of Stability

Each isotope is a dot Up to Z = 20

Ratio N/Z = 1

As Z , #N > Z and N/Z as Z

By Z = 82, N/Z ~1.5 N = # neutrons

Z = # protons

N

Z

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

How To Predict if Nuclei is Stable


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How To Predict if Nuclei is Stable

1)Atomic Mass = weighted average of masses ofnaturally occurring isotopes, i.e. most stable ones

2) CompareAtomic mass of element toA(atomic

mass number) of given isotope and see if it ismoreor less

At. Mass >A too light to be stable

At. Mass 83 are radioactive

190.2

126.9

Too light, n poor

Too heavy, n Rich

More Patterns of Stability


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If we look at stable and unstable nuclei, otherpatterns emerge

283 stable nuclides (out of several thousand knownnuclides)

If we look at which have evenand odd Z and N;patterns emerge

Z N # stablenuclides

even even 165

even odd 56odd even 53

odd odd 9 2H, 6Li, 10B, 14N



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Clearly NOT random: even must implygreater stability

Not too surprising

Same is true of electrons in molecules

Most molecules have an evennumber of

electrons, as electrons pair up in orbitals Odd e molecules, radicals, are very

unstable, i.e. very reactive!!

Magic Numbers


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Magic Numbers

Look at binding energies, see certainnumbersof protons and neutrons result inspecial stability

Called Magic Numbers

1nand 1pin separate shells

Magic numbers (for both 1nand 1p) are2, 8, 20, 28, 50, 82, 126

For e pattern of stability is:

2, 10, 18, 36, 54, 86(Noble gases)

Magic Numbers


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Magic Numbers

Special stability of Noble Gases due toclosed shells of occupied orbitals

Structure of nucleus can also beunderstood in terms of shell structure

With filled shells of 1nand 1p having added

stabilityAt some point adding more 1nto higher

energy neutron shells decreases stability

of nuclei with too high n/pratio

Your Turn!


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ou u

Isotopes above the band of stability are morelikely to:

A. emit alpha particles

B. emit gamma rays

C. capture electronsD. emit beta particles

53

Radioactive Nuclei Found in Nature


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Non-naturally occurring elements (man-madeunstable) are denoted by having atomic mass inparentheses

All nuclei with Z > 83 are radioactive Yet some elements with Z between 83 and 92 occur

naturally

Atomic weight is NOT in parenthesesHow can this be?

There are 3 heavy nuclei, which have very long half-lives

Long enough to have survived for billions of years

Each parent of natural decay chain

Decay Chains


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y238

U half-life () = 4.5 billion years emitter

Daughter 234Th is also radioactive

emitter

Half-life much shorter

Long sequence of emissions, and Recall that emission changesAby 4, while

emission A= 0

Result: every member of chain hasA= (4n+ 2)where n= some simple integer

238Uranium Decay Chain


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y

238U 234Th5109 y

25 d

, 234Pa 7 hr

, 234U

5.7105 y

230Th,

8104 y226Ra

,

2103 y222Rn

4 d

218Po

3 m

214Pb ,

27 m

214Bi

20 m

, 214Po

1.6104

s

210Pb

22 y,

210Bi

5 d210Po

138 d206Pb

92 90 91 92

90888684

82 83 84 82

82 84 83

A stableisotope


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Decay Chains


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y

Final stable member of sequence is 206Pb Some intermediatenuclides have reasonably

short half-lives

Still found in nature because they are constantlybeing replenishedby decay of nuclei further upchain

Uranium-containing minerals (pitchblende ismost famous) contain many radioactive

elements

Your Turn!


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When the reaction, , occurs,the particle emitted is:

A. an alpha particle

B. a beta particle

C. an electronD. a gamma ray

59

222 218

86 84Rn Po + Xn

m

Transmutation


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Change of one isotope for another

Caused by1. 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 targetnucleus to form compound nucleus

NON-SPONTANEOUS Nuclear


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Processes Fusion

Occurs in starsright now

How elements formed

Induced Fission

Bombard heavy nuclei with neutron

OHeC 168

4

2

12

6 HeHH 3

2

1

1

2

1

NpC 1371

1

12

6 oo MnM 9942

10

9842

nKrBanU 10

92

36

142

56

1

0

235

922

Compound Nucleus


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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 ejectssomething

Neutron Proton

Electron Gamma radiation

Decay leaves new nucleus differentfrom original

Ex. Transmutation


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p1

1

17

8

18

9

14

7

4

2OFNHe ++

Compoundnucleus

Newnucleus

Targetnucleus

Bombard-ing particle

Highenergyparticle

Transmutation


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Can synthesize given nucleus in many ways:

Once formed, compound nucleus has nomemory of how it was made

Only knows how much energy it has

AlNaHe 27132311

42

+

AlMg2713

2612

11

+p

AlMgH 27132512

21

+

Transmutation


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Decay pathway depends on how much energy+ 00

2713 Al

p112612 Mg +

n102613 Al +

pn 1110

2512 Mg ++

HeNa

4

2

23

11 +

Al2713

Transmutation


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Used to synthesize new isotopes > 900 total

Most not on band of stability

All elements above 93 (neptunium) are manmade

Includes Actinides above 93 + 104 112 + 114 Heavier elements made by colliding two

larger nuclei

Also known as fusion

n10269110

270110

20882

6228 DsDsPbNi ++

Your Turn!


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What would be the element produced from thefusion of with ? The species would

be in a high energy state and in time would

undergo decay to other species.

A. NoB. Lr

C. UD. Hs

67

59

27Co

192

76Os

Measuring Radioactive Decay


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Atomic radiation = ionizing radiation Creates ions by knocking off electrons

Geiger Counter

Detects and radiation with enough E topenetrate window

Inside tube, gas at low P, form ions when radiationenters

Ions cause current to flow

Amount of current relates to amount of radiation

Measuring Radioactive Decay


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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

Activity


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Number of disintegrations per second Used to characterize radioactive material

A = kN

k= 1st order decay constant in terms of number ofnuclei rate than concentration

N= number of radioactive nuclides

Law of radioactive decay Radioactive decay is 1st order kinetics process

kNt

N=

=A

Units of Activity


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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 x 1010 dps = 3.7 x 1010 Bq

Activity in 1.0 g226

Ra = 1 Ci

Half-Life


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Time it takes for number of nuclides present attime t, Nt , 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 calculateAof known mass of radioisotope

kkt 693.02ln

21 ==

Ex. 3


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What is the activity of 1.0 g of strontium-90?The half-life = 28.1 years

Step 1. Convert t to seconds

Step 2. Convert t to ks

8

1086.8

min1

s60

hr1

min60

day1

hr24

yr1

day365yr28.1

=

1108

1082.71086.8

693.02ln

21

=

== sst

k

Ex. 3 (cont)90


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Step 3. Convert g90

Sr to number of atoms (N)

Step 4. Calculate Activity = kN

Sr1069.6

Sr1

Sr1002.6

Sr90

Sr1Sr0.1

21

23

atoms

mol

atoms

g

molg

=

Sr1069.61082.7 21110 atomssA =

A = 5.23 1012 atoms Sr/s 1 disintegration/atom

A = 5.23 1012 dps or 5.23 1012 Bq

Ex. 4


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3H, tritium, is a emitter with a half-lifet= 12.26 yrs. MW = 3.016 g/mol. Howmany grams of 3H are in a 0.5 mCi sample?

Step 1. Convert half-life to sas Ci is indisintegrations per second (dps)

Step 2. Convert t

to k

shr

sdhr

yrdyr 81087.336002425.365)26.12(

21 =

=

198

1079.11087.3

2ln2ln

21

=

== sst

k

Ex. 4 (cont)


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Step 3. Convert Ci to dps

Step 4. Calculate g 3H to get this activity

Step 5. Convert atoms to g

dpsCi

dps

mCi

CimCiA 7

10

1085.1107.3

1000

15.0 =

=

atoms1003.11079.1

1085.1 1619

7

=

== s

dps

k

A

N

= molg

moleatomsatomsg 016.3

/10022.61003.1H 23

16

31

= 5.2 108 g

Exposure Units


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Not all materials equally absorb radiation, thusactivity doesnt describe effect of exposure

1 gray (Gy) = 1 J absorbed energy/kgmaterial

SI unit of absorbed radiation

1 rad = absorption of 102 J/ kilogram of tissue

Older unit

1 Gy = 100 rad

These units dont take into account type of

radiation

Exposure Units


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Sieverts (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

1Rem = 102 Sv

Still used in medicine

Exposure to Radiation Typically X-ray = 0.007rem or 7mrem


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yp y y

0.3 rem/week is maximum safe exposure set byUS govt. 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

Your Turn!


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Workers cleaning up the Fukushima reactors wereexposed to as much as 400 mSv units of radiation per

hour. How many rems of exposure does this

correspond to?

A. 4000 remB. 400 rem

C. 40 rem

D. 4 rem

80

100 mrem 1 rem400 mSv 40 rem

mSv 1000 mrem

=

Why is Radiation Harmful? Not heat energy


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gy

Ability of ionizing radiation to form unstable ionsor 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 +

+ 0

e-1

Which Types are Most Harmful?


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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 Dont penetrate skin

Background Radiation


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Presence of natural radionuclides means we cantescape 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 Earths crust

Top 40 cm of soil hold 1 g radium () /sq kilometer 40K emit particles

Total average exposure 360 mrem/year 82% natural radiation 18 % man made

Radiation Intensity


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Intensity of radiation varies with distancefrom 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

21

22

2

1

d

d

I

I=

Ex. 5


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If the activity of a sample is 10 units at 5meters 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?

unitsm

munits

5.2)10(

)5(10

d

dI

I 2

2

22

211

2 =

==

ydyd

units

ydunit

8.420

0.05

)(11

I

dId

2

2

2

211

2

==

=

=

Your Turn!


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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 m

B. 100 m

C. 200 mD. 1000 m

86

( ) ( )22

40 rem 10 m100 m

0.4 remd = =

Radioactive DecayKinetics

S d f lid f ll


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Spontaneous decay of any nuclide followsfirst order kinetics

May be complicated by decay of daughter nuclide

For now consider single step decay processes

Rate of reaction for 1st order process

A products In nuclear reaction, consider rate based on

number of nuclei Npresent

kNt

N=

d

dReactionofRate

Radioactive DecayKinetics The integrated form is:

l N l N kt


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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 lineindicative of

1st order kinetics

Plot of Nvs. time gives anexponential decay.

l n

N

t

t

N

kt

oeNN =

Ex. 6131

I i d t b li t i


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I is used as a metabolic tracer inhospitals. It has a half-life, = 8.07days. How long before the activity falls to

1% of the initial value?kt

oeNN =

2

1

2lnln

== t

ktN

N

o

2ln

100

1ln)07.8(

2ln

ln2

1

=

=

daysN

N

t o

t = 53.6 days

Your Turn!

H h ill it t k di i t ith


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How many hours will it take a radioisotope witha half-life of 10.0 hours to drop to 12.5% of its

original activity?

A. 30.0 hrs

B. 20.0 hrsC. 40.0 hrs

D. 63.2 hrs

12.5% of original activity is 3 half-lives or30.0 hrs.

90

Radioisotope DatingHow old is an object?


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Fields Geology, Archeology, and Anthropology

Nature provides us with natural clocks or

stopwatchesA) Radiocarbon Dating (Willard LibbyNobel Prize in 1960)

Cosmic rays (from space) enter atmosphere Some react with N in atmosphere forming

radioisotope 14C

emitter with t = 5730 yr

HCnN 11146

10

147 ++

14C Dating 14C becomes incorporated into atmospheric

CO i ll titi


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CO2 in very small quantities 14C/12C ratio in air is slightly greater than Earths

crust because of ongoing enrichment

Living organisms breath, eat, etc 14C/12C equilibrate with atmosphere

Radioactive14

C is uniformly distributedaround globe Tested experimentally

Checked vs. counting tree rings, etc. For precise work, use correction based on

alternate methods

HOW? Freshly cut wood samples havef l b

14C Dating


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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

Wooden implement in Egyptian tomb (~3000

BC)

14C Dating


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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 paintings

2. Linen wraps on Dead Sea scrolls

Ex. 7

Geologists examine shells found in cliffsSh ll C CO d d b li i


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Geologists examine shells found in cliffs.Shells are CaCO3 and are made by livingorganisms. The activity of the shells is

found to be 6.24 cpm/g total C. How oldis the cliff formation?

21

2lnln

t

tkt

A

A

o

==A = 6.24 cpm/g total C

Ao = 15.3 cpm/g total Ct = 5730 yr

Can use N/No and

A/Ao interchangeablyas

A = kN

Since ratio, k cancels

Ex. 7 (continued)

2lnln tA


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Rearranging and solving for t

21

2lnln=

tAA

o

2ln

ln2

1

oA

At

t =

2ln3.15

24.6ln)5730(

= cpm

cpmyr

t

t= 7414 yr

B) Other Isotopes Provide NaturalClocks

Mi l ( k ) d t d i i t


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Minerals (moon rocks) dated using isotopeswith much longer half-lives

t = 1.27 x 109 yr

Compare ratios in rock

t = 4.5 x 109 yr

Rock with no other source of Pb can bedated using ratios

Pb

U206

238

++ 01

42

20682

23892 68 HePbU

4018014019 AreK ++

Ar

K40

40

Ex. 8

A sample of rock is found to contain 0 232gf 206Pb d 1 605 238U A i h ll


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A sample of rock is found to contain 0.232gof 206Pb and 1.605g 238U. Assuming that allthe 206Pb now present came from the decay

of 238U, calculate the time since thesolidification of this rock.

Step 1. Mass of 238U that decayed =

= 0.268 g 238U decayed

( )PbgPbmolg

Umolg 206206

238

232.0/206

/238

Ex. 8 (continued) Step 2. Mass of 238U in rock initially (t = 0)

N 1 605 g + 0 268 g 1 873 g


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No = 1.605 g + 0.268 g = 1.873 g

21

*2ln

ln t

t

N

No

=

yr105.4

*2ln

605.1

873.1

ln 9=

t

yr105.4

*693147.0

1544.0 9=

t

6931470yr105.41544.0

9

.t =

t= 1.0 x 109yr

Your Turn!A wooden bowl fragment found at an old camp site

thought to be approximately 11 000 years old was


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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 actualage of the sample?

A. 4260 yrs

B. 3347 yrs

C. 9810 yrs

D. 2523 yrs

t = {5730 yrs x ln(4.67/15.3)}/ln2 = 9810 yrs

100

Fission

Induce by bombardingnstable n cle s ith


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Induce by bombardingunstable nucleus witha slow neutron

Nuclear chain reaction Neutrons generated

keep going

With small mass of235U reactioncontinues, but easily

controlled Some neutrons are

lost to environment

Fission

Critical massT h 235U i


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Critical mass Too much 235U in one

place

Too many neutronsabsorbed

Too few lost

Uncontrollable fission

Leads to explosion

Use control rods toabsorb excessneutrons and keepreaction fromgoing critical

Nuclear Reactor No chance of nuclear explosion

Critical mass requires pure 235U


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Critical mass requires pure U

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 explosion

What happened at Chernobyl

Nuclear Reactors Could it happen at U.S. reactors?

Extremely unlikely


104/108


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

Nuclear Reactor


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Use heat from nuclear reaction to heat steam turbine Use to generate electricity

Your Turn!Which of the following fission reactions is

balanced?


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balanced?

A.

B.

C.

D.

106

235 1 142 92 192 0 56 36 0U + n Ba + Kr + 2 n

235 1 142 92 1

92 0 56 36 0

U + n Ba + Kr + n

235 1 141 92 1

92 0 56 36 0U + n Ba + Kr + 2 n

235 1 142 92 1

92 0 55 36 0U + n Cs + Kr + 2 n

Nuclear Fusion

Occurs when light nuclei join to form heavierl


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g jnucleus

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

107

Thermonuclear Fusion Uses high temperatures to overcome electrostatic

repulsions between nuclei


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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 (>200g/cm3

Brings nuclei within 2 fm = 21015 m

Pressures = several billion atms

Not there yet, major problem Containment of high temperature and pressures

Magnetic field current approach

108

Brdy 6Ed Ch21 NuclearReactions

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Transcript of Brdy 6Ed Ch21 NuclearReactions