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1 1 7/18/2007 7/18/2007 General Physics (PHY 2140) Lecture 18 Lecture 18 ¾ Modern Physics Nuclear Physics Nuclear properties Binding energy Radioactivity The Decay Process Natural Radioactivity Chapter 29 http://www.physics.wayne.edu/~alan/2140Website/Main.htm

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Page 1: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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General Physics (PHY 2140)

Lecture 18Lecture 18Modern Physics

Nuclear PhysicsNuclear propertiesBinding energyRadioactivityThe Decay ProcessNatural Radioactivity

Chapter 29

http://www.physics.wayne.edu/~alan/2140Website/Main.htm

Page 2: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

227/18/20077/18/2007

Lightning ReviewLightning Review

Last lecture:

1.1. Quantum physicsQuantum physicsElectron Clouds (Electron Clouds (OrbitalsOrbitals))The Pauli Exclusion PrincipleThe Pauli Exclusion PrincipleCharacteristic XCharacteristic X--RaysRaysAtomic Energy LevelsAtomic Energy LevelsLasers and HolographyLasers and Holography

2 4

2 2

1 , 1,2,3,2e e

nm k eE n

n⎛ ⎞= − =⎜ ⎟⎝ ⎠

Review Problem: Why do lithium, sodium, and potassium exhibit similar chemical properties?

Because they have similar outer orbitals. The outer orbitals determine the chemical bonding characteristics, therefore elements with similar valence orbital structure will have similar chemical properties.

2 4

3 2 2

1 14

e e

f i

m k e cfn nπ λ

⎛ ⎞= − − =⎜ ⎟⎜ ⎟

⎝ ⎠

All have one valence electron.

Page 3: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Modern understanding: the ``onionModern understanding: the ``onion’’’’ picturepicture

Let’s see what’s inside!

Atom

Page 4: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Modern understanding: the ``onionModern understanding: the ``onion’’’’ picturepicture

Let’s see what’s inside!

Nucleus

Page 5: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

557/18/20077/18/2007

Modern understanding: the ``onionModern understanding: the ``onion’’’’ picturepicture

Let’s see what’s inside!

Protons and neutrons

Next chapter…

Page 6: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Introduction: Development of Nuclear PhysicsIntroduction: Development of Nuclear Physics18961896 –– the birth of nuclear physicsthe birth of nuclear physics

Becquerel discovered radioactivity in uranium compoundsBecquerel discovered radioactivity in uranium compoundsRutherford showed the radiation had three typesRutherford showed the radiation had three types

Alpha (He nucleus)Alpha (He nucleus)Beta (electrons)Beta (electrons)Gamma (highGamma (high--energy photons)energy photons)

19111911 Rutherford, Geiger and Rutherford, Geiger and MarsdenMarsden performed scattering experimentsperformed scattering experimentsEstablished the point mass nature of the nucleusEstablished the point mass nature of the nucleusNuclear forceNuclear force was a new type of forcewas a new type of force

19191919 Rutherford and coworkers first observed nuclear reactions in whRutherford and coworkers first observed nuclear reactions in which ich naturally occurring alpha particles bombarded nitrogen nuclei tonaturally occurring alpha particles bombarded nitrogen nuclei to produce produce oxygenoxygen19321932 Cockcroft and Walton first used artificially accelerated protonCockcroft and Walton first used artificially accelerated protons to s to produce nuclear reactionsproduce nuclear reactions19321932 Chadwick discovered the neutronChadwick discovered the neutron19331933 the Curies discovered artificial radioactivitythe Curies discovered artificial radioactivity19381938 Hahn and Hahn and StrassmanStrassman discovered nuclear fissiondiscovered nuclear fission19421942 Fermi achieved the first controlled nuclear fission reactorFermi achieved the first controlled nuclear fission reactor

Page 7: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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29.1 Some Properties of Nuclei29.1 Some Properties of Nuclei

All nuclei are composed of protons and neutronsAll nuclei are composed of protons and neutronsException is ordinary hydrogen with just a protonException is ordinary hydrogen with just a proton

The The atomic numberatomic number, , ZZ, equals the number of protons in the nucleus, equals the number of protons in the nucleusThe The neutron numberneutron number, , NN, is the number of neutrons in the nucleus, is the number of neutrons in the nucleusThe The mass numbermass number, , AA, is the number of nucleons in the nucleus, is the number of nucleons in the nucleus

A = Z + NA = Z + NNucleon is a generic term used to refer to either a proton or a Nucleon is a generic term used to refer to either a proton or a neutronneutronThe mass number is not the same as the massThe mass number is not the same as the mass

NotationNotation

Example:Example:

Mass number is 27Mass number is 27Atomic number is 13Atomic number is 13Contains 13 protonsContains 13 protonsContains 14 (27 Contains 14 (27 –– 13) neutrons13) neutrons

The Z may be omitted since the element can be used to determine The Z may be omitted since the element can be used to determine ZZ

XAZ

where X is the chemical symbol of the elementwhere X is the chemical symbol of the element

Al2713

Page 8: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Charge and massCharge and mass

Charge:Charge:The electron has a single negative charge, The electron has a single negative charge, --e (e (e = 1.60217733 x 10e = 1.60217733 x 10--1919 CC))The proton has a single positive charge, +e The proton has a single positive charge, +e

Thus, charge of a nucleus is equal to Thus, charge of a nucleus is equal to ZeZeThe neutron has no chargeThe neutron has no charge

Makes it difficult to detectMakes it difficult to detect

Mass:Mass:It is convenient to use It is convenient to use atomic mass units,atomic mass units, u, to express massesu, to express masses

1 u = 1.660559 x 101 u = 1.660559 x 10--2727 kgkgBased on definition that the Based on definition that the mass of one atom of Cmass of one atom of C--12 is exactly 12 u12 is exactly 12 u

Mass can also be expressed in MeV/cMass can also be expressed in MeV/c22

From EFrom ERR = m c= m c22

1 u = 931.494 MeV/c1 u = 931.494 MeV/c22

Page 9: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Summary of MassesSummary of Masses

Masses

Particle kgkg uu MeV/cMeV/c22

Proton 1.6726 x 101.6726 x 10--2727 1.0072761.007276 938.28938.28

Neutron 1.6750 x 101.6750 x 10--2727 1.0086651.008665 939.57939.57

Electron 9.101 x 109.101 x 10--3131 5.486x105.486x10--44 0.5110.511

Page 10: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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What is the order of magnitude of the number of protons in your body? Of the number of neutrons? Of the number of electrons? Take your mass approximately equal to 70 kg (154 lbs).

Quick problem: protons in your bodyQuick problem: protons in your body

An iron nucleus (in hemoglobin) has a few more neutrons than protons, but in a typical water molecule there are eight neutrons and ten protons. So protons and neutrons are nearly equally numerous in your body, each contributing 35 kg out of a total body mass of 70 kg.

Same amount of neutrons and electrons.

2827

1 nucleon35 10 protons1.67 10

N kgkg−

⎛ ⎞= ≈⎜ ⎟×⎝ ⎠

Page 11: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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The Size of the NucleusThe Size of the Nucleus

First investigated by Rutherford in First investigated by Rutherford in scattering experimentsscattering experimentsHe found an expression for how He found an expression for how close an alpha particle moving close an alpha particle moving toward the nucleus can come toward the nucleus can come before being turned around by the before being turned around by the Coulomb forceCoulomb forceThe KE of the particle must be The KE of the particle must be completely converted to PEcompletely converted to PE

2

2

4 ek Zedmv

=( )( )2 1 2 21

2 e e

e Zeq qmv k kr d

= = or

For gold: d = 3.2 x 10For gold: d = 3.2 x 10--1414 m, for silver: d = 2 x 10m, for silver: d = 2 x 10--1414 mmSuch small lengths are often expressed in Such small lengths are often expressed in femtometersfemtometers where 1 fm = 10where 1 fm = 10--1515 m m

(also called a (also called a fermifermi))

Page 12: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Size of NucleusSize of Nucleus

Since the time of Rutherford, Since the time of Rutherford, many other experiments have many other experiments have concluded the followingconcluded the following

Most nuclei are approximately Most nuclei are approximately sphericalsphericalAverage radius isAverage radius is

rroo = 1.2 x 10= 1.2 x 10--1515 mm

31

oArr =

Page 13: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Density of NucleiDensity of Nuclei

The volume of the nucleus (assumed to be spherical) is The volume of the nucleus (assumed to be spherical) is directly proportional to the total number of nucleonsdirectly proportional to the total number of nucleonsThis suggests that This suggests that all nuclei haveall nuclei have nearly the same densitynearly the same densityNucleons combine to form a nucleus as though they were Nucleons combine to form a nucleus as though they were tightly packed spherestightly packed spheres

Page 14: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Nuclear StabilityNuclear Stability

There are very large There are very large repulsive electrostatic forcesrepulsive electrostatic forces between protonsbetween protonsThese forces should cause the nucleus to fly apartThese forces should cause the nucleus to fly apart

The nuclei are stable because of the presence of another, shortThe nuclei are stable because of the presence of another, short-- range force, called the range force, called the nuclear (or strong) forcenuclear (or strong) force

This is an This is an attractive forceattractive force that acts between all nuclear particlesthat acts between all nuclear particlesThe nuclear attractive force is stronger than the Coulomb repulsThe nuclear attractive force is stronger than the Coulomb repulsive ive force at the short ranges within the nucleusforce at the short ranges within the nucleus

Page 15: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Nuclear Stability chartNuclear Stability chart

Light nuclei are most stable if Light nuclei are most stable if N = ZN = ZHeavy nuclei are most stable when Heavy nuclei are most stable when N > ZN > Z

As the number of protons increase, As the number of protons increase, the Coulomb force increases and so the Coulomb force increases and so more nucleons are needed to keep more nucleons are needed to keep the nucleus stablethe nucleus stable

No nuclei are stable when No nuclei are stable when Z > 83Z > 83(the metal Bismuth)(the metal Bismuth)

Page 16: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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IsotopesIsotopes

The nuclei of all atoms of a particular element must contain The nuclei of all atoms of a particular element must contain the same number of protonsthe same number of protonsThey may contain varying numbers of neutronsThey may contain varying numbers of neutrons

IsotopesIsotopes of an element have the same Z but differing N and A valuesof an element have the same Z but differing N and A values

Example: Example: C116 C14

6C136C12

6

All have the same number of protons: Z=6

Page 17: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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29.2 Binding Energy29.2 Binding Energy

The total energy of the The total energy of the bound system (the bound system (the nucleus) is less than the nucleus) is less than the combined energy of the combined energy of the separated nucleonsseparated nucleons

This difference in energy is This difference in energy is called the called the binding energybinding energy of of the nucleusthe nucleus

It can be thought of as the It can be thought of as the amount of energy you need amount of energy you need to add to the nucleus to to add to the nucleus to break it apart into break it apart into separated protons and separated protons and neutronsneutrons

Binding Energy per NucleonBinding Energy per Nucleon

Page 18: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Problem: binding energyProblem: binding energy

Calculate the average binding energy per nucleon of 9341 Nb

Page 19: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Calculate the average binding energy per nucleon of

Given:

mp = 1.007276u1.007276umn = 1.008665u1.008665u

Find:

Eb = ?

In order to compute binding energy, let’s first find the mass difference between the total mass of all protons and neutrons in Nb and subtract mass of the Nb:

41pN =

Thus, binding energy is (per nucleon or nuclear particle)

( ) ( )( )2 0.865028 931.58.66 nucleon

93b

m c u MeV uE MeV

= = =

9341 Nb

93 41 52nN = − =Number of protons: Number of neutrons:

Mass difference:

41 52p n Nbm m m mΔ = + −( ) ( ) ( )41 1.007825 52 1.008665 92.9063768

0.865028u u u

u= + −

=

Page 20: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Binding Energy NotesBinding Energy NotesExcept for light nuclei, the Except for light nuclei, the binding energy is about 8 MeV binding energy is about 8 MeV per nucleonper nucleonThe curve peaks in the vicinity of The curve peaks in the vicinity of A = 60A = 60

Nuclei with mass numbers Nuclei with mass numbers greater than or less than 60 are greater than or less than 60 are not as strongly bound as those not as strongly bound as those near the middle of the periodic near the middle of the periodic tabletable

The curve is slowly varying at A The curve is slowly varying at A > 40 > 40

This suggests that the nuclear This suggests that the nuclear force saturatesforce saturatesA particular nucleon can interact A particular nucleon can interact with only a limited number of with only a limited number of other nucleonsother nucleons

Page 21: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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29.3 Radioactivity29.3 Radioactivity

RadioactivityRadioactivity is the spontaneous emission of radiation is the spontaneous emission of radiation Experiments suggested that radioactivity was the result Experiments suggested that radioactivity was the result of the decay, or disintegration, of unstable nucleiof the decay, or disintegration, of unstable nucleiThree types of radiation can be emittedThree types of radiation can be emitted

AlphaAlpha particlesparticlesThe particles are The particles are 44He nuclei (2 neutrons and 2 protons)He nuclei (2 neutrons and 2 protons)

BetaBeta particlesparticlesThe particles are either electrons or positronsThe particles are either electrons or positrons

A positron is the A positron is the antiparticleantiparticle of the electronof the electronIt is similar to the electron except its charge is +eIt is similar to the electron except its charge is +e

GammaGamma raysraysThe The ““raysrays”” are high energy photonsare high energy photons

Page 22: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Distinguishing Types of RadiationDistinguishing Types of Radiation

The gamma particles carry no The gamma particles carry no chargechargeThe alpha particles are deflected The alpha particles are deflected upwardupwardThe beta particles are deflected The beta particles are deflected downwarddownward

A positron would be deflected A positron would be deflected upwardupward

Page 23: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Penetrating Ability of ParticlesPenetrating Ability of Particles

Alpha particlesAlpha particlesBarely penetrate a piece of paperBarely penetrate a piece of paper

Beta particlesBeta particlesCan penetrate a few mm of aluminumCan penetrate a few mm of aluminum

Gamma raysGamma raysCan penetrate several cm of leadCan penetrate several cm of lead

Page 24: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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The Decay ConstantThe Decay Constant

The number of particles that decay in a given time is proportionThe number of particles that decay in a given time is proportional to al to the total number of particles in a radioactive samplethe total number of particles in a radioactive sample

λλ is called the is called the decay constantdecay constant and and determines the rate at which the determines the rate at which the material will decaymaterial will decay

The The decay decay raterate or or activityactivity, R, of a sample is defined as the number , R, of a sample is defined as the number of decays per secondof decays per second

NR Nt

λΔ= =

Δ

( )N N tλΔ = − Δ

Page 25: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Decay CurveDecay Curve

The decay curve follows the The decay curve follows the equationequation

The The halfhalf--lifelife is also a useful is also a useful parameterparameterThe halfThe half--life is defined as the life is defined as the time it takes for half of any time it takes for half of any given number of radioactive given number of radioactive nuclei to decaynuclei to decay

λ=

λ=

693.02lnT 21

0tN N e λ−=

Page 26: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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UnitsUnits

The unit of The unit of activityactivity, R, is the , R, is the Curie, CiCurie, Ci1 Ci = 3.7 x 101 Ci = 3.7 x 101010 decays/seconddecays/second

The SI unit of The SI unit of activityactivity is the is the Becquerel, BqBecquerel, Bq1 Bq = 1 decay / second1 Bq = 1 decay / second

Therefore, 1 Ci = 3.7 x 10Therefore, 1 Ci = 3.7 x 101010 BqBq

The most commonly used units of The most commonly used units of activityactivity are the mCi are the mCi and the and the µµCiCi

Page 27: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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QUICK QUIZ

What fraction of a radioactive sample has decayed after two half- lives have elapsed?

(a) 1/4 (b) 1/2 (c) 3/4 (d) not enough information to say

(c). At the end of the first half-life interval, half of the original sample has decayed and half remains. During the second half-life interval, half of the remaining portion of the sample decays. The total fraction of the sample that has decayed during the two half-lives is: 1 1 1 3

2 2 2 4⎛ ⎞+ =⎜ ⎟⎝ ⎠

Page 28: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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29.4 The Decay Processes 29.4 The Decay Processes –– General RulesGeneral Rules

When one element changes into another element, the process When one element changes into another element, the process is called is called spontaneous decayspontaneous decay or or transmutationtransmutationThe The sum of the mass numberssum of the mass numbers, , AA, must be , must be the samethe same on both on both sides of the equationsides of the equationThe The sum of the atomic numberssum of the atomic numbers, , ZZ, must be , must be the samethe same on both on both sides of the equationsides of the equationConservation of massConservation of mass--energy and conservation of momentum energy and conservation of momentum must holdmust hold

Page 29: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Alpha DecayAlpha Decay

When a nucleus emits an When a nucleus emits an alpha particlealpha particle it loses two it loses two protons and two neutronsprotons and two neutrons

N decreases by 2N decreases by 2Z decreases by 2Z decreases by 2A decreases by 4A decreases by 4

SymbolicallySymbolically

X is called the X is called the parent nucleusparent nucleusY is called the Y is called the daughter nucleusdaughter nucleus

HeYX 42

4A2Z

AZ +→ −

Page 30: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Alpha Decay Alpha Decay ---- ExampleExample

Decay of Decay of 226226RaRa

Half life for this decay is 1600 Half life for this decay is 1600 yearsyearsExcess mass is converted into Excess mass is converted into kinetic energykinetic energyMomentum of the two particles Momentum of the two particles is equal and oppositeis equal and opposite

HeRnRa 42

22286

22688 +→

Page 31: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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QUICK QUIZ

If a nucleus such as 226Ra that is initially at rest undergoes alpha decay, which of the following statements is true? (a) The alpha particle has more kinetic energy than the daughter nucleus. (b) The daughter nucleus has more kinetic energy than the alpha particle. (c) The daughter nucleus and the alpha particle have the same kinetic energy.

(a). Conservation of momentum requires the momenta of the two fragments be equal in magnitude and oppositely directed. Thus, from KE = p2/2m, the lighter alpha particle has more kinetic energy that the more massive daughter nucleus.

Page 32: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Beta DecayBeta Decay

During beta decay, During beta decay, the daughter nucleus has the same the daughter nucleus has the same number of nucleons as the parent, but the atomic number of nucleons as the parent, but the atomic number changes by +1 or number changes by +1 or --11In addition, an electron (positron) was observed In addition, an electron (positron) was observed The emission of the electron is The emission of the electron is from the nucleusfrom the nucleus

The nucleus contains protons and neutronsThe nucleus contains protons and neutronsThe process occurs when a neutron is transformed into a proton The process occurs when a neutron is transformed into a proton and an electronand an electronEnergy must be conservedEnergy must be conserved

Page 33: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Beta Decay Beta Decay –– Electron EnergyElectron EnergyThe energy released in the decay The energy released in the decay process should process should almost allalmost all go to go to kinetic energy of the electronkinetic energy of the electronExperiments showed that Experiments showed that fewfew electrons had this amount of electrons had this amount of kinetic energykinetic energyTo account for this To account for this ““missingmissing”” energy, in 1930 Pauli proposed the energy, in 1930 Pauli proposed the existence of another particleexistence of another particleEnricoEnrico Fermi later named this Fermi later named this particle the particle the neutrinoneutrinoProperties of the neutrinoProperties of the neutrino

Zero electrical chargeZero electrical chargeMass much smaller than the Mass much smaller than the electron, but not zeroelectron, but not zeroSpin of Spin of ½½Very weak interaction with matterVery weak interaction with matter

Page 34: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Beta Decay Beta Decay

SymbolicallySymbolically

νν is the symbol for the is the symbol for the neutrinoneutrinois the symbol for the is the symbol for the antineutrinoantineutrino

To summarize, in beta decay, the following pairs of To summarize, in beta decay, the following pairs of particles are emittedparticles are emitted

An electron and an antineutrinoAn electron and an antineutrinoA positron and a neutrinoA positron and a neutrino

ν++→

ν++→+

−+

eYXeYX

A1Z

AZ

A1Z

AZ

ν

Page 35: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Gamma DecayGamma Decay

Gamma rays are given off when an excited nucleus Gamma rays are given off when an excited nucleus ““fallsfalls”” to a lower to a lower energy stateenergy state

Similar to the process of electron Similar to the process of electron ““jumpsjumps”” to lower energy states and giving off to lower energy states and giving off photonsphotons

The excited nuclear states result from The excited nuclear states result from ““jumpsjumps”” made by a proton or neutronmade by a proton or neutronThe excited nuclear states may be the result of violent collisioThe excited nuclear states may be the result of violent collision or more n or more likely of an alpha or beta emissionlikely of an alpha or beta emissionExample of a decay sequenceExample of a decay sequence

The first decay is a beta emissionThe first decay is a beta emissionThe second step is a gamma emissionThe second step is a gamma emission

The C* indicates the Carbon nucleus is in an excited stateThe C* indicates the Carbon nucleus is in an excited stateGamma emission doesnGamma emission doesn’’t change either A or Zt change either A or Z

γ+→

ν++→ −

C*Ce*CB

126

126

126

125

Page 36: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Uses of RadioactivityUses of Radioactivity

Carbon DatingCarbon DatingBeta decay of Beta decay of 1414C is used to date organic samplesC is used to date organic samplesThe ratio of The ratio of 1414C to C to 1212C is usedC is used

Smoke detectorsSmoke detectorsIonization type smoke detectors use a radioactive source to ioniIonization type smoke detectors use a radioactive source to ionize the ze the air in a chamberair in a chamberA voltage and current are maintained A voltage and current are maintained When smoke enters the chamber, the current is decreased and the When smoke enters the chamber, the current is decreased and the alarm soundsalarm sounds

Radon pollutionRadon pollutionRadon is an inert, gaseous element associated with the decay of Radon is an inert, gaseous element associated with the decay of radiumradiumIt is present in uranium mines and in certain types of rocks, brIt is present in uranium mines and in certain types of rocks, bricks, etc icks, etc that may be used in home buildingthat may be used in home buildingMay also come from the ground itselfMay also come from the ground itself

Page 37: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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29.5 Natural Radioactivity29.5 Natural Radioactivity

Classification of nucleiClassification of nucleiUnstable nuclei found in natureUnstable nuclei found in nature

Give rise to Give rise to natural radioactivitynatural radioactivityNuclei produced in the laboratory through nuclear reactionsNuclei produced in the laboratory through nuclear reactions

Exhibit Exhibit artificial radioactivityartificial radioactivityThree series of natural radioactivity existThree series of natural radioactivity exist

UraniumUraniumActiniumActiniumThoriumThorium

Page 38: Lecture 18 - Physics and Astronomyalan/2140Website/Lectures/Lecture18.pdfCalculate the average binding energy per nucleon of . Given: m. p = 1.007276u m. n = 1.008665u. Find: E. b

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Decay SeriesDecay Series of of 232232ThTh

Series starts with Series starts with 232232ThThProcesses through Processes through a series of alpha a series of alpha and beta decaysand beta decaysEnds with a stable Ends with a stable isotope of lead, isotope of lead, 208208PbPb