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CHEM 312: Lecture 2Nuclear Properties

• Readings:§ Modern Nuclear Chemistry:

Chapter 2 Nuclear Properties§ Nuclear and Radiochemistry:

Chapter 1 Introduction, Chapter 2 Atomic Nuclei

• Nuclear properties§ Masses§ Binding energies§ Reaction energetics

à Q value§ Nuclei have shapes

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Nuclear Properties• Systematic examination of measurable data to determine nuclear

properties§ masses § matter distributions

• Size, shape, mass, and relative stability of nuclei follow patterns that can be understood and interpreted with models § average size and stability of a nucleus described by average

nucleon binding in a macroscopic model§ detailed energy levels and decay properties evaluated with a

quantum mechanical or microscopic modelSimple example: Number of stable nuclei based on neutron and proton number

N even odd even oddZ even even odd oddNumber 160 53 49 4

Simple property dictates nucleus behavior. Number of protons and neutron important

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Which are the 4 stable odd-odd nuclei?

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• Evaluation of Mass Excess• Difference between actual mass of nucleus and expected mass

from atomic number§ By definition 12C = 12 amu

à If mass excess negative, then isotope has more binding energy the 12C

• Mass excess==M-A§ M is nuclear mass, A is mass number§ Unit is MeV (energy)

à Convert with E=mc2

• 24Na example§ 23.990962782 amu§ 23.990962782-24 = -0.009037218 amu§ 1 amu = 931.5 MeV

à -0.009037218 amu x (931.5 MeV/1 amu)à -8.41817 MeV= Mass excess= for 24Na

Data from Mass

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Masses and Q value• Atomic masses

§ From nuclei and electrons• Nuclear mass can be found from atomic mass

§ m0 is electron rest mass, Be (Z) is the total binding energy of all the electrons

§ Be(Z) is small compared to total mass• Energy (Q) from mass difference between parent and daughter

§ Mass excess values can be used to find Q (in MeV)• β- decay Q value

§ AZA(Z+1)+ + β- + n + Qà Consider β- mass to be part of A(Z+1) atomic mass (neglect

binding)à Q=D AZ-DA(Z+1)

§ 14C14N+ + β- + n + Qà Energy =Q= mass 14C – mass 14N

* Use Q values (http://www.nndc.bnl.gov/wallet/wccurrent.html)

à Q=3.0198-2.8634=0.156 MeV

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Q value • Positron Decay

§ AZA(Z-1)- + β+ + n + Q§ Have 2 extra electrons to consider

à β+ (positron) and additional atomic electron from Z-1 daughter* Each electron mass is 0.511 MeV, 1.022 MeV total from the electrons

§ Q=DAZ – (DA(Z-1)- + 1.022) MeV§ 90Nb90Zr- + β+ + n + Q§ Q=D 90Nb – (D 90Zr + 1.022) MeV§ Q=-82.6632-(-88.7742+1.022) MeV=5.089 MeV

• Electron Capture (EC)§ Electron comes from parent orbital

à Parent can be designated as cation to represent this behavior§ AZ+ + e- A(Z-1) + n + Q§ Q=DAZ – DA(Z-1)§ 207Bi207Pb + n + Q§ Q=D 207Bi – D 207Pb MeV§ Q= -20.0553- -22.4527 MeV=2.3947 MeV

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Q value• Alpha Decay

§ AZ(A-4)(Z-2) + 4He + Q§ 241Am237Np + 4He + Q

à Use mass excess or Q value calculator to determine Q value

§ Q=D241Am-(D 237Np+D4He)§ Q = 52.937-(44.874 + 2.425)§ Q = 5.638 MeV§ Alpha decay energy for

241Am is 5.48 and 5.44 MeV

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Q value determination• For a general reaction

§ Treat Energy (Q) as part of the equationà Solve for Q

• 56Fe+4He59Co+1H+Q§ Q= [M56Fe+M4He-(M59Co+M1H)]c2

* M represents mass of isotopeà Q=-3.241 MeV (from Q value calculator)

• Mass excess and Q value data can be found in a number of sources§ Table of the Isotopes§ Q value calculator

à http://www.nndc.bnl.gov/qcalc/§ Atomic masses of isotopes

à http://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl

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Q value calculation examples• Find Q value for the Beta decay of 24Na

§ 24Na24Mg+ +b- + +n Q§ Q= 24Na-24Mg§ M (24Na)-M(24Mg)

à 23.990962782-23.985041699 à 0.005921 amu

* 5.5154 MeV§ From mass excess

à -8.417 - -13.933 à 5.516 MeV

• Q value for the EC of 22Na§ 22Na+ + e- 22Ne + +n Q§ Q= 22Na - 22Ne § M (22Na)-M(22Ne)§ 21.994436425-21.991385113 § 0.003051 amu

à 2.842297 MeV§ From mass excess

à -5.181 - -8.024 à 2.843 MeV

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Terms from Energy• Binding energy

§ Difference between mass of nucleus and constituent nucleonsà Energy released if nucleons formed

nucleus§ Nuclear mass not equal to sum of constituent

nucleonsBtot (A,Z)=[ZM(1H)+(A-Z)M(n)-M(A,Z)]c2

§ average binding energy per nucleon

à Bave(A,Z)= Btot (A,Z)/A

à Some mass converted into energy that binds nucleus

à Measures relative stability• Binding Energy of an even-A nucleus is generally higher than adjacent odd-A

nuclei• Exothermic fusion of H atoms to form He from very large binding energy of 4He• Energy released from fission of the heaviest nuclei is large

§ Nuclei near the middle of the periodic table have higher binding energies per nucleon

• Maximum in the nuclear stability curve in the iron-nickel region (A~56 through 59)§ Responsible for the abnormally high natural abundances of these elements§ Elements up to Fe formed in stellar fusion

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Mass Based Energetics Calculations

• Why does 235U undergo neutron induced fission for thermal energies while 238U doesn’t?

• Generalized energy equation§ AZ + n A+1Z + Q

• For 235U§ Q=(40.914+8.071)-42.441§ Q=6.544 MeV

• For 238U§ Q=(47.304+8.071)-50.569§ Q=4.806 MeV

• For 233U§ Q=(36.913+8.071)-38.141§ Q=6.843 MeV

• Fission requires around 5-6 MeV§ Does 233U from thermal

neutron?

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Binding-Energy Calculation: Development of simple nuclear model

• Volume of nuclei are nearly proportional to number of nucleons present§ Nuclear matter is incompressible§ Basis of equation for nuclear radius

• Total binding energies of nuclei are nearly proportional to numbers of nucleons present§ saturation character

à Nucleon in a nucleus can apparently interact with only a small number of other nucleons

à Those nucleons on the surface will have different interactions •Basis of liquid-drop model of nucleus

§ Considers number of neutrons and protons in nucleus and how they may arrange

§ Developed from mass dataà http://en.wikipedia.org/wiki/Semi-empirical_mass_formula

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Liquid-Drop Binding Energy:

• c1=15.677 MeV, c2=18.56 MeV, c3=0.717 MeV, c4=1.211 MeV, k=1.79 and =11/A1/2

• 1st Term: Volume Energy

§ dominant term

à in first approximation, binding energy is proportional to the number of nucleons

§ (N-Z)2/A represents symmetry energy

àbinding E due to nuclear forces is greatest for the nucleus with equal numbers of neutrons and protons

124

3/123

23/2

2

2

1 11 AZcAZcA

ZNkAc

A

ZNkAcEB

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Liquid drop model

• 2nd Term: Surface Energy§ Nucleons at surface of nucleus have unsaturated forces§ decreasing importance with increasing nuclear size

• 3rd and 4th Terms: Coulomb Energy§ 3rd term represents the electrostatic energy that arises from the Coulomb

repulsion between the protonsà lowers binding energy

§ 4th term represents correction term for charge distribution with diffuse boundary

• term: Pairing Energy§ binding energies for a given A depend on whether N and Z are even or odd

à even-even nuclei, where =11/A1/2, are the most stable§ two like particles tend to complete an energy level by pairing opposite

spinsà Neutron and proton pairs

124

3/123

23/2

2

2

1 11 AZcAZcA

ZNkAc

A

ZNkAcEB

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Magic Numbers: Data comparison

• Certain values of Z and N exhibit unusual stability§ 2, 8, 20, 28, 50, 82,

and 126• Evidence from different

data § masses, § binding energies, § elemental and

isotopic abundances

• Concept of closed shells in nuclei§ Similar to electron

closed shell• Demonstrates limitation

in liquid drop model• Magic numbers

demonstrated in shell model

§ Nuclear structure and model lectures

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Mass Parabolas

• Method of demonstrating stability for given mass constructed from binding energy§ Values given in

difference, can use energy difference

• For odd A there is only one -stable nuclide

§ nearest the minimum of the parabola Friedlander & Kennedy, p.47

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Even A mass parabola

• For even A there are usually two or three possible -stable isobars

§ Stable nuclei tend to be even-even nucleià Even number of protons, even number of neutron for these cases

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R=roA1/3

Nuclear Shapes: Radii

• Nuclear volumes are nearly proportional to nuclear masses§ nuclei have approximately same density

• nuclei are not densely packed with nucleons§ Density varies

• ro~1.1 to 1.6 fm for equation above

• Nuclear radii can mean different things§ nuclear force field§ distribution of charges§ nuclear mass distribution

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Nuclear Force Radii• The radius of the nuclear force field must be less than the

distance of closest approach (do)

§ d = distance from center of nucleus

§ T’ = particle’s kinetic energy

§ T = particle’s initial kinetic energy

§ do = distance of closest approach in a head on collision when T’=0

• do~10-20 fm for Cu and 30-60 fm for U

od

ZeTT

22'

T

Zedo

22

http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html#c1

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Measurement of Nuclear Radii

• Any positively charged particle can be used to probe the distance§ nuclear (attractive) forces become significant

relative to the Coulombic (repulsive force)• Neutrons can be used but require high energy

§ neutrons are not subject to Coulomb forcesà high energy needed for de Broglie

wavelengths small compared to nuclear dimensions

§ at high energies, nuclei become transparent to neutronsà Small cross sections

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Electron Scattering• Using moderate energies of electrons, data is compatible

with nuclei being spheres of uniformly distributed charges • High energy electrons yield more detailed information about

the charge distribution § no longer uniformly charged spheres

• Radii distinctly smaller than indicated by methods that determine nuclear force radii

• Re (half-density radius)~1.07 fm

• de (“skin thickness”)~2.4 fm

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Nuclear potentials

• Scattering experimental data have has approximate agreement the Square-Well potential

• Woods-Saxon equation better fit

§ Vo=potential at center of nucleus

§ A=constant~0.5 fm

§ R=distance from center at which V=0.5Vo (for half-potential radii)

§ or V=0.9Vo and V=0.1Vo for a drop-off from 90 to 10% of the full potential

ARro

e

VV

/)(1

• ro~1.35 to 1.6 fm for Square-Well

• ro~1.25 fm for Woods-Saxon with half-potential radii,

• ro~2.2 fm for Woods-Saxon with drop-off from 90 to 10%• Nuclear skin thickness

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Nuclear SkinNucleus Fraction of nucleons in the “skin”12C 0.9024Mg 0.7956Fe 0.65107Ag 0.55139Ba 0.51208Pb 0.46238U 0.44

]/)[(1)(

eee aRro

er

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Spin

• Nuclei possess angular momenta Ih/2§ I is an integral or half-integral number known as nuclear spin

à For electrons, generally distinguish between electron spin and orbital angular momentum

• Protons and neutrons have I=1/2• Nucleons in nucleus contribute orbital angular momentum

(integral multiple of h/2 ) and their intrinsic spins (1/2)§ Protons and neutrons can fill shell (shell model)

à Shells have orbital angular momentum like electron orbitals (s,p,d,f,g,h,i,….)

§ spin of even-A nucleus is zero or integral § spin of odd-A nucleus is half-integral

• All nuclei of even A and even Z have I=0 in ground state

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Magnetic Moments• Nuclei with nonzero angular momenta have

magnetic moments§ From spin of protons and neutrons

• Bme/Mp is unit of nuclear magnetic moments

§ nuclear magneton•Measured magnetic moments tend to differ from calculated values§ Proton and neutron not simple structures

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Methods of measurements• Hyperfine structure in atomic spectra

• Atomic Beam method

§ Element beam split into 2I+1 components in magnetic field

• Resonance techniques

§ 2I+1 different orientations• Quadrupole Moments: q=(2/5)Z(a2-c2), R2 = (1/2)(a2 + c2)= (roA1/3)2

§ Data in barns, can solve for a and c

• Only nuclei with I1/2 have quadrupole moments

§ Non-spherical nuclei

§ Interactions of nuclear quadrupole moments with the electric fields produced by electrons in atoms and molecules give rise to abnormal hyperfine splittings in spectra

• Methods of measurement: optical spectroscopy, microwave spectroscopy, nuclear resonance absorption, and modified molecular-beam techniques

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Parity

• System wave function sign change if sign of the space coordinates change§ system has odd or even parity

• Parity is conserved• even+odd=odd, even+even=even, odd+odd=odd

§ allowed transitions in atoms occur only between an atomic state of even and one of odd parity

• Parity is connected with the angular-momentum quantum number l

§ states with even l have even parity§ states with odd l have odd parity

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Topic review

• Understand role of nuclear mass in reactions§ Use mass defect to determine energetics§ Binding energies, mass parabola, models

• Determine Q values• How are nuclear shapes described and

determined§ Potentials§ Nucleon distribution

• Quantum mechanical terms§ Used in description of nucleus

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Study Questions

• What do binding energetics predict about abundance and energy release?

• Determine and compare the alpha decay Q values for 2 even and 2 odd Np isotopes. Compare to a similar set of Pu isotopes.

• What are some descriptions of nuclear shape?• Construct a mass parabola for A=117 and A=50• What is the density of nuclear material?• Describe nuclear spin, parity, and magnetic

moment

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Question

• Comment in blog• Respond to PDF Quiz 2