2-1

RFSS and NFSS: Lecture 2Nuclear Properties

• Readings:§ Modern Nuclear Chemistry: Chapter 2

Nuclear Properties§ Nuclear and Radiochemistry: Chapter 1

Introduction, Chapter 2 Atomic Nuclei § Nuclear Forensics Analysis, Chapter 2

• Nuclear properties§ Masses§ Binding energies§ Reaction energetics§ Nuclear shapes

2-2

Nuclear Properties• Systematic examination of general nuclear properties

§ masses § matter distributions

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

average binding of the nucleons in a macroscopic model§ detailed energy levels and decay properties evaluated with

a quantum mechanical or microscopic model

Number of stable nuclei based on neutron and proton number

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

2-3

Mass excess• Difference between actual mass and atomic

number§ Expression of binding energy relative to 12C

à By definition 12C = 12 amu* If mass excess negative, then isotope

has more binding energy the 12C• Mass excess==M-A

§ 24Na § 23.990962782 amu§ 23.990962782-24 = -0.009037218 amu

• 1 amu = 931.5 MeV • -8.41817 = Mass excess=

2-4

Masses• Atomic masses

§ 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 difference in mass between parent and daughter• Consider beta decay of 14C

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

• Positron decay

• Electron Capture

2-5

Masses• Electron Capture

§ Electron comes from parent orbitalà Parent designated as cation to

represent this behavior• Alpha Decay

§ 241Am237Np + 4He + Qà Use mass excess or Q value

calculator to determine Q value§ Q = 52.937 – (44.874 + 2.425)§ Q = 5.638 MeV§ Alpha decay energy for 241Am is

5.48 and 5.44 MeV

2-6

Masses• For a general reaction

§ Treat Energy (Q) as part of the equationà 56Fe+4He59Co+1H+Qà Q= [M56Fe+M4He-(M59Co+M1H)]c2

* M represents mass of isotope

For coursework please note if 1.022 MeV added to positron Q value

2-7

Q value calculation• 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

• Can also use Q-value calculator§ http://www.nndc.bnl.gov/qcalc/

2-8

Terms• Binding energy

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

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

§ average binding energy per nucleonà Bave(A,Z)= Btot (A,Z)/Aà Measures relative stability

• Binding Energy of an even-A nucleus is generally higher than values for adjacent odd-A nuclei

• Exothermic nature of the fusion of H atoms to form He from the 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)

§ Thought to be responsible for the abnormally high natural abundances of these elements

2-9

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 fission from

thermal neutron

2-10

Binding-Energy Calculation• Volume of nuclei are nearly proportional to the number of

nucleons present§ Nuclear matter is incompressible§ Basis of equation for nuclear radius

• Total binding energies of nuclei are nearly proportional to the 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

2-11

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 energyàbinding 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 AZcAZcAZNkAc

AZNkAcEB

2-12

Liquid drop model

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

• 3rd and 4thTerms: 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

oddà even-even nuclei, where =11/A1/2, are the stablest

§ two like particles tend to complete an energy level by pairing opposite spins

++-

-

--

-

- -- 124

3/123

23/2

2

2

1 11 AZcAZcAZNkAc

AZNkAcEB

2-13

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 b-stable nuclide§ nearest the

minimum of the parabola

Friedlander & Kennedy, p.47

2-14

Even A mass parabola• For even A there are usually two or three possible b-stable isobars

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

2-15

Magic Numbers

• Certain values of N and Z--2, 8, 20, 28, 50, 82, and 126 --exhibit unusual stability

• Evidence from masses, binding energies, elemental and isotopic abundances, numbers of species with given N or Z, and -particle energies

• Concept of closed shells in nuclei

• Demonstrates limitation in liquid drop model

2-16

R=roA1/3

Nuclear Shapes: Radii

• Nuclear volumes are about 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• Nuclear radii can mean different things

§ nuclear force field§ distribution of charges§ nuclear mass distribution

2-17

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

odZeTT

22' -

TZedo

22

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

2-18

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

2-19

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

2-20

Nuclear potentials

• Scattering experiments 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

eVV /)(1 -+

2-21

Square-Well and Woods-Saxon Potentials• 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%,

skin thickness, of the full potential

2-22

Nuclear Skin• charge density give information on protons distribution in

nuclei§ no experimental techniques exist for determining total

nucleon distributionà generally assumed that neutrons and protons are

distributed in same way§ nuclear-potential radii are about 0.2 fm larger than radii

of the charge distributionsNucleus 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 -+

2-23

Spin• Nuclei possess angular momenta Ih/2

§ I is an integral or half-integral number known as the nuclear spinà For electrons, generally distinguish between electron spin

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

2-24

Magnetic Moments• Nuclei with nonzero angular momenta have magnetic moments

§ From spin of protons and neutrons• Bme/Mp is used as the unit of nuclear magnetic moments and

called a nuclear magneton• Magnetic moments are often expressed in terms of gyromagnetic

ratios § g*I nuclear magnetons, where g is + or - depending upon

whether spin and magnetic moment are in the same direction•Measured magnetic moments tend to differ from calculated values

§ Proton and neutron not simple structures à Neutron has charge distribution

* Negative (from negative mesons) near edge

2-25

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

2-26

Parity• System wave function sign change if the 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

2-27

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

2-28

Study Questions

• What do binding energetic 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• Describe nuclear spin, parity, and magnetic

moment

2-29

Pop Quiz

• Using the appropriate mass excess calculate the following Q values for 212Bi. Show the reaction§ b- decay§ b+ decay§ EC§ Alpha decay

• Which decay modes are likely?