7. Nuclear modelszanotti/FNSN/FNSN_files/...Lecture3.ppt Author laura fabbietti Created Date...

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7. Nuclear models 7.1 Fermi gas model 7.2 Shell model

Transcript of 7. Nuclear modelszanotti/FNSN/FNSN_files/...Lecture3.ppt Author laura fabbietti Created Date...

  • 7. Nuclear models

    7.1 Fermi gas model

    7.2 Shell model

  • • Free electron gas: protons and neutrons moving quasi-freely within thenuclear volume

    • 2 different potentials wells for protons and neutrons.

    • Spherical square well potentials with the same radius

    Fermi Gas Model

    Statistics of the Fermi distribution

    Given a volume V the numer of states dn goes like:

    dn =4 p

    2dpV

    (2 h)3

    n =4 p

    2dpV

    (2 h)3

    =VpF

    3

    62h3

    0

    pF

    If T=0 the nucleus is in the ground state andpF (Fermi Momentum) is the maximum possiblemomentum of the ground state.

    N = 2n =VpF ,n

    3

    32h3

    Z =VpF ,p

    3

    32h3

    Spin 1/2 particles

    N= number of neutrons

    Z= number of protons

  • V =4

    3R3

    =4

    3R0

    3A R

    0=1.21 fm

    Z = N = A /2A

    2=4

    3

    Ro3ApF

    3

    32h3

    pF = pF ,n = pF ,p =h

    R0

    9

    8

    1

    3

    250MeV /c

    The nucleon moves in thenucleus with a large momentum

    Fermi Energy

    Binding Energy: BE/A= 7-8 MeV V0=EF+B/A~ 40MeV

    EF =pF2

    2mN33 MeV

    Nucleons are rather weakly bound inthe nucleus

  • Potential well in the Fermi-gas model

    The neutron potential well is deeper that the proton well because of themissing Coulomb repulsion. The Fermi Energy is the same, otherwise the p-->ndecay would happen spontaneously. This implies that they are more neutronsstates available and hence N>Z the heavier the nuclei become.

  • < EKin >=

    EKin p2dp

    0

    pF

    p2dp

    0

    pF=3

    5

    pF2

    2mN20MeV

    EKin (N,Z) = N < EKin,N > +Z < EKin,Z >=3

    10mN(NpF ,N

    2+ ZpFZ

    2)

    pFN =N 3

    2h3

    V

    =

    N 92h3

    4 R03A

    1

    3

    pFZ =Z 3

    2h3

    V

    =

    Z 92h3

    4 R03A

    1

    3

    EKin =3

    19mN

    h2

    R0

    92h3

    4 R03

    2

    3 Z

    5

    3 + N

    5

    3

    A

    2

    3

    EKin =h2

    R02

    3

    10mN

    9

    4

    2

    3 N

    5

    3 + Z

    5

    3

    A

    2

    3

    =3

    10mN

    h2

    R02

    9

    4

    2

    3

    A +5

    9

    (N Z)2

    A+ ...

    N-Z

    EKin Decay

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    luigiernestoTypewritten Textp , se Z = N, di ciascun nucleone

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  • L’espressione precedente si ottiene da: D = N − Z

    N= A2+

    N − Z2

    "

    #$

    %

    &'=12A+D( )

    Z= A2−

    N − Z2

    "

    #$

    %

    &'=12A−D( )

    sostituendo in:

    N 5 3 + Z 5 3

    A2 3!

    "#

    $

    %&=

    A+D2

    !

    "#

    $

    %&5 3

    +A−D2

    !

    "#

    $

    %&5 3

    A2 3

    !

    "

    ####

    $

    %

    &&&&

    e sviluppando i binomi arrestandosi al secondo ordine:

    N 5 3 + Z 5 3

    A2 3!

    "#

    $

    %&=

    12!

    "#$

    %&

    53 A

    5 3 +53A2 3D+10

    18A−1 3D2 + A5 3 − 5

    3A2 3D+10

    18A−1 3D2

    A2 3

    !

    "

    ###

    $

    %

    &&&

    N 5 3 + Z 5 3

    A2 3!

    "#

    $

    %&=

    12!

    "#$

    %&

    532A+10

    9D2

    A!

    "#

    $

    %&=

    12!

    "#$

    %&

    23A+ 5

    9D2

    A!

    "#

    $

    %&

    NOTA BENE: A) Il risultato prova che l’energia cinetica totale ha un MINIMO quando N = Z, e dato che E(totale) = T –V, con T proporzionale ad A, E(totale), che e’ negativa ed e’ anche la BINDING ENERGY totale, e’ proporzionale ad A (primo termine dell’equazione semiempirica delle masse) B) Aggiungendo il secondo termine, come nell’equazione semiempirica delle masse, la BINDING ENERGY diminuisce in modulo. A parte I valori dei coefficienti, il modello a gas di Fermi predice correttamente

    EB = T −V = −cVolA+ c

    Sym

    N − Z( )2

    A

  • Calculation of the state density

    h2

    2m=

    h2

    2m

    2

    x2

    +2

    y2

    +2

    z2

    = E

    h2

    2m

    2X(x)

    x2

    = EX X(x)..... (r) = X(x)Y (y)Z(z)

    YZEX X + XZEyY + XYEZZ = EXYZ

    2X(x)

    x2

    = kX k =1

    h2mE

    X = A eik x

    + B eik x

    Boundary Conditions : at x = y = z =a

    2: X =Y = Z = 0

    X+

    =2

    2acosk

    +x X =

    2i

    2asink x k

    +=

    +

    a, =1,3,5.. k =

    a, = 0,2,4...

    EX =h2

    2mk2

    =1

    2m

    h

    a

    2

    E =1

    2m

    h

    a

    2

    ( x2

    + y2

    + z2) p

    2= 2mE =

    h

    a

    2

    ( x2

    + y2

    + z2)

    3a3p3

    a

    Volume inwhich thestate densityis calculated

    a

    Considering the approximation that the nuclear potential shold have a sharpedge in correspondence of the nuclear radius, one can approximate that toparticles trapped in the pot potential

  • We count how many states we have in a spherical volume of radius between and +

    2= x

    2+ y

    2+ z

    2d = 4

    2

    dn =1

    8d =

    2

    2=a

    hp d =

    a

    hdp

    dn =2

    a2

    2h2p2 a

    hdp =

    a3

    22h3p2dp =

    4 p2dpv

    (2 h)3

    Momentum volume

    Space volume

    Phase-space occupied by one state

    p2

    = 2mE p2dp = 2m

    3EdE

    dn = m

    3

    2 ( 22h3)1v dE E = C

    1EdE

    n = 2dn

    dEdE = 2C

    1v EdE = ( 8m

    3

    2 )(32h3)1v EF

    2

    3

    0

    EF

    0

    EF

    Number of spin 1/2 particles which sit in the well:

    EF= maximal energy of theparticle in the well

    dn/dE

    EF

    E

    EF

    =1

    2m

    3

    2

    3

    4

    3h2 n

    v

    2 / 3

    30MeV

    in Cu atoms :n

    v= 6 10

    23 9

    64= 8 10

    22cm

    3EF

    7eV

  • Hierarchy of energy eigenstatesof the harmonic oscillator potential

    N = 2(n-1) + l

    E = h (N + 3/2)

  • Shell Model Potential

    Skin thickness

  • Spin Orbit Coupling

    It changes the hierarchy of the energy levels:

    V (r) =Vcentral (r) +Vls(r)< ls >

    h2

    < ls >

    h2

    =1

    2( j( j +1) l(l +1) s(s+1)) =

    l /2 j = l +1/2

    (l +1) /2 j = l 1/2

    Els =2(l +1)

    2

    Moving belowMoving above

    The energy levels transform into nlj levels.

  • Shell model

    Single particle level calculated in theshell model.

    Spin-orbitcoupling

    Saxon-Wood Potential

    1g58

    1s

    1p

    2s

    2p

    1d

    1f

    40

    92

    2d3s

    1h

    2(2l+1)

    2

    6

    102

    14

    18

    6

    10

  • Single-Particle-Spectrum of the Wood-Saxon Potentialfor a heavy nucleus

  • Energy of first excited nuclear level in gg-nuclei

  • Energy levels of some nucleiExcita

    tion

    Ene

    rgy

    PbCaCaOHe208

    82

    48

    20

    40

    20

    16

    8

    4

    2

  • Nuclear Magnetic Moments in Shell Model

    μnucl = μN (gl li + gssi) /hi=1

    A

    gl =1 for p

    0 for n

    gs =5.58 for p

    3.83 for n

    < μnucl >= μN < nucl gl li + gssi nucl > /hi=1

    A

    Wigner-Eckart Theorem: The expectation value of any vector operator of a systemis equal to the projection onto ist angular momentum

    < μnucl >= gnuclμN< J >

    hgnucl =

    < JMJ gl li + gssi JMJ >

    < JMJ J2JMJ >i=1

    A

    In the case of a single nucleon in addition to the close shell the angularmomentum of the closed shell nuclei couples to 0. The nuclear magneticMomentum of is equal to the nuclear magnetic mom. of the valence nucleon.

    gnucl =gl ( j( j +1) + l(l +1) s(s+1)) + gs( j( j +1) + s(s+1) l(l +1))

    2 j( j +1)

    μnuclμN

    = gnuclJ = gl ±gs gl

    2l +1

    J = l ±1/2

  • Magnetic moments:Shell model calculations and experimental values