Nuclear Physics and Nuclear · PDF fileNuclear Energetics: Liquid Drop Model Rohlf P303 The nucleon-nucleon potential looks similar to the atom-atom potential, but on a different scale

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  • Nuclear Physics and Nuclear Energy

  • The Nuclear Force.Rohlf Ch. 11. p296

    Homework: Ch. 11: 4,5,11,12,42Due Nov. 10

    Additional homework: Due Nov. 13

    Assume a 238U fissions exactly into two equal nuclei.1. What nuclei are the fission products.2. What is the difference in the total binding energy before and after fission?

    Assume this is the energy released by fission.3. Assume the 238U and its fission products are uniformly charged spheres. which repel. What will be their kinetic energy when they fly apart? Compare

    with your answer in part. 2 above.

  • Nuclear Energetics: Liquid Drop ModelRohlf P303

    The nucleon-nucleon potential looks similar to the atom-atom potential, but on a different scale.Thus, conglomerations of nucleons should have propertied similar to those of atoms. In particularthey should be rather incompressible, with rather uniform densities within the volume.

  • Nuclear binding energy = energy required to separate the nucleus into free neutrons and protons.

    Eb = Zmpc2 + Nmnc2 M (Z,N )c2

    Nuclear Binding Energy

    Note: It is nearly constant except for the lightest nuclei.

  • EbV V R3 =C1A EbS R

    2 = C2A2/3 EbC R

    1 = C3Z2A1/3

    Eb = EbV + EbS + EbC =C1A C2A2/3 C3Z

    2A1/3

    EbA

    =C1 C2A1/3 C3Z

    2A4/3

    The constants can be obtained by fitting to the empirical data.

  • Add more terms to binding equation.

    Pauli energy: All things being equal, nucleons tend to have an equal number of protons and neutrons due to the Pauli exclusion principle.

    EbP (N Z )2 / A. EbP = C4 (A 2Z )

    2 / A

    Odd-even energy: Nucleon energies are lower when their spins can pair off in the same spatial state. Even-even are more stable than even-odd, which in turn are more stable than odd-odd.

    EbEE = EbOO = C5 / A1/2 . EbOE = EbEO = 0.

    Weizsaecker semi-empirical binding energy formula:

    Eb =15.8A 17.8A2/3 0.711Z2A1/3 23.7(A 2Z )2 / A+11.2A1/2 EE

    0 EO,OE

    11.2A1/2 OO

    MeV

  • Weizsaecker semi-empirical binding energy formula:

    Eb =15.8A 17.8A2/3 0.711Z2A1/3 23.7(A 2Z )2 / A+11.2A1/2 EE

    0 EO,OE

    11.2A1/2 OO

    MeV

  • , and radioactivity

    Lifetime: N = N0et / When t = NN0

    = e1 0.368

    Half life: NN0

    =12= et1/2 / ln 1

    2

    .693 = t1/2 / t1/2 .693

  • Alpha Decay 4He has the highest binding energy for a few nucleon system. If EB/A for A-4 + 4He > EB/A for nucleus with A nucleons, it will decay

    EB(A 4)+EB(4)>EB(A)EB(4He) = 28MeV

  • Decay

    n p + e +

    Lifetime: = 0.9 103 s = 15 min

    n p

    e-

    G

    A z,n( ) A z +1,n 1( ) + e +

    A z,n( ) A z 1,n +1( ) + e+ +

    varies from s to years

  • - radiation

    N N +

  • E 2(8.5)(115) (7.6)(230) 200MeV

    Energetics of Fission Rohlf P319 (brief)

    Why dont isotopes with A>200 fission?

  • d

    8.5 MeV

    7.6 MeV

    neutron capture

    increased surface energy

    decreased Coulomb energy

    d

    Spontaneous fission Z2 / A > 46

    d ddn

  • Valley of stability

    235U

    118Pd

  • Distribution of fission products.

    Fission products are far off the stability curve. Thus there are extra neutrons emitted and numerous beta decays until the products are back in the valley of stability. Some isotopes are very long lived beta emitters.

  • Exercise: From the A -Z stability curve, estimate the maximum element number Z which can exist, and above which any element would spontaneously and instantly fission.

    Spontaneous fission Z2 / A > 46

  • Pairing energy determines which are fissionable materials

    n + 235U

    236U

    n + 238U

    239U

    E E

  • Energetics of Nuclear Fusion in the SunRohlf P316

    Begin with 2 separate protons: 2mpc2 = 2(938.27) = 1877.54 MeV

    End with deuteron + + + :

    = mpc2 +mnc2 EBd( ) +mec2 +mc2 = mdc2 +mec2 + 0

    = 938.27 + 939.57 - 2.22( ) + 0.51= 1876.14 MeV

    Net energy release: 2mpc2 mdc2 +mec2 +mc2( ) =1.39 MeV.

  • 2 proton-proton fusion reactions: 2 p + p d + + + ( ) release 2(1.39) MeV

    2 proton-deuteron fusion reactions: p + d 3He + release 2(5.5) MeV

    3He- 3He fusion reaction: 3He+ 3He 4He + p + p releases 12.9 MeV

    Total energy release: 26.7 MeV

    Net result: 4 protons are converted to 4He + 2+ + 2 + 2

    4p 4He + 2+ + 26.7 MeV.

    Proton-Proton Cycle

  • Helium burning.

    4He + 4He 8Be 4He + 4He

    8Be+ 4He 12C

    4He + 12C 16O 4He + 16O 20Ne etc.

  • Carbon cycle requires higher core temperatures than p-p.

    p + 12C 13N + 13N 13C + e+ + e

    p + 13C 14N +

    p + 14N 15O + 15O 15N + e+ + e

    p + 15N 12C + 4He

  • Nuclear fusion in the sun and stars.

    radiation pressure

    radiation pressure

    radiation pressure

    radiation pressure surface: 6x103 K

    hydrogen fusion: 10x107 K

    helium core