Upload
others
View
19
Download
0
Embed Size (px)
Citation preview
Nuclear Physics and Nuclear Energy
The Nuclear Force. Rohlf Ch. 11. p296
Homework: Ch. 11: 4,5,11,12,42 Due 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 Model Rohlf 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 particular they 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 ∝ R2 =− C2A2/3 EbC ∝ R−1 =− C3Z
2A−1/3
Eb = EbV + EbS + EbC =C1A − C2A2/3 − C3Z
2A−1/3
EbA
=C1 − C2A−1/3 − C3Z
2A−4/3
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 − 0.711Z2A−1/3 − 23.7(A − 2Z )2 / A+11.2A−1/2 EE
0 EO,OE
−11.2A−1/2 OO
⎧
⎨⎪
⎩⎪⎪
MeV
Weizsaecker semi-empirical binding energy formula:
Eb =15.8A −17.8A2/3 − 0.711Z2A−1/3 − 23.7(A − 2Z )2 / A+11.2A−1/2 EE
0 EO,OE
−11.2A−1/2 OO
⎧
⎨⎪
⎩⎪⎪
MeV
α,β and γ radioactivity
Lifetime: N = N0e−t /τ When t = τ NN0
= e−1 ≈ 0.368
Half life: NN0
=12= e−t1/2 /τ ln 1
2⎛⎝⎜
⎞⎠⎟≈ −.693 = −t1/2 / τ t1/2 ≈ .693τ
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 α decay
EB(A − 4) + EB(4) > EB(A) EB(4He) = 28 MeV
β Decay
n→ p + e− + ν
Lifetime: τ = 0.9 ×103 s = 15 min
n p
e- ν
G
A z,n( )→ A z +1,n −1( ) + e− + ν
A z,n( )→ A z −1,n +1( ) + e+ + ν
τ varies from µs to years
γ- radiation
N∗ → N + γ
ΔE 2(8.5)(115)− (7.6)(230) ≈ 200MeV
Energetics of Fission Rohlf P319 (brief)
Why don’t 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 d d n
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
ΔE ΔE
Energetics of Nuclear Fusion in the Sun Rohlf P316
Begin with 2 separate protons: 2mpc2 = 2(938.27) = 1877.54 MeV
End with deuteron + β+ + ν :
= mpc2 +mnc2 − EBd( ) +mec2 + mνc2 = mdc2 +mec2 + 0
= 938.27 + 939.57 - 2.22( ) + 0.51= 1876.14 MeV
Net energy release: 2mpc2 − mdc2 +mec2 + mνc2( ) =1.39 MeV.
2 proton-proton fusion reactions: 2 p + p→ d + β+ + ν( ) release 2(1.39) MeV
2 proton-deuteron fusion reactions: p + d→ 3He + γ 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 + 2β+ + 2ν + 2γ
4p→ 4He + 2β+ + 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 + γ 13N → 13C + e+ + νe
p + 13C→ 14N + γ
p + 14N→ 15O + γ 15O→ 15N + e+ + νe
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