nuclear physics and nuclear energy

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