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

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

Eb = Zm p c2 + Nmn c2 − M (Z, N )c2

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

EbV ∝ V ∝ R 3 = C1 A

EbS ∝ R 2 = − C2 A 2/3       EbC ∝ R −1 = − C3 Z 2 A −1/3

Eb = EbV + EbS + EbC = C1 A − C2 A 2/3 − C3 Z 2 A −1/3

Eb A

= C1 − C2 A −1/3 − C3 Z 2 A −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: ⎧ 11.2A −1/2   EE ⎪ 2/3 2 −1/3 2 EO,OE Eb =15.8A − 17.8A − 0.711Z A − 23.7(A − 2Z) / A + ⎨ 0 ⎪ −1/2  OO ⎪⎩ −11.2A

MeV

Weizsaecker semi-empirical binding energy formula: ⎧ 11.2A −1/2   EE ⎪ 2/3 2 −1/3 2 EO,OE Eb =15.8A − 17.8A − 0.711Z A − 23.7(A − 2Z ) / A + ⎨ 0 ⎪ −1/2  OO ⎪⎩ −11.2A

MeV

α,β and γ radioactivity

Lifetime: N = N 0 e−t /τ

Half life:

When t = τ

N 1 = = e−t1/2 /τ N0 2

N = e−1 ≈ 0.368 N0

⎛ 1⎞ ln ⎜ ⎟ ≈ −.693 = −t1/2 / τ ⎝ 2⎠

t1/2 ≈ .693τ

Alpha Decay 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 4He

 E B (A − 4)  +  E B (4)  >    E B (A)                E B ( 4 He) = 28 MeV

β Decay

n → p + e− + ν Lifetime: τ = 0.9 × 10 3 s = 15 min n

p G e-

ν

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

Energetics of Fission Rohlf P319 (brief)

ΔE  2(8.5)(115) − (7.6)(230) ≈ 200 MeV Why don’t isotopes with A>200 fission?

increased surface energy

d

decreased Coulomb energy

neutron capture 8.5 MeV

7.6 MeV

d

n

d

d

Spontaneous fission Z 2 / A > 46

d

235U

Valley of stability

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 Z 2 / A > 46

Pairing energy determines which are fissionable materials

n

+

235U

ΔE

236U

ΔE

n

+

238U

239U

Energetics of Nuclear Fusion in the Sun Rohlf P316

Begin with 2 separate protons: 2m p c2 = 2(938.27) = 1877.54 MeV

End with deuteron + β + + ν :

(

                         = m p c2 + mn c2 − E Bd

)

+ me c2  +   mν c2 = md c2 + me c2 + 0

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

(

)

Net energy release: 2m p c2 − md c2 + me c2 +   mν c2 =1.39 MeV.

Proton-Proton Cycle

(

)

2 proton-proton fusion reactions: 2 p + p → d + β + + ν release 2(1.39) MeV 2 proton-deuteron fusion reactions: p + d → 3 He + γ release 2(5.5) MeV 3

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

Total energy release: 26.7 MeV Net result: 4 protons are converted to 4 He + 2 β + + 2ν + 2γ 4p → 4 He + 2 β + + 26.7 MeV.

Helium burning. 4

He + 4 He → 8 Be

→ 4 He + 4 He  8

4

He + 12C → 16O

Be + 4 He → 12C

4

He + 16O →

20

Ne etc.

Carbon cycle requires higher core temperatures than p-p.

p + 12C → 13 N + γ

     13 N → 13C + e+ + ν e

p + 13C → 14 N + γ p + 14 N → 15O + γ p + 15 N → 12C + 4 He

     15O → 15 N + e+ + ν e

Nuclear fusion in the sun and stars. radiation pressure

surface: 6x103 K hydrogen fusion: 10x107 K radiation pressure

radiation pressure

helium core

radiation pressure