nuclear reactions the chain reaction

Nuclear reactionsThe chain reaction

Nuclear reactions

For power applications, want a self-sustained chain reaction.

The chain reaction

Natural U: 0.7% of 235U and 99.3% of 238U

Natural U: 0.7% of 235U and 99.3% of 238UFission neutrons have KE ≈ 2 MeV; and 238U can absorb these high-energy neutrons (not by fission), but 238U does not absorb slow neutrons.

Natural U: 0.7% of 235U and 99.3% of 238UFission neutrons have KE ≈ 2 MeV; and 238U can absorb these high-energy neutrons (not by fission), but 238U does not absorb slow neutrons. Want to slow neutrons to allow fission of 235U and avoid absorption by 238U.

Natural U: 0.7% of 235U and 99.3% of 238UFission neutrons have KE ≈ 2 MeV; and 238U can absorb these high-energy neutrons (not by fission), but 238U does not absorb slow neutrons. Want to slow neutrons to allow fission of 235U and avoid absorption by 238U. Also need to enrich the uranium to several percent of 235U.

Natural U: 0.7% of 235U and 99.3% of 238UFission neutrons have KE ≈ 2 MeV; and 238U can absorb these high-energy neutrons (not by fission), but 238U does not absorb slow neutrons. Want to slow neutrons to allow fission of 235U and avoid absorption by 238U. Also need to enrich the uranium to several percent of 235U. Finally, want each fission even to produce enough neutrons to cause another fission event.

Natural U: 0.7% of 235U and 99.3% of 238UFission neutrons have KE ≈ 2 MeV; and 238U can absorb these high-energy neutrons (not by fission), but 238U does not absorb slow neutrons. Want to slow neutrons to allow fission of 235U and avoid absorption by 238U. Also need to enrich the uranium to several percent of 235U. Finally, want each fission even to produce enough neutrons to cause another fission event.

Using “moderator” to slow down neutrons—light nuclei are best since a collision by a neutron will transfer more kinetic energy to the nucleus that is initially at rest.

Natural U: 0.7% of 235U and 99.3% of 238UFission neutrons have KE ≈ 2 MeV; and 238U can absorb these high-energy neutrons (not by fission), but 238U does not absorb slow neutrons. Want to slow neutrons to allow fission of 235U and avoid absorption by 238U. Also need to enrich the uranium to several percent of 235U. Finally, want each fission even to produce enough neutrons to cause another fission event.

Using “moderator” to slow down neutrons—light nuclei are best since a collision by a neutron will transfer more kinetic energy to the nucleus that is initially at rest. Graphite (carbon) was used by Fermi et al. in 1942.

Natural U: 0.7% of 235U and 99.3% of 238UFission neutrons have KE ≈ 2 MeV; and 238U can absorb these high-energy neutrons (not by fission), but 238U does not absorb slow neutrons. Want to slow neutrons to allow fission of 235U and avoid absorption by 238U. Also need to enrich the uranium to several percent of 235U. Finally, want each fission even to produce enough neutrons to cause another fission event.

Using “moderator” to slow down neutrons—light nuclei are best since a collision by a neutron will transfer more kinetic energy to the nucleus that is initially at rest. Graphite (carbon) was used by Fermi et al. in 1942. Most modern reactors use heavy water D2O.

Natural U: 0.7% of 235U and 99.3% of 238UFission neutrons have KE ≈ 2 MeV; and 238U can absorb these high-energy neutrons (not by fission), but 238U does not absorb slow neutrons. Want to slow neutrons to allow fission of 235U and avoid absorption by 238U. Also need to enrich the uranium to several percent of 235U. Finally, want each fission even to produce enough neutrons to cause another fission event.

Using “moderator” to slow down neutrons—light nuclei are best since a collision by a neutron will transfer more kinetic energy to the nucleus that is initially at rest. Graphite (carbon) was used by Fermi et al. in 1942. Most present-day reactors use heavy water D2O.

(Cd)

Nuclear fusion (the sun’s source of power)

Mostly fusion of protons in the core of the sun

Nuclear fusion (the sun’s source of power)

Mostly fusion of protons in the core of the sun

Density of protons must be high enough to ensure a high rate of collision

Nuclear fusion (the sun’s source of power)

Mostly fusion of protons in the core of the sun

Density of protons must be high enough to ensure a high rate of collision, and proton KE (T ~ 107 K) must be high enough so Coulomb repulsion is overcome—allowing strong attractive nuclear force to take over.

Nuclear fusion (the sun’s source of power)

Mostly fusion of protons in the core of the sun

Density of protons must be high enough to ensure a high rate of collision, and proton KE (T ~ 107 K) must be high enough so Coulomb repulsion is overcome—allowing strong attractive nuclear force to take over.

e21

11

11 eHHH ν++→+ +

The proton-proton cycle

Nuclear fusion (the sun’s source of power)

Mostly fusion of protons in the core of the sun

Density of protons must be high enough to ensure a high rate of collision, and proton KE (T ~ 107 K) must be high enough so Coulomb repulsion is overcome—allowing strong attractive nuclear force to take over.

e21

11

11 eHHH ν++→+ +

The proton-proton cycleelectron neutrino

Nuclear fusion (the sun’s source of power)

Mostly fusion of protons in the core of the sun

Density of protons must be high enough to ensure a high rate of collision, and proton KE (T ~ 107 K) must be high enough so Coulomb repulsion is overcome—allowing strong attractive nuclear force to take over.

e21

11

11 eHHH ν++→+ +

The proton-proton cycle

HeHH 32

21

11 →+

electron neutrino

Nuclear fusion (the sun’s source of power)

Mostly fusion of protons in the core of the sun

Density of protons must be high enough to ensure a high rate of collision, and proton KE (T ~ 107 K) must be high enough so Coulomb repulsion is overcome—allowing strong attractive nuclear force to take over.

e21

11

11 eHHH ν++→+ +

The proton-proton cycle

HeHH 32

21

11 →+

( )H2HeHeHe 11

42

32

32 +→+

electron neutrino

Nuclear fusion (the sun’s source of power)

Mostly fusion of protons in the core of the sun

Density of protons must be high enough to ensure a high rate of collision, and proton KE (T ~ 107 K) must be high enough so Coulomb repulsion is overcome—allowing strong attractive nuclear force to take over.

e21

11

11 eHHH ν++→+ +

The proton-proton cycle

HeHH 32

21

11 →+

( )H2HeHeHe 11

42

32

32 +→+

two parts+

two parts+

one part

electron neutrino

Nuclear fusion (the sun’s source of power)

Mostly fusion of protons in the core of the sun

Density of protons must be high enough to ensure a high rate of collision, and proton KE (T ~ 107 K) must be high enough so Coulomb repulsion is overcome—allowing strong attractive nuclear force to take over.

e21

11

11 eHHH ν++→+ +

The proton-proton cycle

HeHH 32

21

11 →+

( )H2HeHeHe 11

42

32

32 +→+

two parts+

two parts+

one part

electron neutrino

( ) e42

11 2e2HeH4 ν++→ +

( ) e42

11 2e2HeH4 ν++→ +

The two e+s will annihilate with two electrons

γ→+ −+ 2ee

( ) e42

11 2e2HeH4 ν++→ +

The two e+s will annihilate with two electrons

γ→+ −+ 2eeSo overall reaction is

( ) s'2Hee2H4 e42

11 γ+ν+→+ −

( ) e42

11 2e2HeH4 ν++→ +

The two e+s will annihilate with two electrons

γ→+ −+ 2eeSo overall reaction is

( ) s'2Hee2H4 e42

11 γ+ν+→+ − (CAPA Set #13, Prob. #6)

( ) e42

11 2e2HeH4 ν++→ +

The two e+s will annihilate with two electrons

γ→+ −+ 2eeSo overall reaction is

( ) s'2Hee2H4 e42

11 γ+ν+→+ − (CAPA Set #13, Prob. #6)

≈ 2% of energy output carried by neutrinos.

( ) e42

11 2e2HeH4 ν++→ +

The two e+s will annihilate with two electrons

γ→+ −+ 2eeSo overall reaction is

( ) s'2Hee2H4 e42

11 γ+ν+→+ − (CAPA Set #13, Prob. #6)

≈ 2% of energy output carried by neutrinos.Physicists & astronomers interested in measuring neutrino output from sun.

( ) e42

11 2e2HeH4 ν++→ +

The two e+s will annihilate with two electrons

γ→+ −+ 2eeSo overall reaction is

( ) s'2Hee2H4 e42

11 γ+ν+→+ − (CAPA Set #13, Prob. #6)

≈ 2% of energy output carried by neutrinos.Physicists & astronomers interested in measuring neutrino output from sun.

Overall energy output = Q ≈ 25 MeV

The binding energy curve

EB

The binding energy curve

Fusing light nuclei (low EB) produces nuclei with larger EB. Thus energy left over.E

B

The binding energy curve

Fusing light nuclei (low EB) produces nuclei with larger EB. Thus energy left over.E

B

Fission

Elementary particles

In 1960s there was a very large number and variety of subatomic particles. Today, only electrons, photons and a few other particles are elementary. The rest, such as protons and neutrons, are systems of smaller particles called quarks. The quark model reduced the large number of particles to a reasonable value and has successfully predicted new quark combinations that have been subsequently observed.

Fundamental forces in nature

Let’s look at the electromagnetic interaction using the photon-mediating-particle concept.

∆t

π≈∆⋅∆

4h

tE

Violation of energy conservation? No, if ∆t is short enough.

Eph

Invoke uncertainty principle

If ∆t is short enough, then Eph < ∆E—no violation of E conservation

Feynmandiagram

Classification of particles (& antiparticles)

Leptons: truly elementary particles—have no structure or size. They interact only through the weak & electromagnetic forces. There are six leptons:

Classification of particles (& antiparticles)

Leptons: truly elementary particles—have no structure or size. They interact only through the weak & electromagnetic forces. There are six leptons:

ννντµ

ννντµ

τµ+++

τµ−−−

,,,,,e

,,,,,e

e

e

antiparticles

particles

Classification of particles (& antiparticles)

Leptons: truly elementary particles—have no structure or size. They interact only through the weak & electromagnetic forces. There are six leptons:

ννντµ

ννντµ

τµ+++

τµ−−−

,,,,,e

,,,,,e

e

e

antiparticles

particles

Hadrons: have size and structure. They interact primarily through the strong force, but electromagnetic force is the next most important one. Two types: baryons and mesons

Classification of particles (& antiparticles)

Leptons: truly elementary particles—have no structure or size. They interact only through the weak & electromagnetic forces. There are six leptons:

ννντµ

ννντµ

τµ+++

τµ−−−

,,,,,e

,,,,,e

e

e

antiparticles

particles

Hadrons: have size and structure. They interact primarily through the strong force, but electromagnetic force is the next most important one. Two types: baryons and mesons

ππ

ππ

−

+

...,

...,

...n,p...n,p

0

0particles

antiparticles

Classification of particles (& antiparticles)

Leptons: truly elementary particles—have no structure or size. They interact only through the weak & electromagnetic forces. There are six leptons:

ννντµ

ννντµ

τµ+++

τµ−−−

,,,,,e

,,,,,e

e

e

antiparticles

particles

Hadrons: have size and structure. They interact primarily through the strong force, but electromagnetic force is the next most important one. Two types: baryons and mesons

ππ

ππ

−

+

...,

...,

...n,p...n,p

0

0particles

antiparticles

charge, baryon number (B), lepton number(L) & strangeness(S)

Conservation laws: (empirical) Which reactions are allowed?

0np π+→+γ

charge, baryon number (B), lepton number(L) & strangeness(S)

Conservation laws: (empirical) Which reactions are allowed?

0np π+→+γ charge not conserved

charge, baryon number (B), lepton number(L) & strangeness(S)

Conservation laws: (empirical) Which reactions are allowed?

0np π+→+γ charge not conserved

B = +1 B = +1 + 0 = 1 P

charge, baryon number (B), lepton number(L) & strangeness(S)

Conservation laws: (empirical) Which reactions are allowed?

0np π+→+γ charge not conserved

B = +1 B = +1 + 0 = 1 P

L = 0 L = 0 P

charge, baryon number (B), lepton number(L) & strangeness(S)

Conservation laws: (empirical) Which reactions are allowed?

0np π+→+γ charge not conserved

B = +1 B = +1 + 0 = 1 P

L = 0 L = 0 P

pppppp +++→+

charge, baryon number (B), lepton number(L) & strangeness(S)

Conservation laws: (empirical) Which reactions are allowed?

0np π+→+γ charge not conserved

B = +1 B = +1 + 0 = 1 P

L = 0 L = 0 P

pppppp +++→+ charge conserved P

charge, baryon number (B), lepton number(L) & strangeness(S)

Conservation laws: (empirical) Which reactions are allowed?

0np π+→+γ charge not conserved

B = +1 B = +1 + 0 = 1 P

L = 0 L = 0 P

pppppp +++→+ charge conserved P

B = +1 + 1 = +2 B = +1 + 1 – 1 +1 = +2 P

charge, baryon number (B), lepton number(L) & strangeness(S)

Conservation laws: (empirical) Which reactions are allowed?

0np π+→+γ charge not conserved

B = +1 B = +1 + 0 = 1 P

L = 0 L = 0 P

pppppp +++→+ charge conserved P

B = +1 + 1 = +2 B = +1 + 1 – 1 +1 = +2 P

L = 0 L = 0 P

charge, baryon number (B), lepton number(L) & strangeness(S)

Conservation laws: (empirical) Which reactions are allowed?

Must conserve all three lepton numbers: Le, Lµ, and Lτ

eepn ν++→ −

Must conserve all three lepton numbers: Le, Lµ, and Lτ

eepn ν++→ − charge conserved P

Must conserve all three lepton numbers: Le, Lµ, and Lτ

eepn ν++→ − charge conserved P

B = +1 B = +1 P

Must conserve all three lepton numbers: Le, Lµ, and Lτ

eepn ν++→ − charge conserved P

B = +1 B = +1 P

Le = 0 Le = +1 – 1 = 0 P

Must conserve all three lepton numbers: Le, Lµ, and Lτ

eepn ν++→ − charge conserved P

B = +1 B = +1 P

Le = 0 Le = +1 – 1 = 0 P

Lµ = Lτ = 0 Lµ = Lτ = 0 P

Must conserve all three lepton numbers: Le, Lµ, and Lτ

eepn ν++→ − charge conserved P

B = +1 B = +1 P

Le = 0 Le = +1 – 1 = 0 P

Lµ = Lτ = 0 Lµ = Lτ = 0 P

µ−− ν+ν+→µ ee

Must conserve all three lepton numbers: Le, Lµ, and Lτ

eepn ν++→ − charge conserved P

B = +1 B = +1 P

Le = 0 Le = +1 – 1 = 0 P

Lµ = Lτ = 0 Lµ = Lτ = 0 P

µ−− ν+ν+→µ ee charge conserved P

Must conserve all three lepton numbers: Le, Lµ, and Lτ

eepn ν++→ − charge conserved P

B = +1 B = +1 P

Le = 0 Le = +1 – 1 = 0 P

Lµ = Lτ = 0 Lµ = Lτ = 0 P

µ−− ν+ν+→µ ee charge conserved P

B = 0 B = 0 P

Must conserve all three lepton numbers: Le, Lµ, and Lτ

eepn ν++→ − charge conserved P

B = +1 B = +1 P

Le = 0 Le = +1 – 1 = 0 P

Lµ = Lτ = 0 Lµ = Lτ = 0 P

µ−− ν+ν+→µ ee charge conserved P

B = 0 B = 0 PLe = 0 Le = +1 – 1 = 0 P

Must conserve all three lepton numbers: Le, Lµ, and Lτ

eepn ν++→ − charge conserved P

B = +1 B = +1 P

Le = 0 Le = +1 – 1 = 0 P

Lµ = Lτ = 0 Lµ = Lτ = 0 P

µ−− ν+ν+→µ ee charge conserved P

B = 0 B = 0 PLe = 0 Le = +1 – 1 = 0 P

Lµ = +1 Lµ = +1 P