nuclear reactions the chain reaction
TRANSCRIPT
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Nuclear reactionsThe chain reaction
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Nuclear reactions
For power applications, want a self-sustained chain reaction.
The chain reaction
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Natural U: 0.7% of 235U and 99.3% of 238U
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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.
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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.
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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.
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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.
![Page 8: Nuclear reactions The chain reaction](https://reader031.vdocument.in/reader031/viewer/2022021208/6206340d8c2f7b173005635d/html5/thumbnails/8.jpg)
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.
![Page 9: Nuclear reactions The chain reaction](https://reader031.vdocument.in/reader031/viewer/2022021208/6206340d8c2f7b173005635d/html5/thumbnails/9.jpg)
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.
![Page 10: Nuclear reactions The chain reaction](https://reader031.vdocument.in/reader031/viewer/2022021208/6206340d8c2f7b173005635d/html5/thumbnails/10.jpg)
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.
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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)
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Nuclear fusion (the sun’s source of power)
Mostly fusion of protons in the core of the sun
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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
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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.
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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
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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
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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
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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
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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
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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++→ +
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( ) e42
11 2e2HeH4 ν++→ +
The two e+s will annihilate with two electrons
γ→+ −+ 2ee
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( ) e42
11 2e2HeH4 ν++→ +
The two e+s will annihilate with two electrons
γ→+ −+ 2eeSo overall reaction is
( ) s'2Hee2H4 e42
11 γ+ν+→+ −
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( ) 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)
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( ) 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.
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( ) 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.
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( ) 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
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The binding energy curve
EB
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The binding energy curve
Fusing light nuclei (low EB) produces nuclei with larger EB. Thus energy left over.E
B
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The binding energy curve
Fusing light nuclei (low EB) produces nuclei with larger EB. Thus energy left over.E
B
Fission
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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.
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Fundamental forces in nature
Let’s look at the electromagnetic interaction using the photon-mediating-particle concept.
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∆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
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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:
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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
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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
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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
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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
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charge, baryon number (B), lepton number(L) & strangeness(S)
Conservation laws: (empirical) Which reactions are allowed?
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0np π+→+γ
charge, baryon number (B), lepton number(L) & strangeness(S)
Conservation laws: (empirical) Which reactions are allowed?
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0np π+→+γ charge not conserved
charge, baryon number (B), lepton number(L) & strangeness(S)
Conservation laws: (empirical) Which reactions are allowed?
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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?
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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?
![Page 43: Nuclear reactions The chain reaction](https://reader031.vdocument.in/reader031/viewer/2022021208/6206340d8c2f7b173005635d/html5/thumbnails/43.jpg)
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?
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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?
![Page 45: Nuclear reactions The chain reaction](https://reader031.vdocument.in/reader031/viewer/2022021208/6206340d8c2f7b173005635d/html5/thumbnails/45.jpg)
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?
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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?
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Must conserve all three lepton numbers: Le, Lµ, and Lτ
eepn ν++→ −
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Must conserve all three lepton numbers: Le, Lµ, and Lτ
eepn ν++→ − charge conserved P
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Must conserve all three lepton numbers: Le, Lµ, and Lτ
eepn ν++→ − charge conserved P
B = +1 B = +1 P
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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
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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
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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
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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
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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
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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
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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
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