0100020003000 energy consumption fossil fuel contribution to global energy demand year
TRANSCRIPT
0 1000 2000 3000
En
ergy
Con
sum
pti
on
Fossil Fuel Contribution to Global Energy Demand
Year
The constraints of limited/vanishing fossils fuels in the face of an exploding population
…together with undeveloped or under-developed new technologies
The constraints of limited/vanishing fossils fuels
Nuclear
will renew interest in nuclear power
)()2/(
)1(
2
3/13/2
AA
ZAa
AZZaAaAaB
sym
csv
volume termgrows as
nuclei added
surface termcorrects for
surface nuclei
Coulomb termrepulsion due
to protons
symmetry termdrives number of
protons ≈ neutrons
pairing energy termincreased stabilityfor paired nuclei
The fusion of 2 protons actually involves positron decay which converts one of them into a neutron.
eHHH 211 Q=0.42 MeV
The sun’s longevity is due to the fact that the accompanying positron identifies this as a weak, i.e., rare process. It is only slowly consuming its hydrogen. Since this allowed the evolution of life on earth - the weakness of the weak interaction is of significant to us!
The fusing of 2 protons might seem to exemplify the basic process,
but there is no 2-proton state!
This is, however, the first step of hydrogen burning in the sun.
t triton nucleus of tritium 3H
d deuteron nucleus of deuterium 2H
alpha nucleus of helium 4He
The lightest nuclides
And their binding energies
NuclideNumberof bonds Total
Binding energy(MeV)per particle per bond
2H3H4He
136
2.2 8.528.0
1.1 2.8 7.0
2.2 2.8 4.7
The Deuteron, 2H
Charge +1.610-19 C = 1e
mass 1875.5803 MeV = 2.013553 u
spin 1
magnetic moment +0.85742N
total angular momentum, J 1
Does not exist in an excited state ( of higher ℓ ).
What does this tell us about the nuclear force?
What does this tell us about the orbital angular momentum?
Odd – Odd Isotopes
Isotope Spin, I
Na
Na
Na
N
N
B
B
Li
H
24
11
22
11
20
11
16
7
14
7
10
5
8
5
6
3
2
1
4
3
2
2
1
3
2
1
1
The total wave function describing 2 identical fermionsmust be
anti-symmetric to particle exchange.
)()()()(2121
2
1rrrr
ABBA
)()( Sr
)(),()(
SrRm
YSpace Part
Spin Part
Any part will contribute an independent symmetry of its own.
B
A
B
r1
r2
Particle Exchange:
A
r2
r1
Parity Operation:
AB
r1r2
B
A
B
r1
r2
Particle Exchange:
A
r2
r1
Parity Operation:
AB
r1r2
r A
B
r1
r2
r 1
r 2
r
r
r 1r 2
r 1r 2
The Spherical Harmonics Y ,ℓ m(,)
ℓ = 0
ℓ = 1
ℓ = 2
ℓ = 34
100 Y
ieY sin
8
311
cos4
310
Y
ieY 2
2
sin2
15
4
122
ieY cossin
8
1521
2
12cos2
3
4
1520
Y
ieY 3
3
sin4
35
4
133
ieY 2
cos2
sin2
105
4
132
ieY 12cos5sin4
21
4
131
cos
2
33cos2
5
4
730Y
The total wave function describing 2 identical fermionsmust be
anti-symmetric to particle exchange.
The space part of an ℓ = 0 state has no change in sign when particle positions
are exchanged!
so the spin part of this wavefunction must beanti-symmetric.
Which would allow only a total spin of 0.
41
00Y
cos
4
310Y
sin8
311
ieY
1cos34
3 2
20
Y
cossin8
1521
ieY
HnH 2
1
1
0
1
1 Q=2.23 MeV
The simplest fusion event to consider: forming a deuteron from a proton and neutron.
but doesn’t exemplify the typical fusion reaction very well:
eHHH 211 Q=0.42 MeV
Since
is suppressed by its (weak) cross section
In the fusion of light elements into medium nuclei the coulomb barrier has to be penetrated from the outside.
In fission an elongated neck develops, prolonging scission. In fusion approaching nuclei are flattened by coulomb repulsion, delaying contact. Because of this repulsion between nuclei, fusion is not a
naturally occurring process on earth. In order to bring nuclei close enough together for fusion to occur requires kinetic energy.
2
4
21
2
11
2
1HeHH Q=23.8 MeV
but this energy release is larger than that required to remove a proton or neutron from 4He!
eHHH1
2
10
1
10
1
1 Q=0.42 MeV
The sun can make deuterium only through the weak (slow) process:
Once there is sufficient deuterium, could consider:
nHeHH 2
3
21
2
11
2
1Q=3.3 MeV
So there are more likely products of deuteron fusion (d-d):
pHHH 2
3
11
2
11
2
1Q=4.0 MeV
pHHH 2
3
11
2
11
2
1 Q=4.0 MeV
liberating ~1 MeV/nucleon
can be performed in the laboratory with a beam of deuterons on a deuterium target.
Assuming R1.5 fm,the electrostatic potential:
R
e
2
2
041
JmC
mN C 14
2
1067.7)105.1(2
1099.8
15
19
2
29 )1060.1(
0.5 MeV.
A (typical) A beam (even if every deuteron struck a deuteriumnucleus and fused) would produce the net rate of energy:
MeVeC sec 5.3)/1)(/10( 6
WVsecC 5.3))(/5.4(
Could a hot deuterium gas have enough thermal kinetic energy to overcome the coulomb barrier and allow nuclei to fuse?
With 0.25 MeV thermal KE, 2 deuterium atoms meeting in a head-on collision would have the 0.5 MeV KE needed.
1 c of heavy water D2O has the potential of liberating 51012 Jeven if extracted over the course of 24 hours
50 MW!
Ordinary water ~0.015% D2O Fusion energy available in 1 liter, ordinary water
chemical energy available in 300 liters of gasoline=
What temperature would provide a mean kinetic energy of 0.5 MeV?
MeVTeV/K.kT 5.0)106178( 5
2
3
2
3
KT 9109.3
By comparison, the temperature of the surface
of the sun 6000 K.
nHeHH 2
4
22
3
11
2
1Q=17.6 MeV
A fusion reaction involving the heavier hydrogen isotope, tritium is
This deuterium-tritium or D-T reaction is currentlythe reaction of choice in fusion reactor designs.
In the center of momentum frame, the 4He and n share the final energy with equal but opposite momenta:
nHem
p
m
pGeV
226.17
2
4
2
nHe
He
n
n mm
mQ
m
pT
4
4
2
2
Tn=14.1 MeV
From the mass difference: this should release 26.7 MeV of energy.
Energy production in stars
Stars form out of gas clouds which collapse under mutual gravitational attraction. The gravitational potential energy is converted into kinetic energy of the gas molecules - the gas heats up (and its density increases).
About 99.9% of all nuclei in the Universe are hydrogen or helium.
1H is dominant, accounting for 92.5% of the nuclei. Only 7.4% are 2He.
He nuclei are ~4 more massive than hydrogen. So by mass of the Universe is 24% helium.
In stars the 1st fusion processes which can occur must involving protons.
The first step in the process is of turning 4 protons into an .