Nuclear Fusion and Radiation
Lecture 3 (Meetings 5 & 6)
Eugenio Schuster
[email protected]
Mechanical Engineering and Mechanics
Lehigh University
Nuclear Fusion and Radiation – p. 1/39
Fusion Reactions
It was previously noted (lecture 1) that for the D − T fusion
reaction to take place a relative kinetic energy of
∼ 0.36M eV would be required to overcome the Coulomb
barrier. However, experimental observations indicate that
substantial fusion of D − T takes place at energies well
below the barrier height. Classical mechanics cannot
explain these observations and we must turn to quantum
mechanics to understand what is happening. Consider
particles approaching a rectangular potential barrier:
Figure 1: Potential barrier
Nuclear Fusion and Radiation – p. 2/39
Fusion Reactions
According to classical mechanics the incident particles
would be totally reflected. However, according to quantum
mechanics there can be partial reflection and partial
transmission of the incident beam.
Figure 2: Potential barrier
Using a graphical wave representation:
Figure 3: Potential barrier
Nuclear Fusion and Radiation – p. 3/39
Fusion Reactions
Calculation of the transmission probability is complicated
and will not be considered here. It can be shown that the
probability for transmission, PT , has the dependencies,
PT = f (E, U, a)
The theory predicts significant transmission at energies well
below the barrier height.
The probability for a given fusion reaction to take place
depends on the probability for transmission (with formation
of a compound nucleus) and the probability for the given
exit channel to occur.
Nuclear Fusion and Radiation – p. 4/39
Fusion Reactions
Figure 4: Head-on nuclear fusion reaction (HSMK)
Nuclear Fusion and Radiation – p. 5/39
Fusion Reactions
For example,
2
1D
+31 T → (52 He)∗
exhibits two exit channels:
Nuclear Elastic Scattering:
(52 He)∗ →21 D +31 T
Fusion:
(52 He)∗ →42 He +10 n
The combined probabilities for compound nucleus
formation and de-excitation by a given exit channel
represents the reaction probability.
Nuclear Fusion and Radiation – p. 6/39
Cross Sections
Historically reaction probabilities have been designated as
reaction “cross sections.”
We first give a functional definition for the cross section σ .
Consider a test particle of species 1 (e.g., D) moving in a
medium of target particles of species 2 (e.g., T ). For this
case we make the following definition:
Reaction Probability per Test Particle per Unit Length Traveled in the Medium
≡ σn2
where σ is the cross section for reaction, and n2 is the
number density of the medium. Note that σ has dimensions
of l2 (e.g., cm2 , m2 ) and n2 has dimensions of #/l3 (e.g.,
#/cm3 , #/m3 ), and σn2 has units of (#)/l or l−1 , that is,
probability/unit length traveled by the test particle in the
medium.
Nuclear Fusion and Radiation – p. 7/39
Cross Sections
We now give a geometrical interpretation for the cross
section.
Figure 5: Cross section geometrical interpretation
Nuclear Fusion and Radiation – p. 8/39
Cross Sections
If each target particle has “cross sectional area,” σ , then
n2 l 3
P = σ 2 = σn2 l
l
and the probability per unit length traveled by the test
particle in the medium is
P
= σn2
l
just as we had before.
If we have n1 test particles per unit volume of the medium
(in addition to n2 target particles per unit volume) and if the
velocity of the test particles (with respect to the target
particles) is v , then the total path length traveled by the n1
particles per sec per unit volume is n1 v .
Nuclear Fusion and Radiation – p. 9/39
Cross Sections
Thus, the reaction rate, the number of reactions between
particles 1 and 2 per unit volume of medium per sec is
Reaction Rate
= (σn2 )(n1 v)
σn2 = the reaction probability per unit length traveled by
one particle.
n1 v = the total path length traveled by all particles of
type 1 per unit volume per sec.
In the CGS system the reaction rate, R, has the following
units:
cm
1
#
1
2
=
R = (σn2 )(n1 v) = (cm )
3
3
cm
cm
sec
cm3 sec
Note that the quantity n1 v is designated the “flux” φ in
neutron transport calculations.
Nuclear Fusion and Radiation – p. 10/39
Cross Sections
The determination of σ as a function of E is the work of
nuclear physicists. Experimental values of cross sections
are needed in general and theory only provides a guide.
Sometimes we can get an order of magnitude estimate for a
cross section by use of the geometric argument σ ∼ πR2 .
For example, consider the D − T reaction for which
RDT ∼ 4 × 10−13 cm and, thus,
σ ∼ πR2 ≈ 0.5 × 10−24 cm2
This, of course, is NOT an accurate cross section value but
indicates the order of magnitude of nuclear cross sections,
10−24 cm2 . For this reason this dimension has been given a
name as follows
10−24 cm2 ≡ 1barn
Nuclear Fusion and Radiation – p. 11/39
Cross Sections
Consider first the fusion reaction
1
1H
+11 H →21 D + β + + ν
(Q = 1.44M eV )
The cross section for this reaction is too small to be
observed in the laboratory. However, the cross section can
be inferred from stellar processes and can be estimated
based on theory (∼ 10−9 b). The two reactions
2
1
H
+
1D
1
3
1
H
+
1T
1
→
→
3
2 He + γ
4
2 He + γ
(Q = 5.5M eV )
(Q = 19.8M eV )
have small, but observed cross sections (∼ 10−6 b). These
cross sections are too small to be of interest for fusion
power.
Nuclear Fusion and Radiation – p. 12/39
Cross Sections
Next consider the reactions
2
2
D
+
1D
1
2
2
D
+
1D
1
3
2
D
+
1T
1
3
2
D
+
2 He
1
→
→
→
→
1
3
He
+
0n
2
1
3
T
+
1H
1
1
4
He
+
0n
2
1
4
He
+
1H
2
(Q = 3.3M eV )
(Q = 4.03M eV )
(Q = 17.6M eV )
(Q = 18.3M eV )
The cross section versus deuteron energy for these
reactions is given in Fig. 6. The following points are noted:
σDT is by far the largest cross section.
σDT is significant well below 0.36M eV (360keV ), the
approximate Coulomb barrier height.
σDT at ∼ 110keV is ∼ 5b, which is approximately an
order of magnitude larger than πR2 .
Nuclear Fusion and Radiation – p. 13/39
Cross Sections
Figure 6: D − D, D − T and D − He3 cross sections.
Nuclear Fusion and Radiation – p. 14/39
Fueling Strategy
As we shall see later in the course, under the proper
conditions some of the fusion reactions listed above have
large enough cross sections to yield reaction rates which
are interesting for fusion power production. However, we
must first address a more fundamental question:
“Are there sufficient reserves of fusion fuels for a fusion energy
economy?”
Let us examine this question in terms of a D − T based
fusion power economy. What are the “fuel” requirements?
Deuterium is a stable isotope and can be recovered
from water ( 0.015 atom %) relatively easily.
Tritium, on the other hand, is radioactive decaying by
beta decay to helium-3,
3T
1
β
−→ 32 He, T1/2 ∼ 12.4 years.
Nuclear Fusion and Radiation – p. 15/39
Fueling Strategy
Because of the short half-life of T , there are not sufficient
amounts of T in nature to sustain a D − T fusion power
economy. Therefore, T for D − T fusion power must be
manmade. Fortunately this appears possible using the
following strategy:
1
4
3
2
He
+
T
→
D
+
0n
2
1
1
6
1
n
+
3 Li(7.5%)
0
7
1
n
+
3 Li(92.5%)
0
→
→
3
4
He
+
1T
2
3
4
1
He
+
n
+
1
2
0
T
As we shall show later, we can design a fusion reactor so
that it will breed the tritium it consumes by using the D − T
generated neutrons and lithium. Thus, the fuel requirement
for T becomes a resource requirement for lithium. Lithium
and deuterium resources are sufficient for 1000′ s of years of
energy production via D − T fusion reactors.
Nuclear Fusion and Radiation – p. 16/39
Other Energy Units
In examining energy resources it is useful to introduce a
quantity of energy called the terawatt–yr (T W − Y R)
defined as
1T W −Y R ≡ 1012 (J/sec)(365)(3600)(24)(sec/yr) ∼ 3.15×1019 J
Consider the following energy consumption points:
1975 : US consumed ∼ 2.6T W − Y R
WORLD consumed ∼ 8.5T W − Y R
2025 : US consumption ∼ 4.5T W − Y R
WORLD consumption ∼ 27T W − Y R
Nuclear Fusion and Radiation – p. 17/39
Other Energy Units
List of countries by electricity consumption
http://en.wikipedia.org/wiki/List_of_countries_by_electricity_consumption
List of countries by total primary energy consumption
http://en.wikipedia.org/wiki/List_of_countries_by_total_primary_energy_consumption_and_production
Energy in the United States
http://en.wikipedia.org/wiki/Energy_in_the_United_States
Nuclear Fusion and Radiation – p. 18/39
Fusion Gain in Ideal Reactor
So far everything looks promising for a D − T fusion reactor
so let us try to construct an “idealized” fusion reactor.
Consider the following system:
Figure 7: Idealized fusion reactor.
Nuclear Fusion and Radiation – p. 19/39
Fusion Gain in Ideal Reactor
The probability for a D − T fusion reaction per cm of D+
path length in the target, P/l, is given by
0.2g/cm3
P/l =
(6.02 × 1023 #/mol)(5 × 10−24 cm2 ) = 0.2#/cm
3g/mol
On this basis we can easily make the target thick enough to
achieve a reaction probability approaching unity. Let us
define an energy gain, Go , for our system:
Fusion Energy/Incident D 17.6M eV /D-T Fusion
=
= 176
Go =
Energy of Incident D
0.1M eV /Incident D
THIS IS A VERY HIGH GAIN. THE “REACTOR” LOOKS
VERY ATTRACTIVE. IS THERE ANYTHING WRONG
WITH THIS ANALYSIS?
Nuclear Fusion and Radiation – p. 20/39
Competing Processes
We neglected the effect of “Coulomb scattering” in our
calculation of Go .
Coulomb or Rutherford scattering (elastic scattering
between charged particles) is the primary source of
competition for fusion.
In Coulomb scattering the incident D+ looses energy
and because the fusion cross section decreases as the
incident particle energy decreases, the probability for
fusion decreases.
If the Coulomb scattering cross section is large with
respect to the fusion cross section, the D+ will be
“down-scattered” in energy before it can engage in
significant fusion reaction.
Nuclear Fusion and Radiation – p. 21/39
Coulomb Scattering Cross Sections
The detailed derivation of the Coulomb scattering cross
section can be found in the textbook (Principles of Fusion
Energy, Chapter 3). We shall use some simple arguments
here to estimate σCS .
We would expect significant Coulomb scattering when two
nuclei “feel” the presence of the Coulomb potential relative
to their mutual kinetic energy. Consider the following
schematic representation.
Nuclear Fusion and Radiation – p. 22/39
Coulomb Scattering Cross Sections
Figure 8: Coulomb potential.
Nuclear Fusion and Radiation – p. 23/39
Coulomb Scattering Cross Sections
If the kinetic energy of particle 2 corresponds to point a,
particle 2 will not “feel” the Coulomb potential, that is, since
Ta >> Ec (R) the Coulomb potential is not strong enough to
deflect (scatter) particle 2 significantly.
On the other hand, if the kinetic energy of particle 2
corresponds to point b, Tb ≈ EC (R) and particle 2 will “feel”
the Coulomb potential. In this case, particle 2 will be
deflected significantly.
Based on these arguments we can define the effective
“range,” Ref f , of the Coulomb potential of particle 1 with
respect to particle 2 by the following condition:
E1,2 (Ref f ) = T1,2
Z1 Z2
⇐⇒
= T1,2
4πǫo Ref f
Nuclear Fusion and Radiation – p. 24/39
Coulomb Scattering Cross Sections
Taking Z1 = Z2 = 1.6 × 10−19 C and ǫo = 8.85 × 10−12 F/m, we
obtain
1.44 × 10−12
2.3 × 10−28
m=
m
Ref f =
T (J)
T (keV )
If we employ a geometric interpretation for σCS :
σCS ∼ π
Ref f
2
2
π (1.44 × 10−12 )2 2 1.6 × 104
m = 2
barns
=
2
4
T (keV )
T (keV )
The above expression gives a very good estimate for large
angle (significant deflection) Coulomb scattering events
when T is the “center of mass energy” of the two particles:
1
1 M1 M2
2
V2
T ≡ Mo V =
2
2 M1 + M2
(V ≡ relative velocity)
Nuclear Fusion and Radiation – p. 25/39
Coulomb Scattering Cross Sections
It is emphasized that this expression is only an estimate. lt
does, however, give reasonable values for large angle
scattering and it gives the correct energy dependence,
σCS ∼ 1/T 2 , for all Coulomb scattering events (large and
small angle).
Nuclear Fusion and Radiation – p. 26/39
Energetic Deuterons on “Cold” Targets
Consider the following situation:
Figure 9: Energetic vs. cold deuteron collision
Nuclear Fusion and Radiation – p. 27/39
Energetic Deuterons on “Cold” Targets
For this case
2
MD
MD MD
MD
Mo =
=
=
MD + MD
2MD
2
1 MD 2
1
1
2
VD = T D
T ≡ Mo V =
2
2 2
2
Thus,
D−D(cold)
σCS
1.6 × 104
6.4 × 104
=
b
b=
2
TD 2
TD
( 2 )
For our idealized reactor, TD = 100keV ,
D−D(cold)
σCS
6.4 × 104
=
b = 6.4b
2
100
Note that σfD−D ≈ 0.03b at 100keV . Therefore, σCS /σf ≈ 200.
Nuclear Fusion and Radiation – p. 28/39
Energetic Deuterons on “Cold” Targets
Let us now consider D on cold T . For this case,
2 /2
3MD
MD MT
MD (3MD /2)
3MD
Mo =
=
=
=
MD + MT
MD + (3MD /2)
5MD /2
5
1
1 3MD 2
3
2
T ≡ Mo V =
VD = T D
2
2 5
5
Thus,
D−T (cold)
σCS
4.4 × 104
1.6 × 104
b=
= 3TD
b
2
TD
( 5 )2
For our idealized reactor, TD = 100keV ,
4
4.4
×
10
D−T (cold)
b = 4.4b
σCS
=
2
100
Note that σfD−T ≈ 5b at 100keV . Therefore, σCS /σf ≈ 1.
Nuclear Fusion and Radiation – p. 29/39
Energetic Deuterons on “Cold” Targets
We will now estimate the energy gain of our idealized fusion
reactor when D − T Coulomb scattering is taken into
account. Consider the following:
Figure 10: Incident D beam on T target
Nuclear Fusion and Radiation – p. 30/39
Energetic Deuterons on “Cold” Targets
As the beam penetrates into the target, beam particles can
engage in two types of interactions with target particles:
(1) D − T fusion
(2) D − T Coulomb scattering
If fusion takes place, a deuteron is removed from the
incident beam.
Figure 11: D − T fusion reaction
Nuclear Fusion and Radiation – p. 31/39
Energetic Deuterons on “Cold” Targets
If scattering takes place, the scattered deuteron is still
available for fusion or additional scattering.
Figure 12: D − T scattering
Nuclear Fusion and Radiation – p. 32/39
Energetic Deuterons on “Cold” Targets
After Coulomb scattering, a beam particle has been
degraded in energy and:
(1) the probability for fusion has decreased
(2) the probability for Coulomb scattering has increased
Figure 13: Fusion vs. scattering probabilities
Nuclear Fusion and Radiation – p. 33/39
Energetic Deuterons on “Cold” Targets
The following table gives specific values
E(keV ) σf (b) σCS (b) σf /σCS
100
5
4.4
1.13
50
1.5
17.6
0.085
0.15
70.4
0.002
25
Thus, as the beam penetrates into the target, the beam
energy is degraded and the probability for fusion decreases
dramatically.
Let us try to estimate the effect of beam degradation on the
fusion energy gain, G.
Nuclear Fusion and Radiation – p. 34/39
Energetic Deuterons on “Cold” Targets
Figure 14: Beam degradation
The number of fusion reactions at Eo in dx about x is given
by (we neglect fusion reactions at ED < EO ):
#
R(x)dx = I(x)σf ndx
[I] = [nv] =
cm2 s
The total number of fusion reactions in the target per cm2 s
is given by
Z t
R(x)dx
0
Nuclear Fusion and Radiation – p. 35/39
Energetic Deuterons on “Cold” Targets
The fusion energy gain, G, is thus,
Rt
0 QR(x)dx
G=
Io Eo
Now,
Z
t
R(x)dx =
0
Z
0
t
I(x)σf ndx = σf n
Z
t
I(x)dx
0
To calculate I(x) consider the reduction in beam intensity,
dI , in dx about x,
dI = −I(x)σf ndx − I(x)σCS ndx
Nuclear Fusion and Radiation – p. 36/39
Energetic Deuterons on “Cold” Targets
Figure 15: Beam degradation
The Coulomb scattering term assumes that each scattering
event degrades the incident deuteron energy below the
point at which it could engage in a fusion reaction. A
“better” assumption would be:
σCS
dI = −I(x)σf ndx − I(x)
ndx
s
where s is the number of Coulomb scattering collisions
required for “substantial” degradation of ED .
Nuclear Fusion and Radiation – p. 37/39
Energetic Deuterons on “Cold” Targets
Thus,
dI
σCS dI
1
= −I(x)n σf +
⇐⇒ I(x) = −
dx
s
dx n σf +
Now,
Z t
0
R(x)dx = σf n
Z
0
t
σCS
s
Z t
σf n
dI
dx
−
I(x)dx =
σCS
dx
n σf + s
0
lf we take the thickness, t, large enough to totally attenuate
Io ,
Z t
σf Io
R(x)dx =
σCS
σf + s
0
Nuclear Fusion and Radiation – p. 38/39
Energetic Deuterons on “Cold” Targets
Thus,
σf Io
σf Q
Q
G=
σCS I E =
σCS
σf + s
E o σf + s
o o
Now we return to the D on T (cold) case and make the
most pessimistic assumption about s, i.e., s = 1. Then,
5
17.6
G=
= 0.53Go ≈ 93
4.4 0.1
5+ 1
Thus, D − T scattering degrades G by only about a factor of
2 and the resulting gain is still substantial.
WHAT MUST BE CONSIDERED IN ADDITION TO D − T
(COLD) COULOMB SCATTERING?
Nuclear Fusion and Radiation – p. 39/39