Basdevant J.-L., Rich J., Spiro M. Fundamentals in Nuclear Physics: From Nuclear Structure to Cosmology

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6.9 The Oklo prehistoric nuclear reactor 323

In 100 kg of initial fuel, i.e. 3 kg of

235

U and 97 kg of

238

U, there remains,

after three years of running:

• 0.9kg of

235

U (2 kg have ﬁssioned)

• 97 kg of

238

U (2 kg have been transformed into

239

Pu)

• 0.6kg of

239

Pu (in the above 2 kg, 1 kg ﬁssioned)

• 1.7 kg of ﬁssion products.

It is necessary to renew the fuel elements periodically. The reactor is

stopped each year for 3 or 4 weeks. One third or one quarter of the fuel is

renewed.

The fuel re-processing . The re-processing consists of separating, in the

irradiated fuel, the uranium and the plutonium (which can be used again)

from the ﬁssion products (which are essentially useless and are very radioac-

tive, as shown in Fig. 6.15). The separation is basically a chemical process,

using various solvents.

The storage of ﬁssion products. The volume of liquid containing the ﬁs-

sion products generated in one year by a 900 MW reactor is roughly 20 m

The liquid is ﬁrst stored in special vessels refrigerated by water in order to

evacuate the residual heat. After a few years, it is possible to evaporate the

solvents and to vitrify the waste. The vitriﬁed blocks can be stored under-

ground in ventilated areas.

After several years, the problem of long term storage of the radioactive

vitriﬁed waste arises. Several methods are envisaged, in particular deep un-

derground storage in salt or granite.

Accelerators coupled to ﬁssile matter could have an interesting application

in that they can be designed to destroy long lifetime radioactive waste. Un-

der neutron irradiation, plutonium, trans-plutonium elements such as Ameri-

cium and Curium, and other dangerous long-lived ﬁssion products such as

Technetium can absorb neutrons and either be transformed to short-lived or

stable isotopes or ﬁssion into lighter elements. (as a general rule, the lighter

the elements, the less dangerous they are).

6.9 The Oklo prehistoric nuclear reactor

It is interesting to note that if civilization had taken 2.5 billion years to de-

velop on Earth rather than 4.5 billion years, the use of nuclear reactors would

have been much simpler because at that time the abundance of

235

Uwassuf-

ﬁciently high that enrichment would have been unnecessary. In fact, on at

least one occasion, the necessary conditions for stable reactor operation were

united apparently without intelligent intervention. These conditions include

a suﬃciently high concentration of uranium (> 10% by weight), a suﬃciently

low concentration of nuclei with high neutron-absorption cross-sections, and

suﬃcient water (> 50% by weight) to serve as the moderator.

324 6. Fission

−4

−2

10 10

Years after processing

137mBa

90Y

137Cs

243Am

239Pu

99Tc

213Po

214Bi

222Rn = 210Po

218Po

229Th

thermal power (W)

original U

Fig. 6.15. Radioactivity of the ﬁssion products and trans-uranium elements in fuel

that has produced 100 Mw-yr of electrical energy [65]. It is assumed that 99.5%

of the uranium and plutonium was removed for reprocessing. For comparison, also

shown is the activity of the original uranium.

These conditions were present in a uranium-ore deposit in Oklo, Gabon

[67]. As illustrated in Fig. 6.16, the uranium ore in this deposit is depleted

235

U (as little as 0.42% instead of 0.72%) and enriched in ﬁssion products.

It is believed that the reactor operated ∼ 1.8 × 10

years ago for a period of

∼ 10

yr.

One interesting result of studies of the Oklo reactor is a very stringent

limit on the time variation of the fundamental constants [68], even stronger

than those derived from the study of

187

Re decay in Sect. 5.5.2. The limit

comes from the observation that the nuclide

149

Sm has an Oklo abundance

that is typical of reactor wastes, i.e. about 40 times less than the natural

isotopic abundance of 13.8%. This low abundance comes about because of a

resonance for thermal neutron capture (Fig. 6.17) that transforms

149

Sm to

150

Sm in the high neutron ﬂux.

6.9 The Oklo prehistoric nuclear reactor 325

0.72

0.62

0.52

0.42

U−235 / U−238 (%)

U−238/ all (% by weight)

1.5 2.0

transverse postion (meters)

142

143

144

145 146 148

150

Oklo

Terrestrial

isotopic abundance

Fig. 6.16. The composition of a uranium deposit in Oklo, Gabon [67]. The deposit

contains several layers, each about 1 meter thick, containing very rich ore, about

50% uranium by weight. The top panel shows the uranium abundance proﬁle across

the layer (dashed line). The solid line shows that

235

U is highly depleted in the

layer, as little as 0.42% compared to the normal 0.72%. The bottom panel shows

the abundances of Nd isotopes. The ﬁssion products

143

Nd −

150

Nd all have larger

than normal abundances compared to that of

142

Nd which cannot be produced by

ﬁssion (Exercise 6.5).

326 6. Fission

cross−section (barns)

neutron energy (eV)

110

2−4

−2

150

=7.987 MeV

+0.9 eV

+0.09 eV

Fig. 6.17. The radiative neutron capture cross-section on

149

Sm. The absorp-

tion resonances correspond to excited states of

150

Sm that are above the neutron-

separation energy S

=7.987 MeV. The diagram on the right shows the ground

state and the ﬁrst two states responsible for the resonances. The ﬁrst one at S

+0.09

can resonantly absorb thermal neutrons (3kT =0.078 eV for T = 300 K).

Thermal neutron-capture on nuclide (A, Z) are due to highly excited

states of (A +1,Z) that can decay either by photon emission or by emis-

sion of a neutron of energy E

∼ kT ∼ 0.02 eV. This means that, relative

to the ground state of (A +1,Z), the excited state must have an energy

E ∼ S

+0.02 eV where S

∼ 8 MeV is the neutron separation energy. The

fact that the thermal-neutron-capture resonance in

149

Sm was still operative

years ago means that the level has not changed by more than 0.02 eV

over this period. This corresponds to a relative change < 10

−8

The positions of nuclear levels depend on the values of the fundamental

constants, so this limit on the change in the level can be transformed into a

limit on changes in the constants. About 1% of the levels energy is electro-

static, so the limit of 10

−8

on the level change is conservatively interpreted

as a limit of 10

−6

on the change in the ﬁne-structure constant over the last

2 billion years [68].

6.10 Bibliography

1. O. Hahn and F. Strassman, Naturwissenschaften, 27, 11 (1939).

2. L. Meitner and O.R. Frisch, Nature, 143, 239 (1939).

3. N. Bohr and J.A. Wheeler, Phys. Rev., 56, 426, (1939).

Exercises for Chapter 6 327

4. A.M. Weinberg and E.P. Wigner, The Physical Theory of Neutron Chain

Reactors, University of Chicago Press, Chicago, Ill., 1958.

5. L. Wilets, Theories of Nuclear Fission, Clarendon Press, Oxford, 1964.

6. S. Glasstone and A. Sesonke, Nuclear Reactor Engineering, Van Nos-

trand, New York, 1967.

Exercises

6.1 Consider the typical ﬁssion process induced by thermal neutrons

235

U →

140

Xe +

Sr + ν neutrons . (6.76)

What are the values of ν and Q

ﬁs

for this reaction? Calculate the total ﬁssion

energy of 1 kg of

235

U assuming all ﬁssions proceed via this reaction.

6.2 Assume that the average time τ between production and absorption of a

neutron in a reactor is 10

−3

s. Calculate the number of free neutrons present

at any time in the core when the reactor is operating at a power level of

1GW.

6.3 A beam of neutrons of 0.1eVisincidenton1cm

of natural uranium.

The beam ﬂux is 10

neutrons s

−1

. The ﬁssion cross section of

235

Uat

that energy is 250 b. The amount of

235

Uis0.72%. The density of uranium

is 19 g cm

−3

. Each ﬁssion produces 165 MeV in the material. What is the

nuclear power produced?

6.4 Consider a nuclear plant producing an electric power of 900 MW with

thermal neutrons and enriched uranium at 3.32% in

235

U. The total yield of

nuclear energy into electric energy is R =1/3 (including the thermal yield).

The total uranium mass is 70 tons.

1. How many

235

U atoms are burnt per second?

2. What mass of

235

U is used per day?

3. Assuming the plant works at constant full power, how long can it run

before changing the fuel?

6.5 Using the Table in Appendix G, explain why

142

Nd would not be ex-

pected to be abundantly produced in a nuclear reactor, unlike the other stable

Nd isotopes.

6.6 Estimate the amount of uranium needed to create 100 Mw-yr of electrical

energy assuming a thermal-to-electricity eﬃciency of 0.3. This is the amount

328 6. Fission

of uranium considered in Fig. 6.15. In this ﬁgure, translate the thermal power

to decay rate (in Bq) by assuming ∼ 5 MeV per decay. Discuss the origin of

the nuclides shown in the ﬁgure.

7. Fusion

Fusion reactions, taking place in the Sun, have always been the main source

of energy on Earth. Even fossil fuels like coal and petroleum were fabricated

through photosynthesis and should therefore be considered as stored solar en-

ergy. The only exception is ﬁssion energy due to the heavy elements uranium

and thorium, which were synthesized during supernova explosions.

Since the ﬁrst explosive occurrence of fusion on Earth in 1952, mankind

has had the ambition to tame that form of energy. It is cleaner than ﬁssion,

creating much less long-lived radioactive waste. Its resources are unlimited

over historical time scales. In 300 liters of sea water, there is 1 g of deuterium.

Thefusionoftwo

H nuclei yields about 10 MeV ∼ 2 ×10

−12

J. Therefore the

water in the oceans could provide suﬃcient energy for human needs during

time scales of several hundred millions of years. It is particularly frustrating

to see that, unlike ﬁssion which was used industrially a few years after its

discovery, fusion is still in a prospective stage more than 50 years after its

ﬁrst terrestrial use.

The apparent diﬃculty in using fusion reactions comes from the fact that,

unlike neutron-induced ﬁssion reactions, fusion reaction involve positively

charged nuclei that, at normal temperatures and densities, are prevented

from reacting by the Coulomb barrier. The challenge of taming fusion is to

maintain a (non-explosive) plasma that is suﬃciently hot and dense to have

a useful rate of fusion.

In this chapter, we will ﬁrst catalog the possible fusion reactions with the

conclusion that the most promising reaction for terrestrial energy reaction is

deuterium-tritium fusion

dt →

He n + 17.5MeV . (7.1)

We will then calculate fusion rates in a hot plasma as a function of tem-

perature and density. In Sect. 7.2 we derive performance criteria for fusion

reactors, in particular the Lawson criterion for eﬀective energy generation.

The basic problem will be to maintain a plasma at a suﬃciently high temper-

ature and pressure for a suﬃciently long time before it cools down, mostly

through photon emission by electrons (bremsstrahlung) and atomic impu-

rities. Sections 7.3 and 7.4 will then discuss the problems of the two most

widely discussed method of maintaining the plasma, those using magnetic

conﬁnement and those using laser induced inertial implosion.

330 7. Fusion

7.1 Fusion reactions

Because the binding energy per nucleon generally increases with A for A<50,

the fusion of two light nuclei is frequently exothermic. Some examples are :

dd →

He n + 3.25 MeV (7.2)

dd →

H p + 4 MeV (7.3)

dt →

He n + 17.5MeV . (7.4)

Reactions like (7.4) that produce

He are particularly exothermic owing to

the large binding energy of that nucleus. Other examples are reactions that

yield

He from the weakly bound Li-Be-B nuclei:

Li →

He + 4.8 MeV (7.5)

Li → 2

He + 22.4 MeV (7.6)

B → 3

He + 8.8MeV . (7.7)

The above reactions are the “terrestrial ” reactions with which we are

concerned in this chapter. The basic fusion reaction in the Sun is

pp →

+0.42 MeV . (7.8)

This reaction transforms primordial hydrogen into deuterium, which sub-

sequently fuses to

He, and then to the heavier elements of which we are

made. Unlike the terrestrial fusion reactions, (7.8) is due to the weak inter-

actions because of the necessity for transforming a proton into a neutron.

Consequently, it has a tiny cross-section which, forgetting the Coulomb bar-

rier penetration probability, is of order ∼ (G

/(hc)

)

(¯hc)

∼ 10

−47

with E =0.42 MeV. This makes it impossible to observe in a laboratory.

One would have to observe ∼ 10

proton–proton collisions in order to have

a chance of seeing one event of the type (7.8).

In the terrestrial reactions (7.2-7.7), the neutrons are already present in

the initial nuclei, having been produced in the primordial Universe (Chap. 9).

Therefore the reactions can proceed through the strong and electromagnetic

interactions.

The interest of fusion lies in two main facts. For civil uses, the main

advantage over ﬁssion is the large amount of available fuel, i.e. hydrogen

isotopes. For military uses, the advantage is the absence of any critical mass

constraint. The power of a ﬁssion device is limited by the fact that the pre-

ignition masses of its components cannot exceed the critical mass. This is

not the case for fusion devices since the device can explode if and only if

it is brought to suﬃciently high temperatures and densities. The amount

of material is irrelevant. For the same reason, a controlled fusion installation

does not have the same risks of a nuclear accident that a critical ﬁssion system

presents.

If fusion is still not yet exploited, it is due to the fact that there are

tremendous unsolved technical problems in achieving it.

7.1 Fusion reactions 331

The best fuel for terrestrial fusion is the deuterium-tritium mixture gen-

erating energy through reaction (7.4). This is done in association with some

amount of

Li which absorbs neutrons and regenerates the tritium through

reaction (7.5).

Tritium itself is β-unstable,

H →

He e

−

1/2

=12.33 yr , (7.9)

and is consequently not present in large quantities on Earth. It is produced by

neutron irradiation of

Li, exploiting the exothermic reaction (7.5), or by ra-

diative neutron capture by deuterium. The radioactivity of tritium obviously

creates practical problems for its manipulation.

7.1.1 The Coulomb barrier

The physical diﬃculty in achieving fusion reactions is due to the fact that,

unlike neutron absorption in ﬁssion reactions, the nuclei which interact are

charged and, therefore, have a Coulomb repulsion. In order for the nuclei to

have strong interactions, they must approach one another at a distance of the

order of the nuclear forces, i.e. the radius of the nuclei. This is once again a

situation where one must cross an electrostatic potential barrier. The barrier

has a height given by

4π

α¯hc

=1.4MeV × Z

1fm

, (7.10)

where a is the distance within which the attractive nuclear forces become

larger than the Coulomb force. For energies or temperatures less than ∼

1 MeV the barrier is operative.

If E is the energy of the particle impinging on this barrier, the probability

to cross the barrier by quantum tunneling is proportional to the Gamow

factor we have seen in Sect. 2.6:

P ∼ exp

⎡

⎣

−2





2m(V (r) − E)

¯h

⎤

⎦

(7.11)

where m is the reduced mass m = m

/(m

+ m

) of the two interacting

nuclei and b is the classical turning point deﬁned by V (b)=E where V (r)is

the repulsive Coulomb potential.

The integral in the right hand side of (7.11) can be calculated easily, since

we note that

a  b =

4π

α¯hc

= 143 fm × Z

10 keV

, (7.12)

and therefore the radius a can be taken equal to zero in good approxima-

tion. This leads to the following expression due to Gamow (in 1934) for the

tunneling probability:

332 7. Fusion

P ∼ exp



−2πZ

4π

¯hv



=exp





−E



. (7.13)

where v is the relative velocity and E = µv

/2 is the center-of-mass kinetic

energy for a reduced mass µ. The barrier is characterized by the parameter

=2π

µc

= 1052 keV × Z

µc

1GeV

. (7.14)

Table 7.1. Some fusion reactions. The ﬁrst three are used in terrestrial fusion

reactors. The last three make up the “PPI” cycle responsible for most of the energy

generation in the Sun. Note the tiny S(E) for the weak-reaction pp → de

.It

can only be calculated using weak-interaction theory.

reaction QS(10 keV) E

(1 keV) E

(20 keV)

(MeV) (keV b) (keV) (keV) (keV)

dd → n

He 3.25 58.3 987. 5.1 37.5

dd → p

H 4. 57.3 987. 5.1 37.5

dt → n

He 17.5 14000. 1185 6.8 50.1

pp → de

1.442 3.8 × 10

−22

526 5.1 37.5

pd →

He γ 5.493 2.5 × 10

−4

701 5.6 41.2

He → pp

He 12.859 5 × 10

25200. 18.5 136

We note that the argument of the exponential increases in absolute value

with the product of the charges and that it decreases as the inverse of the

velocity. The higher the energy of the nuclei is, the greater the probability to

tunnel through the barrier. Likewise, the larger the product of charges Z

is, the higher the barrier, therefore at a given energy, it is the lighter nuclei

which can undergo fusion reactions. For particles of charge +1 (e.g. d + d)

we have

E =1keV⇒ P ∼ 10

−13

E =10keV ⇒ P ∼ 10

−3

This suggests that ∼ 10 keV is the order of magnitude of the kinetic energy

that the nuclei must have in order for fusion reactions to take place. (Much

below that energy, the cross section vanishes for all practical purposes.) The

energy of the nuclei comes from their thermal motion, therefore from the

temperature of the medium where they are contained. Hence the name ther-

monuclear reactions for fusion reactions.

We must ﬁnd the factors of proportionality between the tunneling prob-

ability and the reaction cross-section. In Sect. 3.6, where we treated absorp-

tion reaction inhibited by potential barriers, we argued that the cross-section

should be of the form