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Unformatted text preview: Chapter 4
NUCLEAR PROCESSES, THE STRONG FORCE
© M. Ragheb
1/29/2010 4.1 INTRODUCTION
To gain a practical understanding of nuclear, plasma and radiological phenomena one
needs to understand the basic nuclear processes involving isotopes, nuclear reactions,
radioactivity, as well as fission and fusion. The goal here is to deal with the strong force holding
the nuclei together in the atomic nucleus and its possible release through the processes of fission
and fusion. We first compare nuclear reactions to chemical reactions, then compare their energy
releases. It is observed that nuclear reactions produce as much as 1 million times the energy per
reaction as compared with chemical ones. We also consider the equivalence of mass to energy
and the corresponding mass and momentum conservation relations governing nuclear reactions. 4.2 CHEMICAL REACTIONS
Chemical reactions as learned in the field of Chemistry involve the combination or
separation of whole atoms. The interaction between the reactants and product nuclei is an atomic
process involving the electronic cloud surrounding the nuclei. As examples, one can consider
first the chemical reaction where carbon in fossil fuels, or in plant and animal metabolic
processes, is burned or oxidized into carbon dioxide. Second, one can consider the reaction in
which hydrogen is combined with oxygen chemically to produce water. Third, one can also
consider the chemical reaction in which uranium dioxide is converted into uranium tetrafluoride
or the Green Salt through the interaction with hydrofluoric acid (HF) in the manufacturing of
uranium fuel for nuclear reactors:
Reactants Products C+ O2 CO2 + 4 eV (1) 2H2 + O2 2 H2O (2) UO2 + 4HF 2 H2O + UF4 (3) In the case of the first reaction one notices a release of energy in the reaction that can be
measured through calorimetry as 4 eV. The electron Volt (eV) unit is the amount of kinetic
energy acquired by a single electron if it is accelerated though a potential drop of 1 Volt. In this
case that unit can be transformed into other units of energy as follows:
1 eV = 1 electron Volt of energy =
=
= 1.5190 x 1022 BTU (British Thermal Unit)
4.4400 x 1026 Kw.hr (Kilowatt.hour)
1.6021 x 1019 Joule (Where 1 Joule = 1 Watt.second) Although the eV energy unit is adequate in chemical applications, in nuclear applications,
a larger unit, the Million electron Volt is used where:
1 MeV = 106 eV. In these chemical reactions, one can observe the following characteristics:
1. Each atom participates as a whole in the reaction and retains its identity after the reaction is
completed,
2. The resulting product molecules are different from those entering the reactions as reactants,
3. A sharing or exchange of valence electrons occurs, and,
4. The nuclei of the participating atoms are unaffected. 4.3 NUCLEAR REACTIONS, TRANSMUTATIONS
Nuclear reactions differ substantially from chemical reactions in that the reactions
involve the nuclei of the atoms rather than their electronic cloud as in chemical reactions.
Consequently, the reactant nuclei do not necessarily show up as products of the reactions and
transmutation as the dream of the Middle Age alchemists is achieved. We could find either
isotopes of the reactants or completely different ones.
In balancing nuclear reactions we conserve nuclear particles rather than whole atoms like
in chemical reactions. As examples the bombardment of aluminum or nitrogen nuclei with alpha
particles, which are helium nuclei, can lead to the following nuclear reactions:
Products Reactants
27
13Al
14
7N + 2He4 + 2He4 30
14Si 17
8O + ∆m (4) + ∆m (5) + Z4VA4 + ∆m (6) + 1H1 + 1H1 In general, one can write:
A1
Z1X + Z2YA2 A3
Z3U where:
X and Y represent the reactant nuclei,
U and V represent the product nuclei,
Z1 and Z2 represent the atomic numbers, or the charges, or the number of protons of the
reactant nuclei,
A1 and A2 represent the mass numbers, or the total number of nucleons in the reactant
nuclei, Z3 and Z4 represent the atomic numbers, or the charges, or the number of protons of the
reactant nuclei,
A3 and A4 represent the mass numbers, or the total number of nucleons in the reactant
nuclei.
In nuclear reactions the number of protons or the total charge is conserved on both sides
of the nuclear reaction:
Z1 + Z2 = Z3 + Z4 (7) In addition, the total number of nucleons is conserved on both sides of the nuclear
reaction:
A1 + A2 = A3 + A4 (8) As an extra feature of nuclear reactions, mass is also conserved. The conservation of
mass process can be expressed through the statement that:
∆m = mass (reactants) – mass (products) (9) In addition, momentum and another nuclear property designated as “parity,” are conserved. 4.4 MASS CONSERVATION IN NUCLEAR REACTIONS
Mass conservation in nuclear reactions allows us to calculate the energy releases from
these nuclear reactions. To measure atomic and nuclear masses, a standardized unit designated
as the atomic mass unit (amu) is used. Since carbon is abundant on earth in hydrocarbons and
living matter, the amu unit of mass has been chosen as 1/12th the mass of the most common
isotope of carbon: 6C12. Thus we can define the amu as:
m( 6 C12 )
1 amu =
12 (10) In older scientific papers and reports, the amu was defined as 1/16th the mass of the O16 isotope.
If we now want to express the mass of the 6C12 isotope, we can write that:
m(6C12) = 12 amu
We can also express the masses of the isotopes involved in the previous nuclear reactions as:
m(13Al27) = 26.981541
m(2He4) = 4.002603
m(14Si30) = 29.973772
m(1H1) = 1.007825
m(7N14) = 14.003074
m(8O17) = 16.999131 amu
amu
amu
amu
amu
amu EXAMPLE
We can apply the conservation of mass law in terms of the nuclear masses in the
previously listed nuclear reaction involving 13Al27 as:
∆m =
=
=
=
= m (reactants) – m (products)
[m(13Al27) + m(2He4)] – [m(14Si30) + m(1H1)]
[26.981541 + 4.002603]  [29.973772 + 1.007825]
30.984143  30.981597
+ 0.002540 amu This difference in mass is positive in value implying an “exothermic” reaction.
If we apply the same mass conservation process to the reaction involving the 7N14 isotope, we
get:
∆m =
=
=
=
= m(reactants) – m(products)
[m(7N14) + m(2He4)] – [m(8O17) + m(1H1)]
[14.003074 + 4.002603]  [16.999134 + 1.007825]
18.005677  18.0069567
 0.001279 amu The difference in mass is this case is negative, implying an “endothermic” reaction. This
suggests that we need to supply energy to the reactants in the form of kinetic energy to the
reactants for the reaction to proceed.
Notice that this terminology is the opposite of the one used in chemical reactions. 4.5 AVOGADRO’S LAW
To convert the amu into units of grams, and later into units of energy, we invoke
Avogadro’s law from the field of Chemistry stating that:
“The number of atoms or molecules in a mole of any substance is a constant: Av =
0.602 x 1024 [atoms/mole]”
where: 1 mole = 1 gram molecular weight
= Amount of a substance having a mass, in grams, equal to the atomic
or molecular weight of the substance. For instance:
1 mole of 6C12 = 12 gm of 6C12
and:
48 gm of 6C12 contain : 48/12 = 4 moles of 6C12 In general:
Mass of Element
Atomic or Molecular Weight Number of moles = m( Z X A )
A Here we have approximated the atomic weight M, by the mass number A.
As an example, for 64 grams of oxygen O2: Number of moles 64
64 2
16 +16
32 Since one gram mole contains Avogadro’s number in atoms or molecules, and one gram
mole of 6C12 has a mass of 12 g, and contains NA atoms, thus:
Mass of one atom of 6 C12 = 12
12
=
= 1.99 x 10
A v 0.602x1024  23 gm This can help us establish the value of the amu in grams:
1 amu = m( 6 C12 ) 1.99 x 10
=
12
12  23 = 1.66 x 10  24 gm Avogadro’s Law allows us to estimate the number of nuclei N, atoms, or molecules in a
given mass g in grams of a substance: N= g
Av
M (11) If we are interested instead in the number of nuclei, atoms, or molecules per unit volume,
we use instead the number density form of the previous equation, by dividing by the volume V,
or using the density of the substance instead of its mass resulting in a modified form of
Avogadro’s law as: N
N' =
=
V g
( )
ρ
V A
Av
=
v
M
M 4.6 MASS AND ENERGY EQUIVALENCE (12) A result of Relativity Theory is that mass and energy are equivalent and convertible into
each other. In particular the complete annihilation of a particle of rest mass m in grams releases
an amount of energy E in units of ergs given by the formula:
E = mc2 [ergs] (13) where: c is the speed of light = 2.9979 x 1010 [cm / sec].
If Δm is the decrease or increase in mass when a number of particles combine to form the
nucleus, then an amount of energy is absorbed or released in the process as:
ΔE = Δm c2 [ergs] (14) This energy is the binding energy of the particles in the nucleus.
Thus the total annihilation of 1 gm of matter leads to the release of the following amount
of energy:
E = 1 x (2.9979 x 1010)2
= 8.9874 x 1020
= 8.9874 x 1013 [ergs]
[Joules] 1 erg = 10  7 Joule. since: This is a substantial amount of energy that is equal to about 25 million kilowatt.hours.
Since the mass of the electron is 9.1095 x 10  28 g, the rest mass energy of the electron
can be estimated as:
E = 9.1095 x 10  28 x (2.9979 x 1010)2
= 8.1871 x 10  7
[erg]
= 8.1871 x 10  14 [Joule]
= 8.1871 x 10  14 Joule / (1.6022 x 10 = 0.511 MeV 13 Joule/MeV) where we used the definition of the electron Volt as the kinetic energy of an electron charge (1.6
x 10  19 Cb) accelerated through a one Volt potential drop:
1 eV = 1 electron Volt = 1.6 x 10  19 Coulomb.Volt = 1.6 x 10  19 Joule.
Since we have earlier established that the atomic mass unit (amu) is equal to 1.66 x 10  24
gm, hence the energy equivalent of 1 amu is:
1 amu = 1.66 x 10  24 [gm]
MeV
x 0.511 [,
gm
electron
 28
9.1095 x 10
[
electron or:
1 amu = 931.481 MeV (15) 4.7 THE CURVE OF BINDING ENERGY
The binding energy of a nucleus EB is defined as the difference between the masses of the
Z protons and the N neutrons in the free state and the mass of the nucleus containing the A = Z +
N nucleons. We can use the masses of the neutral atoms instead of the nuclear masses, since the
masses of the electrons will cancel out. Thus:
E B = Z .M p + N. M n  M A
= Z .(M p + m e )+ N. M n  (M A +Z .m e ) (16) = Z .M H + N .M n  M where MH is the atomic mass of the hydrogen atom of mass number 1, Mn is the mass of the
neutron, and M is the mass of the isotope of atomic number Z and mass number A. This binding
energy is expressed in amus or in the MeV units.
The binding energy per nucleon is defined as: E B (Z M H + N M n  M)
=
A
A (17) Fig. 1: The curve of binding energy in MeV per nucleon as a function of the mass number A.
The graph that shows this quantity as a function of the mass number A, is designated as
the “Curve of Binding Energy” and is shown in Fig. 1 for the naturally occurring isotopes. For values of A larger than 20, the binding energy per nucleon increases slowly from 8 to
8.5 MeV/nucleon around the value of A = 60. It then decreases slowly to about 7.5
MeV/nucleon for the heaviest elements. For the values of A from 1 to 20, there are large
variations in the values of the binding energy per nucleon.
The mass defect Δ is defined as the difference between the atomic mass M of an isotope
and its mass number A:
Δ=M–A (18) The values of the mass defect are tabulated in nuclear data libraries and are a useful
quantity to calculate the atomic mass of any radioactive isotope. They are shown in Fig. 1 for
the naturally occurring isotopes and the naturally occurring and artificial isotopes.
The packing fraction is defined as the mass defect of the whole atom per nucleon: ∆
(M  A)
=
A
A 150 MA [MeV] (19) 150 100 50 50 0 0 50 50 100 100
0
20
40
60
80
100
120
140
160
180
200
220
240
260
280 100 Mass Number A 0
20
40
60
80
100
120
140
160
180
200
220
240
260
280 MA [MeV] F= Mass Numbe r A Fig. 2: The mass defect diagram for the naturally occurring (left) and the artificially created and
naturally occurring isotopes (right). For mass numbers A between 20 and 180, the packing fraction is negative with a
minimum at around 60. For the heavy and light elements, the packing fraction is positive.
The peak of the binding energy per nucleon corresponds to nuclei with mass numbers
around 60, containing the iron, cobalt and nickel isotopes. Since the heavy elements like
uranium, have a lower binding energy per nucleon, they can be split or fissioned into other nuclei
near the peak of the curve, with a release of energy from the fission process.
Similarly, the light nuclei such as deuterium with a low binding energy per nucleon, can
undergo a fusion process, leading to more stable nuclei, with an accompanying release of energy
from the fusion process. 4.8 STABLE NUCLIDES OCCURRENCE
Nuclides found on Earth are either stable or are radioactive with long half lives in the
range of billions of years, since they were produced in a distant past supernova that created the
Earth about 4.6 billion years ago. If we plot the number of protons Z against the number of
neutrons N, as shown in Fig. 3, we notice an average line of stability clustered around the
straight line described by N = Z, among the light nuclei. For heavier nuclei, there is a deviation
favoring a larger number of neutrons than protons in the nuclei. This can be attributed to the
increased importance of the Coulomb force. Atomic Number Z 100
90
80
70
60
50
40
30
20
10
0
0 10 20 30 40 50 60 70 80 90 10 11 12 13 14 15 16
0 0 0 0 0 0 0
Neutron Number N Fig. 3: The proton Z neutron N diagram for the naturally occurring isotopes
There are 279 stable isotopes which can be divided into 4 categories depending on the
parities of their atomic number Z, their neutron number N, and their mass number A, as shown in Table 1. It is obvious that the even A, N and Z nuclides predominate as stable nuclides. There
are about as many nuclides with odd A with an even number of protons as there are with an odd
number of protons. This suggests that the nuclear strong force tying the nucleons in the nucleus
does not depend on whether the nucleons are protons or neutrons. Odd Z and odd N stable
nuclei are rather rare. Their scarcity can be attributed to the pairing energy between nuclei in the
same nuclear shell.
Table 1: Stable nuclides distribution as even and odd numbers nuclei.
Number of nuclides 168 57 50 4 Neutron number N
Atomic number Z
Mass number A Even
Even
Even Odd
Even
Odd Even
Odd
Odd Total
279 Odd
Odd
Even In Fig. 3, elements along the diagonals perpendicular to the N = Z line possess the same
mass number A and these nuclides are designated as isobars.
Nuclides with the same number of neutrons N, are designated as isotones.
For odd A, only one stable isobar exists except for A = 113 and 123. For even A, only
even N and even Z stable nuclides exist, except for A = 2, 6, 10, and 14, which have odd Z and
odd N isobars. 4.9 RADIOACTIVE NUCLIDES
The use of nuclear reactors and particle accelerators has led to the generation of more
isotopes of the elements than occur in nature. The terms isomer or metastable states, refers to
isotopes that have different radioactive properties in different longlived energy states. At
present there are 2004 nuclides and 423 isomers known. Of these only 279 are stable or
naturally occurring isotopes.
The known nuclides are classified in the Chart of the Nuclides shown in Fig. 4 in the
same way that the chemical elements are classified in the periodic table of the elements or the
Mendeleev table. The black dots correspond to the naturally occurring isotopes which constitute
a band below the Z = N straight line.
Nucleons above that band are proton rich, and the nuclei try to reach the region of
stability by emitting a positive electron or positron, or by the process of electron capture, where
an inner shell electron is captured by the nucleus.
Nucleons below the stability range are neutron rich, and try to reach the stability region
by emitting a negative electron or beta particle. Some of the heavy elements emit an alpha
particle or a helium nucleus. The radioactive decays can leave the nucleus in an excited state,
which is accompanied by gamma rays emission to reach the ground states.
EXAMPLE
Cesium137 is a product of the fission process and has a halflive of 30.17 years. It decays
by negative beta emission to a shortlived daughter barium137 with a half live of 2.552 minutes. The latter emits a gamma ray photon with energy of 0.66164 MeV, to the stable barium137
isotope. The ground state is a stable element, which can be considered as having an infinite halflife. Fig. 4: The proton neutron diagram for the artificially created and the naturally occurring
isotopes. 4.10 REST MASS ENERGY, KINETIC ENERGY AND TOTAL ENERGY
From the Special Theory of Relativity, for a relativistic particle, the total energy is
expressed as:
Total Energy = Kinetic Energy + Rest Mass Energy (20) Expressed in terms of the mass of the particle m and the square of the speed of light, this
equation can be written as; mc 2 = E k + m0 c 2
From this equation the kinetic energy of a relativistic particle is: (21) Ek = 2 − m0 c 2 = − m0 )c 2
mc
(m (22) The relativistic mass of the particle depends on the ratio of its velocity v to the speed of
light c as β= v
c (23) It can be written as: m= m0 (24) 1− β 2 Substituting for the relativistic mass Eqn. 24 into the expression for the kinetic energy
Eqn. 22, we get: m 0
= − m0 c 2
Ek 1− β 2 (25) 1
= − 1 m0c 2 1− β 2 Using the binomial expansion: (1 + x )n = 1 + nx + n(n − 1) 2 n(n − 1)(n − 2) 3
n!
x +
x + ... +
x r + ...
3!
(n − r )!r!
2! (26) For small values of the ratio of the particle’s speed to the speed of light, we can expand
the square root in Eqn. 25 as:
1
1− β 2 = 1 (1 − β ) 1
2 2 ≅ 1+ 1 2
β + ...
2 (27) Substituting from Eqns. 23 and 27 into Eqn. 25 we obtain for the kinetic energy for a
particle with a velocity that is small compared to the speed of light: 1 E k = 1 + β 2 + ... − 1m0 c 2 2 (28)
2 1v
1
= m0 c 2 = m0 v 2
2c
2
This shows that the kinetic energy of classical mechanics is a special case of the
relativistic kinetic energy when the particle’s energy is much less than the speed of light. 4.11 NUCLEAR REACTIONS Q VALUES
Let us consider a nuclear reaction between two reactants a and b, leading to two products
c and d:
a+bc+d
reactants products.
We can write a massenergy balance involving the kinetic energies, KE of the particles in
the reaction as well as their rest mass energies as:
(KE + rest mass energies)reactants = (KE + rest mass energies)products (29)
Explicitly, we denote the kinetic energies as E, the masses as M, and the rest mass
energies as mc2. The mass energy balance can thus be expressed as:
(Ea + mac2) + (Eb + mbc2) (Ec + mcc2) + (Ed + mdc2) (30) Gathering the kinetic energies on the left hand side (LHS) of the equation and the masses
at the right hand side (RHS) of the equation we can define the Q value of the reaction as the
difference between the kinetic energies of the products minus the kinetic energies of the
reactants, which is also equal to the difference between the masses of the reactants minus the
masses of the products:
Q = (Ec +Ed)  (Ea +Eb) = (mac2 + mbc2)  (mcc2 + mdc2) (31) Since it is easier to access the data about the masses of the nuclei undergoing the reaction
rather than measuring their velocities to calculate their kinetic energies, the Q values for nuclear
reactions are calculated by using the masses of the reactants and products:
Q = [(ma + mb)  (mc + md)] c2 (32) The masse in Eqn, 32 are the masses of the nuclei, whereas the masses that are measured
experimentally using mass spectrometers are the masses of the neutral atoms, or the masses of
the nuclei and the surrounding electrons. To account for the electron masses, we use the
conservation of charge equation for the nuclear reaction considered as:
Za + Zb =Zc +Zd (33) By considering the mass of the electron as me we can rewrite the expression for the Q
value in Eqn. 22 in terms of the neutral atoms masses, not just the nuclei masses as:
Q = {[(ma + Zame)+ (mb + Zbme)] – [(mc + Zcme) + (md + Zdme)]} c2 (34) When the value of Q is positive, the reaction is called exothermic, and would lead to a
positive energy release. When the value Q is negative, the reaction needs to be fed energy to
proceed, normally in the form of kinetic energy of the reactants.
EXAMPLE
For the DT fusion reaction between the two isotopes of hydrogen deuterium (D) and tritium (T):
2
3
1D + 1T 2He4 +0n1 (35) Data mining the appropriate data warehouses such as the Chart of the Nuclides for the
neutral atom masses of the reactants and products yields:
m(1D2) = 2.014102 amu,
m(1T3) = 3.016049 amu,
m(0n1) = 1.008665 amu,
m(2He4) = 4.002604 amu.
Using the mass energy equivalence that we derived earlier where 1 amu is equivalent to
931.481 MeV of energy, we get for the energy release or Q value from this reaction as:
Q = {[(2.014102)+ (3.016049)] – [(4.002604) + (1.008665)]} 931.481
= 0.018882 x 931.481 (amu). (MeV/amu)
= 17.6 MeV. 4.12 REACTIONS INVOLVING POSITRON EMISSION
Positron decay leaves the product nucleus with an atomic number diminished by one unit,
leaving the mass number A unchanged, and is accompanied by a neutrino emission. In general,
the calculation of the Q values is equal to the difference between the masses of the reactants
neutral atom masses, and the products masses. An exception exists in the case of positron
emission. Let us consider a positron emission reaction: 6 C 11 → 5 B 11 + +1 e 0 + ν (36) Ignoring the mass of the neutrino, but not the electron masses, the Q value is written in
terms of the nuclear masses as:
Q = [m(C)  6me] – [m(B) – 5me] me
This can be rewritten as:
Q = m(C)m(B) – 2me (37) Thus two electron masses must be subtracted from the difference in the neutral atom
masses.
For the reaction to occur spontaneously:
Q > 0,
implying that:
m(C)m(B) > 2me > 2 x 0.51 MeV > 1.02 MeV (38) Thus the difference of masses between parent and daughter should be larger than 2
electron masses for positron decay to occur. 4.13 REACTIONS INVOLVING ORBITAL ELECTRON CAPTURE
Orbital electron capture decreases the atomic number of the nucleus by one unit while
leaving the mass number unchanged, just like positron decay. Let us consider the K electronic
shell electron capture reaction:
29 Cu 64 + −1 e 0 → 28 Ni 64 + ν (39) The neutrino acquires the entire energy Q released by the reaction leaving little for the
recoil nucleus. We must notice that the captured electron releases its entire mass when it is
absorbed in the nucleus minus the mass equivalent of the binding energy of the electron in the
atomic shell, EB. Therefore the Q value can be calculated as:
Q = [m(Cu) – 29me] – [m(Ni) –28me] + (me –E) (40) Q = m(Cu) – m(Ni) EB (41) Thus: Since EB is small, for instance 0.024 MeV for the K shell of palladium103 it can be
ignored and the Q value can be calculated on the basis of the neutral atomic masses.
Electron capture competes with positron decay. However positron decay needs at least
the energy equivalent of two electron masses to occur, whereas electron capture needs the energy
equivalent of the binding energy of the electron in the K shell.
Since electron capture leaves an empty inner shell, characteristic x rays of the product
nucleus is always emitted, and is used in addition to Auger electrons and the recoil of the product
nucleus to detect the electron capture process.
Auger electrons are emitted as a result of a radiationless transition, for instance, instead
of a K xray being emitted when an L electron goes to the K shell, the energy is used to eject an
electron from the L level. 4.14 INTERNAL CONVERSION
Internal conversion occurs when the energy of an excited nuclear state is transferred to an
atomic electron from the K or L shell, ejecting it from the atom. Auger electrons are internal
conversion electrons. Internal conversion occurs as an alternative to gamma ray emission from
the nucleus.
The kinetic energy of the ejected electron Ee is equal to the excitation energy of the
nucleus minus the binding energy of the electron in its shell:
Ee = Eexcitation  EB (42) Internal conversion is prevalent in heavy nuclei and increases as Z3 but decreases with
Eexcitation. Gamma decay predominates in light nuclei. Examples of nuclei with internal
conversion are Cs137 and Tc99m. 4.15 REACTIONS INVOLVING NEGATIVE BETA DECAY
These reactions occur in the neutron rich nuclei. The result is a nucleus with an atomic
number increased by one unit, an unchanged mass number. An example is the reaction involving
the decay of the tritium isotope of hydrogen:
T 3 → −1 e 0 + 2 He 3 + ν * 1 (43) The Q value of the reaction is:
Q = [m(T) – me] – [m(He) –2me] – me (44) Q = m(T) – m(He) (45) from which: Thus the neutral atom masses of the reactants and the product can be used ignoring the
mass of the emitted electron. 4.16 REACTIONS INVOLVING ALPHA DECAY
Alpha decay occurs among the heavy nuclei. An example is the reaction that transforms
the manmade Pu239 into the normally naturally occurring U235:
94 Pu 239 → 2 He 4 + 92 U 235 (46) Q = [m(Pu) – 94me] – [m(U) – 92me] – [m(He)  2me] (47) Q = m(Pu) – m(U) m(He) (48) from which: In this case the mass of the emitted alpha particle must be subtracted as a product
nucleus, in contrast to negative beta decay where the mass of the emitted electron is not. 4.17 ENERGY PARTITIONING IN NUCLEAR REACTIONS,
CONSERVATION OF MOMENTUM
It is of interest to be able to calculate how the energy release apportions itself among the
products of a nuclear reaction such as the one described by Eqn. 32. We need to supplement the
conservation of mass energy equation by the equation of conservation of momentum.
Considering the particles as initially at rest, forming a compound nucleus then splitting apart in
two opposite directions, conservation of momentum requires that:
mcvc = mdvd
Squaring both sides yields:
mc 2 vc 2= md 2 vd 2
or:
mc mc vc 2= md md vd 2
Multiplying both sides by ½ yields:
mc. ½ mc vc 2= md ½ md vd 2
Noting that the kinetic energies of the product particles are:
Ec = ½ mc vc 2 (49) Ed = ½ md vd 2
We can write:
mc Ec = md Ed (50) Now, conservation of energy for particles initially at rest implies that:
Q = Ec + Ed (51) Ec = Q  Ed (52) or: Combining Eqns. 50 and 52 yields:
mc (Q – Ed) =md Ed
mc Q = (mc +md) Ed
Which implies that:
Ed = mc
Q
mc + m d (53) Ec = md
Q
mc + m d (54) and similarly: This leads to the interesting result that the energy release apportions itself in inverse
proportion to the masses of the product nuclei.
EXAMPLE
We consider the fusion reaction in Eqn. 35 whose Q value is 17.6 MeV. If we want to
calculate the kinetic energies of the neutron and the helium nucleus, we can use as an
approximation the mass numbers instead of the actual atomic masses to yield from Eqns. 53 and
54: E=
neutron 4
4
80
17.6
17.6
17.6
= =
= 14.08 MeV
4 +1
5
100 Ehelium
= 1
1
20
17.6
17.6
17.6
= =
= 3.52 MeV
4 +1
5
100 These are approximation to the more exact values of 14.06 and 3.54 MeV. It must be
noted that the lighter particle, the neutron, carries 80 percent of the released nuclear energy,
whereas, the heavier particle, the helium nucleus, carries only 20 percent of the energy release
from the reaction. 4.18 THE FISSION PROCESS
The absorption of a neutron in some of the fissile nuclei with even Z and odd A, even
with zero energy can cause them to fission into two fission product and a few neutrons. The two
fission products primarily share the energy release. The neutrons released can leak from the
system, get absorbed in the structural materials, or be available to fission other fissile nuclides
and generate a fission chain reaction as shown in Fig. 5.
A typical reaction is the fission of the U235 isotope, which could yield:
1 0n + 92U235 36Kr97 + 56Ba137 +20n1 +Q (55) Fig. 5: Schematic of a fission chain reaction showing the fission products and the prompt
neutrons.
The fission products yields appear with asymmetric values of the mass number as shown
in the fission yield graph in Fig. 6 for fission with thermal neutrons at a speed of 2,200 m/sec.
As the energy of the neutron increases say to 14.08 MeV as the neutrons originating from the DT
fusion reaction considered earlier, the two humps flatten out, with the yield curve rising up in its
central section. EXAMPLE
To calculate the Q value of the fission reaction in Eqn. 55 we use the neutral atom
masses:
m (0n1)
m (92U235)
m (36Kr97)
m (56Ba137) =
=
=
= 1.00867 amu
235.04390 amu
96.92120 amu
136.90610 amu from which the Q value of the reaction is:
Q = [m (0n1) + m (92U235) – {m (36Kr97) + m (56Ba137)}  2m(0n1)] c2
= [(1.00867 + 235.04390) – (96.92120 + 136.90582) – 2 x 1.00867] x 931.481
= (236.05257 –233.82702 –2.01734) x 931.481
= 0.20821 x 931.481 [amu] [MeV/amu]
= 193.94 MeV
= 193.94 x 1.52 x 1016 [MeV] [BTU/MeV]
= 2.95 x 1014 BTU
(56)
Thus it can be concluded that a single fission event leads to the release of 194 MeV or
about 200 MeV of energy. This is about 10 times the energy release from the DT fusion reaction
at 17.6 MeV. Thus fission can be construed to be energy rich per reaction compared with fusion
as typified by the DT fusion reaction.
It is of great interest from the perspective of fission power generation from fission to
estimate the energy release from a given mass g of fissile material. Applying Avogadro’s law
from Eqn. 11 yields the number of nuclei of U235 in grams as: Fig. 6: The fission yield curve from thermal and fast fissions of U235. N (U 235 ) = g
g
Av =
x0.60225 x10 24 [nuclei]
235.0439
M If we consider that the fission event yields about:
200 [MeV/nucleus]
of which 10 MeV are carried out by the antineutrinos and are not extractable, then we consider
that the extractable energy per fission event is: 200 – 10 = 190 [MeV/nucleus]
The complete fission of g grams of U235 will release the energy E:
E = N (U 235 ) x190[nuclei ][ MeV / nucleus ]
g
=
x 0.60225 x1024 x190[ MeV ]
235.0439
= 0.487 x1024 g[ MeV ] (57) = 2.162 x10 g[kW .hr ]
= 900.6 g[kW .day ]
= 0.9006 g[ MW .day ]
4 An expression for the power release can be deduced for the complete fission of g’
[gm/day] as: P
= E
= 0.9 g '[ MWth ]
t (58) This implies that in a fission reactor the complete fission of 0.9 gm of U235 per day
corresponds to about 1 MWth of thermal power production.
EXAMPLE
A typical fission power reactor produces 3,000 megawatts (MWth) of thermal power. If
it were completely burning its fuel inventory in U235, it would fission:
3,000 [MWth] x 0.9 [gm/(day.MWth)] = 2,700 [gm/day] = 2.7 [kgs/day]
The amount that would be fissionned in a year would be:
2.7 x 365 [kgs/day][days/year] = 9855 [kgs/year] = 0.9885 [metric tonne/year]
Thus a minimum amount of about 1 metric ton of U235 fuel is needed per year for a
typical 3,000 MWth fission power reactor. 4.19 SPONTANEOUS FISSION
Some heavy nuclei decay in a process where the nucleus breaks up into two intermediate
mass fragments and several neutrons. It occurs in with nuclei with mass number A > 230.
Since the maximum binding energy per nucleon occurs at A = 60, nuclides above A >
100 are unstable with respect to spontaneous fission, sine a condition for spontaneous fission is: m( A, Z ) > m( A ', Z ') + m( A − A ', Z − Z ')
in the spontaneous fission reaction:
Z XA→ Z' X A ' + Z − Z ' X A− A ' Because of the high Coulomb barrier for the emission of the fission fragments,
spontaneous fission is only observed in the heaviest nuclei. 4.20 ENERGY DISTRIBUTION OF NEUTRONS FROM FISSION
The average number of neutrons emitted per fission of a U235 nucleus by slow or
thermal neutrons is:
= 2.47 ± 0.03[
ν neutrons
fission (59) These fission neutrons possess a distribution in energy shown in Fig. 7. Fig. 7: Energy distribution of fission neutrons from U235.
The curve has a maximum or most probable energy at 0.75 MeV and an average energy
of 1.98 MeV. It is represented by the semi empirical equation, known as the Watt’s curve as: 1 N ( E ) = C sinh( 2 E ) 2 e − E (60) 4.21 DELAYED NEUTRONS FROM FISSION
Neutron emission from the fission products can occur on a delayed basis, in addition to
those prompt neutrons emitted at the time of fission. Delayed neutron emission occurs if the
parent nucleus had already decayed through beta decay, but is still left in an excited state with an
energy that exceeds the binding energy of a neutron of this nucleus. Since the neutron emitting
nucleus was formed in a beta decay process, the neutron activity will have the same halflife of
the parent nuclide. Examples of delayed neutrons emitting fission products are Br87 and I137,
whose decay diagrams are shown in Fig. 8. Fig. 8: Delayed neutrons emission from the two precursors Br87 and I137.
Delayed neutrons are important in the process of building control systems for controlling
fission chain reactions, and the operation of fission reactors. 4.22 THE LIQUID DROP MODEL FOR NUCLEI
The radius of a nucleus is proportional to its mass number to the 1/3 power and is given
by: R = 1.25 x 1013 A1/3 [cm] (61) The volume of a nucleus can be estimated from the equation of its radius as:
V = 4/3 πR3 = 4/3 π (1.25 x 1013 A1/3)3
= 4/3 π (1.25 x 1013)3 x A [cm3]. (62) This implies that the volume of the nucleus is proportional to A. The density of nuclear matter,
or the number of nucleons per unit volume of the nucleus is given by:
A/V = 1/{4/3 π (1.25 x 1013)3 } = constant, (63) for all nuclei, whether they are large or small. This uniform density suggests that nuclei are
similar to a drop of liquid (Fig. 9). This analogy accounts for many of the physical properties of
nuclei. Fig. 9: Liquid drop model of the neutron fission of a fissile nucleus.
Both the volume and the total binding energy of nuclei are nearly proportional to the total
number of nucleons present A. This implies that nuclear matter is incompressible. It also
implies that nuclear forces must have a saturation character, that is, a nucleon in a nucleus can
interact with only a small number of neighboring nuclei, just as an atom in a liquid is strongly
bound to its neighbor atoms.
The nuclear binding energy of a nucleus is the difference between its mass M expressed
in energy units, and the masses of its constituent nuclei as protons, electrons and neutrons.
Expressed mathematically this is:
E = M – { Z Mp + Z Me +(AZ)Mn} [MeV] = M – { Z MH +(AZ)Mn} [MeV] (64) where: MH is the mass of the hydrogen atom, in energy units,
and:
Mn is the mass of the neutron, in energy units.
Using a semiempirical approach, an equation can relate the binding energy, or total
mass, of any nucleus to its nuclear composition in terms of the atomic number (or number of
protons) Z, and its mass number (or total number of nucleons) A.
For values of A > 40, the energy required to dissociate a nucleus into its constitutive
nucleons is given by:
E = 14.0A – 13.1 A2/3 – 0.585 Z(Z1) A1/3 – 18.1 (A2Z)2A1 + δA1 [MeV] (65) This equation can generate a three dimensional plot of the binding energies of all nuclei
as a function of A and Z.
The term {14.0 A}, is the dominant term, and expresses the fact that the binding energy is
proportional to the total number of nucleons A, and is a consequence of the short range and the
saturation character of the nuclear forces. When four nucleons, 2 protons and 2 neutrons are
joined together, these forces become saturated as in the cases of the nuclei of 2He4, 6C12, and
16
8O .
The term {– 13.1 A2/3} expresses the fact that the surface of a nucleus has unsaturated
forces, and consequently, a reduction in the binding energy proportional to that surface is to be
expected. Since the radius of the nucleus is proportional to A1/3, the surface of the nucleus is
proportional to A2/3. When the size, or radius, of the nucleus increases the surface to volume
ratio of the nucleus:
S / V = (4 π R2) / (4/3 πR3) = 3 / R (65) decreases. Thus this term becomes less dominant as the size of the nucleus increases.
The term {– 0.585 Z(Z1) / A1/3} expresses the Coulomb repulsive forces between the
protons. Each of the Z protons interacts with the other (Z1) protons over a distance of a nucleus
radius, which is in turn proportional to A1/3. This term is mostly dominant for Z > 20, and
accounts for the fact that stable nuclei contain more neutrons than protons in the Chart of the
Nuclides.
The term {– 18.1 (A2Z)2 / A} is a symmetry term accounting for the fact that most
stable nuclei tend to have an equal number of neutrons and protons. Thus a term containing the
absolute value of the difference between the number of neutrons and protons:
 N  Z  =  A – Z – Z  =  A – 2Z (66) is added. The binding energies of unstable light nuclei show that this effect is symmetric around
the value N = Z. The 1/A dependence comes about from the binding energy contribution per
neutronproton pair. This is proportional to the probability of having such a pair within a volume, which is determined by the range of the nuclear forces. This probability is in turn
inversely proportional to the nuclear volume that is proportional to A.
The term { δ / A} expresses the fact that the binding energies depend upon whether N
and Z are even or odd.
For the most stable nuclei,
Z is even, N is even, then: δ = 132.0 Z is even, N is odd,
Ζ is odd, Ν is even
Z is odd, N is odd then:
then:
then: (67) δ = 0.0
δ = 0.0
δ = −132.0 when: The values of δ given above are suitable for the range A > 80. For A < 60, a lower value
in the range of δ = 65 should be used. This term is called the pairing term since it is related to
the tendency of two like particles to complete an energy level in the nucleus by pairing of
opposite nuclear spins.
The difference in the stabilities of these four types of nuclei is expressed in the stable
nuclides. There exists as shown in Table 1: and only: 168
57
50
4 stable even Z, even N nuclei,
stable even Z, odd N nuclei,
stable odd Z, even N nuclei,
stable odd Z, odd N nuclei. The latter category includes the following nuclei:
2
6
10
1D , 3Li , 5B , and 7N14. The greater stability of eveneven nuclei with filled energy states is clear as apparent
from their greater abundance relative to other nuclei. Elements of even Z are 10 times more
abundant than elements of odd Z. For these elements with even Z, the isotopes of even N, and
consequently also even A, account for 70 to 100 percent of each element, with beryllium, xenon,
and dysprosium being exceptions.
The binding energy curve which has a maximum at A = 60, displays an increasing trend
before the maximum, followed by a decreasing trend beyond the maximum. This is caused by
the relative contributions of surface energy, decreasing with A, and the Coulomb and symmetry
energies, which are increasing with A.
It is possible to express the masses of nuclei in terms of the binding energy equation in
the form:
M = { Z MH +(AZ)Mn} E [MeV] (68) By substituting the expression for the binding energy E, we can obtain the semiempirical
mass equation as: M = 925.55A – 0.78Z + 13.1 A2/3 + 0.585 Z(Z1) A1/3 + 18.1 (A2Z)2A1  δA1 [MeV] (69) This equation is a quadratic in Z and can be rewritten as:
M = aZ2 + bZ +c  δA1
where (70) a = 0.585 A1/3 + 72.4 A1 ,
b = 0.585 A1/3 73.18,
c = 943.65A + 13.1 A2/3. Fig. 10: Double parabolas for the mass number A=128. The beta decay energies for the
neighboring isobars are shown compared to the experimental values in parentheses.
For a given constant A, the coefficients a, b and c are constant, and we can generate a
parabola for a given value of δ. For odd value of A, δ = 0, and we obtain a single parabola. For
an even value of A, δ = +/132, and we obtain two parabolas displaced from each other by a
value of (2δ/A) along the energy axis (Fig. 10). The minimum mass, or maximum binding energy for a given value of A is obtained at a
given value of Zmin, by taking a partial derivative of M with respect to Z, with A as a constant,
and equating the derivative to zero:
δM/δZ = 2a Zmin + b = 0,
from which:
Zmin = b/2a. We can use the expression for Zmin to rewrite the mass equation for M as:
M = a(ZZmin)2  δA1 + {c – (b2/4a)} [MeV] (71) This equation is found in the literature to express the total atomic mass of a nucleus according to
the liquid drop model. 4.23 THE SHELL MODEL OF THE NUCLEUS
High stability and high abundance with respect to the neighboring nuclides are
characteristics for nuclei for which the magic numbers for N or Z are:
2, 8, 20, 28, 50, 82, 126.
The magic nuclei are more tightly bound and require more energy to be excited than the
non magic nuclei. These correspond to closed shell in the structure of the nucleus in the same
way that we encounter closed electronic shells in the structure of the atom. Figure 11 shows the
frequency of the stable isotones as a function of the neutron number N. Higher frequencies of
occurrences correspond to the magic numbers. . 8 Stable Isotones Number 7
6
5
4
3
2
1 96 91 86 81 76 71 66 61 56 51 46 41 36 31 26 21 16 11 6 1 0
Neutron Number N Fig. 11: The frequency of the stable isotones as a function of the neutron number N.
Higher frequencies correspond to the magic numbers.
A refinement to the shell model of the nucleus involves the spinorbit coupling model.
This model suggests that a strong interaction should exist between the orbital angular momentum
and the intrinsic spin angular momentum of each nucleon. A natural consequence of the spinorbit model is the existence of the isomeric states near the shell breaks. If the magnitude of the
spinorbit splitting is properly adjusted, the major shell breaks occur at the experimentally
determined magic numbers, which the original shell model does not succeed in predicting.
A limitation of the spinorbit shell model is that it cannot explain why even Z  even N
nuclei do not always possess a zero ground state spin, or why any even number of identical
nucleons couples to zero ground state spin. As the nuclear numbers deviate from the magic
numbers, cooperative effects appear among the nucleons, and these are taken into accounts by
the collective models of the nucleus. In this case rotational motion and vibrational motion are
built into the models. The meson theory of nuclear forces explains more of the nuclear
properties. 4.24 ISLANDS OF STABILITY AND THE SUPERHEAVIES
From A = 200 to 220 an empty area in the Chart of the Nuclides occurs, followed by an
“island” containing the isotopes of the naturally occurring heavy elements which re members of
the actinides series in the periodic table of the elements (Fig. 12). As the atomic number is increased for the heavier nuclides, shorter and shorter halflives are expected. This is caused by
the higher Coulomb repulsion between protons, which overcomes the attractive strong nuclear
force between the nucleons. Spontaneous fission among the heavy nuclides becomes more likely
as their atomic number increases beyond 100. The process of alpha particle decay also increases
as the atomic number increases. The shell model provides a way to suggest that when the
protons and neutron shells are closed, a higher level of stability can be attained. The magic
numbers for which such an extra stability would be attained is Z=114 and N=184. Other
suggested numbers are Z = 126, 116, 124 and 127. Fig. 12: The actinides series in the periodic table of the elements. Fig. 13: Map of the Isotopes expresses the elusive search for an island of stability of
superheavy spherical nuclei beyond the sea of instability.
There has been a prediction started in the sixties that another island of stability could
exist with super heavy elements reaching another region of stability in the shell structure. A
suggestion was made that they could have been created before the earth itself was created, and
that they could still be present if they half lives are long enough. Researchers looked at all kinds
of ores minerals, rocks including manganese nodules at the bottom of the oceans, moon rocks,
gold and platinum ores, meteorites, hot springs water, volcanic lava, and medieval stained glass,
without much success.
Indirect evidence for the one time existence of elements 113 to 115 in the Allende
meteorite has been reported by a group at the University of Chicago. Synthesis of the super
heavy elements has been attempted by bombarding curium248 with a beam of accelerated
calcium48 ions, but met with negative results. The search for the super heavies remains as a
quest for the Holy Grail in the field of nuclear science (Fig. 13). 4.25 COSMOLOGICAL INVENTORY
The rotational speed of the galaxies and other lines of evidence indicate that the universe
contains more mass than the sum of the mass of the luminous stars. Some of the missing mass
designated as dark matter could be ordinary atoms and molecules that are not hot enough to emit
light. If all the mass of dark matter was ordinary matter as neutrons and protons by weight, it
would be inconsistent with the density of ordinary matter inferred from the Big Bang model and
from the relative abundances of light nuclides. A search is ongoing for ordinary matter which
could reside in the outer reaches of galaxies in the Massive Astrophysical Compact Halo Objects (MACHOs) and for more exotic forms of matter such as the Weakly Interacting Massive
Particles (WIMPs) or the axion.
Luminous matter in the universe including the stars and galaxies, accounts for just 0.4
percent of the known universe. Radiation accounts for a mere 0.005 percent. Nonluminous
matter such as intergalactic gas (3.6 percent), neutrinos (0.1 percent), supermassive black holes
(0.004 percent), account together for only 3.7 percent of the cosmological inventory of the
universe. The remaining mass and energy of the universe is divided as about 23 percent dark
matter and about 73 percent dark energy.
What makes these components is subject to many theories suggesting that dark matter is
predominantly a remnant of unknown and new elementary particles, perhaps axions, from the
beginning of the universe at the time of the Big Bang (Table 2).
Table 2: Cosmological inventory of matter and energy.
Component
Dark energy
Dark matter
Luminous matter:
Stars and luminous gas
Radiation
Other nonluminous components:
Intergalactic gas
Neutrinos
Supermassive black holes Percent
73.000
23.000 Properties
Identity unknown
Identity unknown
Known 0.400
0.005
Known
3.600
0.100
0.004 Cosmologists are divided in support of two explanations: cold dark matter or extremely
massive neutrinos other than the three known types of neutrinos. Cold matter would be some
particle that did not acquire much velocity in the early universe. These particles are considered
at the same speed as the stars in the galaxy, which is a much slower speed than the speed of light.
Massive neutrinos would have been moving fast at the birth of the universe and they
would still have velocities close to the speed of light. This is considered unlikely since the
density needed to make up a significant portion of dark matter would have smoothed out
irregularities and prevented the formation of the structures that evolved into the galaxies in the
first billion years of the universe.
If the predominant form of dark matter is cold dark matter, then two elementary particles
are suggested as candidates:
1. A stable Weakly Interacting Massive Particle or WIMP that is about 10 to 100 billion
eV in mass equivalent.
2. An axion, which is theoretically considered to be a very light particle of a mass
equivalent of just 106 to 103 eV with neither electric charge nor spin, thus it interacts hardly at
all. These axions are supposedly so minute that their density is assumed to be 10 [trillions/cm3]
of space in our galaxy. Those occupying a sugar cube volume of space would still weigh less than does half of a proton. Their extreme lightness and nearly nonexistent coupling to radiation
make these particles extremely long lived to maybe 1050 seconds.
The age of the universe itself is estimated at 1018 seconds or 32 billion years old. The
axions could be considered as a stable particle. If they exist, they would constitute the bulk of
dark matter. What they lack in mass each, they make up for in sheer numbers.
The axions would resolve a difficult issue in the Standard Model theory attempting at
explaining fundamental particles and how they interact. It was first suggested as a solution to a
thorny problem of particle physics concerning the absence of Charge Parity (CP) violations in
nuclear strong force interactions. This has to do with the principle in physics that if two systems
are mirror image of each other but are otherwise identical, and if parity is conserved, all
subsequent evolution of these two systems should remain identical except for the mirror
difference. Nature prefers symmetry to asymmetry. It has no preference for right handed versus
left handed behavior as far as particle physics are concerned.
In 1956, Chen Ning Yan and Tsung Dao Lee realized that although many experiments
had been conducted to show that minor symmetry was true for the strong interaction force and
for the electromagnetic interaction force, no experiments had been conducted for the weak
interaction force. Experiments by ChiengShiung Wu showed that contrary to what was
expected, weak interactions did show a preference or handedness in Co60. When the decay of
Co60 is observed in a strong magnetic field, there was a preferred direction for emitting beta
particles in a decay proving that the weak interaction violated the conservation of parity. This
ran counter to all expectations.
In the 1970s, Quantum Electro Dynamics (QED) was developed to describe the strong
force interaction, and it held that large amounts of CP violations. These violations had not been
observed in experiments. In an effect to explain the phenomenon, Roberto Peccei and and Helen
Quinn proposed a new symmetry of nature that resulted in the particle dubbed the axion. Their
predicted mass for the axion was about 50 keV.
As a result of observations done in 1987 on the neutrinos of a supernova, a conclusion
was made that the mass of the axion is about 103 eV, suggesting that the supernova core would
have been cooled not just by the emitted neutrinos, but by the axions as well. If that was the
case, the length of the neutrino burst associated with the supernova would have been much
shorter than was observed.
Axions lighter than a microelectron volt could have been produced in the Big Bang
resulting in the universe being more massive than it is, which would be at variance with current
observations where dark matter is at most ¼ of the closure or the critical density of the universe.
Refined calculations suggest that the axion mass is between 106 to 103 eV.
Several experiments world wide are dedicated to the WIMP search. Two experiments are
tracking down the axion. One experiment in Japan designated as CARRACK2 at Kyoto
University’s Institute for Chemical Research, and the other at the Lawrence Livermore National
Laboratory (LLNL) in the USA. The LLNL experiment is based on the theory that an axion
when it interacts decays into two photons with frequencies in the microwave region of the
electromagnetic spectrum. There is an idea suggesting that an axion could be stimulated to
decay into a single photon in the presence of a large magnetic field threading a microwave
cavity. The setup uses an 8 Tesla 6 tons superconducting magnet coil that is wound around the
outside of a copper plated stainless steel cylinder, the size of an oil drum. A set of tuning rods
inserted into the cylinder’s cavity is moved by stepper motors to tune the cavity’s frequency. Liquid helium cools the cavity and reduces the background noise and amplifiers attempt at
boosting the faint axion signal.
If the axion is not found then something else, maybe another particle, a symmetry, or path
or another new theory must be developed to explain what makes up dark matter. 4.26 OBSERVATION OF DARK MATTER
The Bullet Cluster is thought to have formed after two large clusters of galaxies collided,
and passed trough each other in the most energetic event since the postulated Big Bang. Until
present, the existence of dark matter was inferred by the fact that galaxies had only 1/5 of the
visible matter needed to create the gravity that keeps them intact; implying that the balance must
be invisible to light telescopes. Fig. 14: Bullet Cluster known as galaxy cluster 1E065756. Hot stellar gas is shown in red
and dark matter in blue. NASA photograph.
The observations were conducted using a combination of observations from the orbiting
Hubble Space Telescope and the Chandra x ray Observatory, along with the groundbased
European Southern Observatory Very Large Telescope and the Magellan Telescope. A paper in
the Astrophysical Journal Letters, a journal of the American Astronomical Society, describes the
observations. Fig. 15: Hubble Space Telescope image of a ring of dark matter in the galaxy cluster Cl
0024+17. Source: NASA.
Images taken by NASA's orbiting Hubble Space Telescope in 2007 allowed astronomers
to detect this ring of dark matter created by the collision of two galaxy clusters 5 billion lightyears from Earth. Scientists came across the new evidence while studying the distribution of
dark matter within a galaxy cluster designated as Cl 0024+17. Wondering about the genesis of
this ring, the researchers came across earlier work showing that the galaxy cluster had run into
another cluster 1 billion to 2 billion years ago.
The collision between the two galaxy clusters created a ripple of dark matter that left
distinct footprints in the shapes of the background galaxies. This is like looking at the pebbles
on the bottom of a pond with ripples on the surface. The pebbles' shapes appear to change as the
ripples pass over them. Dark matter previously has been detected in other galaxy clusters, but
has never been previously found to be so distinctly separated from the hot gas and the galaxies
that make up these clusters. REFERENCES
1. F. W. Walker, G. J. Kirouac and F. M. Rourke, “Chart of the Nuclides,” Knolls, Atomic
Power Laboratory, General Electric Company, 1977.
2. G. Friedlander. J.W. Kennedy and J.M. Miller, ‘Nuclear and Radiochemistry,” John Wiley and
Sons, 1955.
3. M. M. ElWakil, “Nuclear Heat Transport,” International Textbook Company, an Intext
Publisher, 1971.
4. W. E. Meyerhof, “Elements of Nuclear Physics,” McGraw Hill Book Company, 1967.
5. H. Semat, “Introduction to Atomic and Nuclear Physics,” Holt, Rinehart and Winston, 1962.
6. A. Parker, “Small Particle may answer Large Physics Questions,” Science and Technology
Review, LLNL, Jan./Feb., 2004. 7. A. Heller, “Uncovering the secrets of the Actinides,” Science and Technology Review, LLNL,
June, 2000. EXERCISES
1. In a possibly future matter/antimatter reactor, use the mass to energy equivalence relationship
to calculate the energy release in ergs, Joules and MeV from the complete annihilation of:
a. An electron/positron pair.
b. An antiproton/proton pair.
Consider the following masses:
m electron = m positron = 9.10956 x 1028 gram
m proton = m antiproton = 1.67261 x 1024 gram.
2. Balance the following nuclear reactions by applying conservation of charge and of nucleons.
Next, apply mass/energy conservation and calculate the energy releases in units of MeV for the
following fusion reactions:
a. 1D2 + 1T3 0n1 + ? + Q1 (DT fusion reaction)
b. 1D2 + 1D2 1H1 + ? + Q2 (Proton branch of the DD fusion reaction)
c. 1D2 + 1D2 0n1 + ? + Q3 (Neutron branch of the DD fusion reaction)
d. 1D2 + 2He3 2He4 + ? +Q4 (Aneutronic or Neutronless DHe3 reaction)
3. For the Fusion reactions listed in the last problem, using conservation of momentum, calculate
how the energy release (Q) apportions itself among the reaction products.
4. Calculate the energy release or the Qvalues of the following fission reactions:
1
235 3 0n1 + 53I137 + 39Y96
0n + 92U
1
235 3 0n1 + 54Xe136 + 38Sr97
0n + 92U
1
235 2 0n1 + 56Ba137 + 36Kr97
0n + 92U
5. Calculate the Q values of the following nuclear reactions:
11
11
0
6 C → 5 B + +1 e + ν (Positron decay reaction)
29 Cu 64 + −1 e 0 → 28 Ni 64 + ν (Orbital electron capture reaction) T 3 → −1 e 0 + 2 He 3 + ν * (Negative beta decay reaction) 1 94 Pu 239 → 2 He 4 + 92 U 235 (Alpha particle decay reaction) ...
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This note was uploaded on 06/16/2010 for the course NPRE 402 taught by Professor Ragheb during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 RAGHEB

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