04 nuclear energetics.pdf

Chapter 4

Nuclear Energetics

In reactions (nuclear, atomic, mechanical) in which some quantity is changed intosome other quantity (e.g., two cars into a mass of tangled metal), energy is usuallyemitted (exothermic reaction) or absorbed (endothermic reaction). This energy,according to Einstein's Special Theory of Relativity comes from a change in therest mass of the reactants. This change in mass AM is related to the reactionenergy AE" by the famous relation

= AMc2. (4.1)

Here AM = Mjnitjai — Mfinaj, i.e., the loss of rest mass between the initial massof the reactants and the mass of the final products. If mass is lost in the reaction^initial > -^finab then AE > 0 and the reaction is exothermic. If mass is gained inthe reaction Mflnaj > Mjnitja], then AE < 0 and the reaction is endothermic.

Any reaction in which reactants ^4, B ... form the products C, D ... can be writ-ten symbolically as

A + B + >c + D + - - - . (4.2)

If Mi represents the mass of the ith component, the mass change in the reaction is

AM = (MA + MB + •••)- (Mc + MD + • • • ) • (4.3)

The energy emitted is then given by Eq. (4.1).In principle, the energy emitted (or absorbed) in any reaction could be computed

from Eq. (4.1). However, only for nuclear reactions is it feasible to calculate the AEfrom AM. To illustrate, consider a familiar exothermic chemical reaction involvingthe rearrangement of atomic electrons:

C + O2 —> CO2, A#293K = -94.05 kcal/mol, (4.4)

This says1 mol C + 1 mol O2 —>1 mol CO2 + 94050 cal. (4.5)

Since one mole contains Na = 6.022 x 1023 entities, the creation of one moleculeof CO2 from one atom of C and one molecule O2 liberates AE = — AH2Q3K/Na =1.56 x 10~19 cal = 4.08 eV. The energy emitted in this chemical reaction is typicalof almost all chemical reactions, i.e., a few eV per molecule.

Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.

How much of a mass change occurs in the formation of a CC>2 molecule? Fromthe energy release of AE = 4.08 eV

( eV \ f /MeV\ K ( eV \AM = 4.08 ^ x 931 x 106 -^-:

\CO2 molec.y [ \amu / \MeV/

amu= 4.4 x 1(

CO2 molec.

One molecule of CO2 has a mass of about 44 amu, and thus about 100(4.4 x10~9/44) = 10~8% of the reactant mass is converted into energy. To measure sucha mass change would require the atomic and molecular masses to be correct toabout 10 significant figures, far more accuracy than present technology allows.

For more macroscopic reactions, such as the deformation of a bullet penetratinga steel plate, even greater, and unrealistic, accuracy in the reactant masses wouldbe needed in order to use Eq. (4.1) to calculate the energy released. However,for reactions in which nuclei are altered, typically a million times more energy isabsorbed (or emitted) compared to a chemical reaction. Thus for nuclear reactions,nuclear masses correct to only 4 or 5 significant figures (well within reach of currenttechnology) are needed.

4.1 Binding EnergyA special reaction is one in which two or more entities A,B,... come together toform a single product C. i.e.,

A + B + > C. (4.6)

The energy emitted in such a reaction is called the binding energy, abbreviated BE.Should the BE turn out to be negative, then energy is absorbed (endothermic reac-tion) in the formation of C, while a positive BE indicates an exothermic reaction.Alternatively, the BE can be viewed as the energy required to divide C into itsconstituents (A B , . . . ) ,

As in any reaction, the BE arises because of a change in the mass of the reactants.The mass defect AA7/ is defined as the difference between the sum of initial massesarid the sum of the final masses. Here

AM = [mass A + mass B + • • • ] - mass C. (4.7)

The binding energy can then be computed as BE = AM c2, provided the masses ofthe reactants are known with sufficient accuracy.

4.1.1 Nuclear and Atomic MassesFirst some notation: The rest mass of an atom is denoted by M(^X) while that ofits nucleus is denoted by ra(^X). A few frequently used symbols are also used forconvenience: the neutron rest mass rnn = ra(gn), the proton rest mass rnp = ra(jH),and the electron rest mass me = m(_°e). NOTE: the data in Appendix B are forthe atomic rest masses M(^X) of all the known isotopes, not the nuclear masses.


An atom of ^X is formed by combining Z electrons with its nucleus to form theneutral atom. As these electrons bind to the nucleus, energy is emitted equal to thebinding energy BEz& of the electrons. To remove all the electrons from the atomwould thus require an ionization energy of BE^e. This electron binding energycomes from a decrease in the mass of the atom compared to the sum of nuclearand electron masses, by an amount BE^e/c

2. Thus, atomic and nuclear masses arerelated by

m(^X) + Zme - 5 . (4.8)

The binding energy term in this relation is often neglected since it is alwaysvery small compared to the other terms. For example, it requires 13.6 eV to ionizethe hydrogen atom. This electron binding energy represents a mass change ofBEie/c2 = 13.6 (eV)/9.315 x 108 (eV/u) = 1.4 x 1(T8 u. This mass change isnegligible compared to the mass of the hydrogen atom and nucleus (each about 1u) and even the electron (about 5.5 x 10~4 u).

4.1.2 Binding Energy of the NucleusTo examine the energies involved with nuclear forces in the nucleus, consider thebinding energy of a nucleus composed of Z protons and N = A — Z neutrons. Theformation of such a nucleus from its constituents is described by the reaction

Z protons + (A - Z) neutrons — > nucleus(^X) + BE. (4.9)

The binding energy is determined from the change of mass between the left andright-hand sides of the reaction, i.e.,

Mass Defect = BE/c2 = Z mp + (A - Z)mn - m(^X) (4.10)

where ra(^X) is the nuclear mass. However, masses of nuclei are not available;only atomic masses are known with great accuracy (see Appendix B). To expressEq. (4.10) in terms of atomic masses, use Eq. (4.8) to obtain

55 = z M(}H) -me + 51 + (A - z)mn

= ZM(}H) + (A- Z)mn - M(^X) + [ZEEle - BEZe]. (4.11)

The last term is the equivalent mass difference between the binding energies ofZ hydrogen electrons and the Z electrons to the nucleus of interest. These electronbinding energies are generally not known. However, the term involving electronbinding energies can be ignored for two reasons: (1) the two electron binding ener-gies tend to cancel, and (2) electron binding energies are millions of times less thanthe nuclear binding energy. Thus, neglect of the last term gives

BE(^X) = [ZM(}H) + (A - Z}mn - (4.12)


Example 4.1: What is the binding energy of the nucleus in ^He? From theatomic mass data in Appendix B, the mass defect is

Mass Defect = BE/c2 = 2M(\H) + 2mn - M(|He)

= 2(1.0078250) + 2(1.0086649) - 4.0026032 = 0.0303766 u

ThusBE(tHe) = Mass Defect (u) x 931.5 (MeV/u) = 28.30 MeV.

Notice the manner in which mass is converted to equivalent energy. Simplymultiply the mass deficit in u by the conversion factor 931.5 MeV/u. This is fareasier than multiplying a mass deficit in kg by c2 and going through the necessaryunits conversion.

4.1.3 Average Nuclear Binding EnergiesThe larger the BE of a nucleus, the more energy is needed to break it up into itsconstituent neutrons and protons. In Fig. 4.1, the average nuclear binding energyper nucleon (total BE divided by the number of nucleons A) is shown for all thenaturally occurring isotopes. The value of BE/A. is a stability measure of thenucleus; the larger BE/A, the more energy is needed, per nucleon, to tear apart thenucleus. From Fig. 4.1, we see that there is a broad maximum in the BE/ A curvearound A ~ 60 (e.g., Cr, Mn, Fe isotopes).

From the expanded portion of the BE/ A curve (left-hand part of Fig. 4.1, wesee that several light nuclei have particularly high average ~BE/A values. These areisotopes whose nuclei are multiples of the |He nucleus (i.e., an alpha particle). Thisobservation implies that inside a nucleus, the nucleons tend to form alpha- particlegroupings.

The BE/A versus A curve immediately suggests two ways to extract energy fromthe nucleus.

FusionIf two light nuclei (A < 25) are joined (or fused) to form a single heavier nucleus,the average BE per nucleon will increase. This increase in BE/A times the numberof nucleons A then equals the energy that must be emitted. For example, considerthe fusion reaction in which two deuterium nuclei are combined to form a heliumnucleus, namely,

The energy released can be computed in several equivalent ways.

Method 1 (from the change in BE): From Eq. (4.12) one finds that BE(?H)=2.225 MeV and BE(fHe) = 28.296 MeV. Thus the increase in the binding energy ofthe 4 nucleons involved is 28.296 - 2(2.225) = 23.85 MeV. This is the energy Efusion

emitted in this particular fusion reaction.


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Method 2 (from the change in mass): The fusion energy released Elusion mustequal the energy equivalent of the decrease in mass caused by the formation of f He.

^fusion = [2m(2H) - m(^He)] (u) x 931.5 (MeV/u).

= [2(M(2H) - me + BEie/c2) - (M(|He) - 2me + BE2e/c

2)] x 931.5.

~ [2M(2H) - M(|He)] x 931.5.

= [2(2.014102) - 4.002603] x 931.5.

= 23.85 MeV.

This release of energy through the fusion of two light nuclei is the mechanismresponsible for energy generation in stars. Moreover, all the light elements abovelithium and below about iron have been formed by fusion reactions during thenormal lifetime of stars. In each exothermic fusion reaction, heavier products areformed that move up the BE/-A curve. The heavier elements above iron have alsobeen formed in stars, but only during the final moments of their cataclysmic deathas an exploding nova. The tremendous energy released during a nova explosionallows intermediate mass nuclides to fuse (an endothermic reaction) to create theelements beyond iron.

FissionAn alternative way of extracting energy from a nucleus is to start at the upper end ofthe BE/^4 curve with a very heavy nucleus, such as "^U, and split (or fission) it intotwo lighter nuclei. To obtain an approximate idea of how much energy is releasedin a fission event, we see from Fig. 4.1 that if a nucleus of mass number 235 is splitinto two nuclei with A ~ 117, the BE per nucleon increases from about 7.7 to 8.5MeV/nucleon. Thus, the total fission energy released is about 235 x (8.5 — 7.7) ~ 210MeV.

Chemical versus Nuclear ReactionsThe formation of a single molecule of CC^ from carbon burning in oxygen releases4.08 eV, typical of chemical reactions. By contrast, a nuclear fusion and fissionevent releases 5 million to 50 million times this energy, respectively. This hugeconcentration of nuclear energy gives nuclear power plants a major advantage overfossil-fuel power plants. For the same power capacity, a nuclear power plant con-sumes only about one millionth the mass of fuel that a fossil-fuel plant consumes.For example, a 1000 MW(e) coal-fired power plant consumes 11 x 106 kg of coaldaily. By contrast, a nuclear power plant of the same capacity consumes only 3.8kg of 235U daily.

4.2 Nucleon Separation EnergyClosely related to the concept of nuclear binding energy is the energy required toremove a single nucleon from a nucleus. Consider the addition of a single neutronto the nucleus of ^X, i.e.,


The energy released in this reaction, Sn(^X), is the energy required to remove (orseparate) a single neutron from the nucleus ^X. This energy is analogous to theionization energy required to remove an outer shell electron from an atom. Theneutron separation energy equals the energy equivalent of the decrease in mass forthe reaction, namely

[ m - X ) + mn - m(^X)]c2 ~ [ M ( X ) + mn - M(^X)]c2. (4.13)

This separation energy can be expressed in terms of nuclear binding energies. Sub-stitution of Eq. (4.12) into this expression yields

Sn(^X) = BE(^X) - BE(^-jX). (4.14)

A similar expression is obtained for the energy required to remove a single pro-ton from the nucleus of ^Y, or equivalently, the energy released when a proton isabsorbed by a nucleus Î^X, i.e.,

The energy equivalent of the mass decrease of this reaction is

JX) + mp - m(^Y)]C2

}X) + M(}H) - M(^Y)]c2 = BE(^Y) - BE^lJX). (4.15)

The variation of Sn and Sp with Z and TV provides information about nuclearstructure. For example, nuclides with an even number of neutrons or protons havehigh values of Sn and Sp, respectively, indicating that stability is increased whenneutrons or protons are "paired" inside the nucleus.

Example 4.2: What is the binding energy of the last neutron in 1gO? This isthe energy released in the reaction

Ô + Jn—îO.

Then, from Eq. (4.13),

Sn = [m(1 |0)+mn-m(1iO)]c2

= [15.0030654 + 1.00866492 - 15.9949146] u x 931.5 MeV/u

= 15.66 MeV.

This is an exceptionally high value for a single nucleon. The average bindingenergy per nucleon for 1|O is 7.98 MeV. This large result for Sn indicates that1gO is very stable compared to 1gO.


4.3 Nuclear ReactionsNuclear reactions play a very important role in nuclear science and engineering,since it is through such reactions that various types of radiation are produced ordetected, or information about the internal structure of a nucleus is gained. Thereare two main categories of nuclear reactions.

In the first category, the initial reactant X is a single atom or nucleus thatspontaneously changes by emitting one or more particles, i.e.,

X —>y l + Y 2 + ...

Such a reaction is called radioactive decay. As we have seen from the Chart of theNuclides, the vast majority of known nuclides are radioactive. We will examinein detail in the next chapter the different types of radioactive decay and how thenumber of radioactive atoms vary with time. For the present, we simply notethat radioactive decay is a particular type of nuclear reaction that, for it to occurspontaneously, must necessarily be exothermic; i.e., mass must decrease in the decayprocess and energy must be emitted, usually in the form of the kinetic energy ofthe reaction products.

In the second broad category of nuclear reactions are binary reactions in whichtwo nuclear particles (nucleons, nuclei or photons) interact to form different nuclearparticles. The most common types of such nuclear reactions are those in whichsome nucleon or nucleus x moves with some kinetic energy and strikes and interactswith a nucleus X to form a pair of product nuclei y and Y, i.e.,1

x + X —> Y + y.

As a shorthand notation, such a reaction is often written as X(x, y}Y where x andy are usually the lightest of the reaction pairs.

4.4 Examples of Binary Nuclear ReactionsFor every nuclear reaction, we can write a reaction equation. These reaction equa-tions must be balanced, just as chemical reactions must be. Charge (the number ofprotons) and mass number (the number of nucleons) must be conserved. The num-ber of protons and the number of neutrons must be the same before and after thereaction. We shall illustrate this with some examples of typical miclear reactions.

(a, p) reaction: The first nuclear reaction was reported by Rutherford. He bom-barded nitrogen in air with alpha particles (helium nuclei) and observed theproduction of protons (hydrogen nuclei),

—^JO + IH or VjNfop)1^.

The product of this reaction is 1gO and there are nine protons and nine neu-trons on both sides of the equation, so the equation is balanced.

1 Although there are a few important nuclear reactions (e.g., fission) in which an interacting pairproduces more than two products, we restrict our attention, for the present, to those producingonly a pair of product nuclei.


(a, n) reaction: In 1932, Chadwick discovered the neutron by bombarding beryl-lium with alpha particles to produce neutrons from the reaction

^He + ^Be — * ^C + Jn or 5Be(a,n)1|C.

(7, n) reaction: Energetic photons (gamma rays) can also interact with a nucleus.For example neutrons can be produced by irradiating deuterium with suffi-ciently energetic photons according to the reaction

7 + ?H-^;H + ;n or ?H(7,n)}H or ?H(7,p)Jn.

(p, 7) reaction: Protons can cause nuclear reactions such as the radiative captureof a proton by 7Li, namely

}H + I.Li — > ^ B e + 7 or

The product nucleus |Be is not bound and breaks up (radioactively decays)almost immediately into two alpha particles, i.e.,

(7, an) reaction: As an example of a reaction in which more than two productsare produced, a high-energy photon can cause 17O to split into 12C, an aparticle and a neutron through the reaction

7 + 8 — + e + n or

(n, p) reaction: Fast neutrons can cause a variety of nuclear reactions. For ex-ample, in a reactor core, fast neutrons can interact with 16O to produce 16N,which radioactively decays (half life of 7.12 s) with the emission of a 6.13-MeV(69%) or a 7.11-MeV (5%) photon. The radionuclide 16N is produced by thereaction

Jn + 'SO— ̂ fN + ip or ^Ofap)1?]*.

4.4.1 Multiple Reaction OutcomesIn general, more than one outcome can arise when a particle x reacts with a nucleusX . Let us consider a number of possible results when a neutron of moderately highenergy (a few MeV) reacts with a f|S nucleus. In the first place, there may be noreaction at all with the neutron being scattered elastically, i.e., the nucleus whichscatters the neutron is left unaltered internally. This elastic scattering reaction iswritten

On + 16S — > '16S + On.

This reaction is written as (n,n). An inelastic scattering reaction may occur, inwhich the fgS is left in a nuclear excited state, namely


This is an (n, n') reaction. The asterisk means that the fgS nucleus is left in anexcited state2 and the prime on the n' means that the neutron has less energy thanit would have if it had been elastically scattered. The incident neutron can beabsorbed and a proton ejected,

!n -L 32Q _, 32p i ITTO n + 16b ~* 151 + 1M'

which is called an (n. p) reaction. The neutron may also be simply captured resultingin the emission of a gamma ray from the product nucleus

!n _l_ 32Q _^ 33Q _1_ ̂O n+ 16S -> i6b + 7,

which is called a radiative capture reaction or simply an (n, 7) reaction in whichthe binding energy of the neutron to the fgS nucleus is emitted as a photon calleda gamma ray.

Elastic scattering and radiative capture are possible at all values of the incidentneutron energy. However, the neutron's incident kinetic energy must exceed certainthreshold energies to make the other reactions possible. If we use neutrons ofsufficient energy, all of the four possibilities, as well as others, can occur when webombard fJS with neutrons.

4.5 Q-Value for a ReactionIn any nuclear reaction energy must be conserved, i.e., the total energy includingrest-mass energy of the initial particles must equal the total energy of the finalproducts, i.e..

mlc2} (4.16)

where EI (E^) is the kinetic energy of the ith initial (final) particle with a rest massmi (m-) .

Any change in the total kinetic energy of particles before and after the reactionmust be accompanied by an equivalent change in the total rest mass of the particlesbefore and after the reaction. To quantify this change in kinetic energy or rest-masschange in a reaction, a so-called Q value is defined as

Q = (KE of final particles) — (KE of initial particles). (4.17)

Thus, the Q value quantifies the amount of kinetic energy gained in a reaction.Equivalently, from Eq. (4.16), this gain in kinetic energy must come from a decreasein the rest mass, i.e.,

Q = (rest mass of initial particles)c2 — (rest mass of final particles)c2

(4.18)

2 An atom whose nucleus is in an excited state has a mass greater than the ground-state atomicmass listed in Appendix B. The mass difference is just the mass equivalent of the nuclear excita-tion energy.


The Q value of a nuclear reaction may be either positive or negative. If the restmasses of the reactants exceed the rest masses of the products, the Q value of thereaction is positive with the decrease in rest mass being converted into a gain inkinetic energy. Such a reaction is exothermic.

Conversely, if Q is negative, the reaction is endothermic. For this case, kineticenergy of the initial particles is converted into rest-mass energy of the reactionproducts. The kinetic energy decrease equals the rest-mass energy increase. Suchreactions cannot occur unless the colliding particles have at least a certain amountof kinetic energy.

4.5.1 Binary ReactionsFor the binary reaction x + X —>• Y + y, the Q value is given by

Q = (Ey + EY) - (Ex + Ex] = [(mx + mx) - (my + mY)}c2. (4.19)

For most binary nuclear reactions, the number of protons is conserved so that thesame number of electron masses may be added to both sides of the reactions and,neglecting differences in electron binding energies, the Q-value can be written interms of atomic masses as

Q = (Ey + EY) - (Ex + Ex) = \(MX + Mx) - (My + MY)}c2. (4.20)

4.5.2 Radioactive Decay ReactionsFor a radioactive decay reaction X —> Y + y, there is no particle x and the nucleusof X generally is at rest so EX = 0. In this case

Q = (Ey + EY) = [mx - (my + my)]c2 > 0. (4.21)

Thus radioactive decay is always exothermic and the mass of the parent nucleusmust always be greater than the sum of the product masses. In some types ofradioactive decay (e.g., beta decay and electron capture), the number of protons isnot conserved and care must be exercised in expressing the nuclear masses in termsof atomic masses. This nuance is illustrated in detail in Section 4.6.1 and in thenext chapter.

4.6 Conservation of Charge and the Calculation of Q-ValuesIn all nuclear reactions total charge must be conserved. Sometimes this is notclear when writing the reaction and subtle errors can be made when calculating theQ-value from Eq. (4.18). Consider, for example, the reaction 1gO(n,p)1fN or

Jn + ^O—^jN + ip. (4.22)

One might be tempted to calculate the Q-value as

Q = {mn + M(1680) - M(!?N) - mp}c2.

However, this is incorrect since the number of electrons is not conserved on bothsides of the reaction. On the left side there are 8 electrons around the :oO nucleus,


while on the right there are only 7 electrons in the product atom ^N. In reality,when the proton is ejected from the 1|O nucleus by the incident neutron, an orbitalelectron is also lost from the resulting ^N nucleus. Thus, the reaction of Eq. (4.22),to conserve charge, should be written as

16i (4.23)

The released electron can now be conceptually combined with the proton to forma neutral atom of }H, the electron binding energy to the proton being negligiblecompared to the reaction energy. Thus, the reaction can be approximated by

and its Q-value is calculated as

Q = {mn + - M(}H)}c2.

(4.24)

(4.25)

In any nuclear reaction, in which the numbers of neutrons and protons are con-served, the Q-value is calculated by replacing any charged particle by its neutral-atom counterpart. Two examples are shown below.

Example 4.3: Calculate the Q value for the exothermic reaction |Be (a, n}l^Cand for the endothermic reaction 1|O(n, a)l^C. In terms of neutral atoms, thesereactions may be written as 466 + ^He —>• 1gC + on and 1|O + on —» 1eC + fHe.The Q-values calculations are shown in detail in the tables below using the atomicmasses tabulated in Appendix B.

Reactants |Be

2He

sum

Products ^C

^sum

Difference of sums

AtomicMass (u)

9.0121824.002603

13.014785

12.0000001.008664

13.008664

0.006121 u

X931.5 MeV/u = 5.702 MeV

16«o

Reactants 1|Olo-

sum

Products X|C

2He

sum

Difference of sums

AtomicMass (u)

15.9949151.008664

17.003579

13.003354

4.002603

17.005957

-0.002378 u

X931.5 MeV/u = -2.215 MeV

4.6.1 Special Case for Changes in the Proton NumberThe above procedure for calculating Q-values in terms of neutral-atom masses,cannot be used when the number of protons and neutrons are not conserved, and


a more careful analysis is required. Such reactions involve the so-called weak forcethat is responsible for changing neutrons into protons or vice versa. These reactionsare recognized by the presence of a neutrino v (or antineutrino 17) as one of thereactants or products. In such reactions, the conceptual addition of electrons toboth sides of the reaction to form neutral atoms often results in the appearance ordisappearance of an extra free electron. We illustrate this by the following example.

Example 4.4: What is the Q-value of the reaction in which two protons fuseto form a deuteron (the nucleus of deuterium fH). Specifically, this reaction inwhich the number of protons and neutrons changes is

iP + l p — > ? d + + ? e + i/. (4.26)

where +?e is a positron (antielectron). This particular reaction is a key reactionoccuring inside stars that begins the process of fusing light nuclei to release fu-sion energy. To calculate the Q- value of this reaction in terms of neutral atomicmasses, conceptually add two electrons to both sides of the reaction and, neglect-ing electron binding energies, replace a proton plus an electron by a }H atom andthe deuteron and electron by iH. Thus, the equivalent reaction is

. (4.27)

The Q-value of this reaction is then calculated as follows:

Q = {2M(jH) - M(?D) - 2me - mv}c

= {2 x 1.00782503 - 2.01410178 - 2 x 0.00054858}(u) x 931.5 (MeV/u)

= 0.0420 MeV.

where we have assumed the rest mass of the neutrino is negligibly small (if notactually zero).

4.7 Q-Value for Reactions Producing Excited NulceiIn many nuclear reactions, one of the product nuclei is left in an excited state,which subsequently decays by the emission of one or more gamma photons as thenucleus reverts to its ground state. In calculating the Q-value for such reactions,it must be remembered that the mass of the excited nucleus is greater than thatof the corresponding ground state nucleus by the amount E* /c2, where E* is theexcitation energy. The mass of a neutral atom with its nucleus in an excited stateis thus

M(^X*) =M(^X) + £*/c2 (4.28)

where M(^X) is the mass of ground state atoms given in Appendix B. An exampleof a Q-value calculation for a reaction leaving a product nucleus in an excited stateis shown below.


Example 4.5: What is the Q-value of the reaction 10B(n, ct)7Li* in which the7Li* nucleus is left in an excited state 0.48-MeV above its ground state? In termsof neutral atomic masses, the Q-value is

Q = [mn + M(105E) - M(^He) - M(sLi*)]c2.

The mass of the lithium nucleus with an excited nucleus M(3Li*) is greater thanthat for the ground state atom M^Li), whose mass is tabulated in Appendix B,by an amount 0.48 (MeV)/931.5 (MeV/u), i.e., M^Li*) = M(^Li)+0.48 MeV/c2.Hence

Q = [mn + M(1°B) - M(|He) - M(3Li)]c2 - 0.48MeV

= [1.0086649 + 10.0129370 - 4.0026032 - 7.0160040] u x 931.5 MeV/u-0.48 MeV

= 2.310 MeV.

BIBLIOGRAPHY

EVANS, R.D., The Atomic Nucleus, McGraw-Hill, New York, 1955; republished by KriegerPublishing Co., Melbourne, FL, 1982.

KAPLAN. I., Nuclear Physics, Addison- Wesley, Reading, MA, 1963.

MAYO, R.M., Nuclear Concepts for Engineers, American Nuclear Society, La Grange Park,IL, 1998.

SATCHLER, G.R., (Ed.), Introduction to Nuclear Reactions, 2nd ed., Oxford UniversityPress, Oxford, 1990.

SHULTIS, J.K. AND R.E. FAW, Radiation Shielding, Prentice Hall, Upper Saddle River,NJ, 1996.

PROBLEMS

1. Complete the following nuclear reactions based on the conservation of nucleons:

(b) ^

(c)

(d) (?

2. Determine the binding energy (in MeV) per nucleon for the nuclides: (a) 1|O,(b) 17A (c) iiFe, and (d) 2iU.

3. Calculate the binding energy per nucleon and the neutron separation energyfor ^O and


4. Verify Eq. (4.15) on the basis of the definition of the binding energy.

5. A nuclear scientist attempts to perform experiments on the stable nuclide le^e.Determine the energy (in MeV) the scientist will need to

1. remove a single neutron.

2. remove a single proton.

3. completely dismantle the nucleus into its individual nucleons.

4. fission it symmetrically into two identical lighter nuclides ifAl.

6. Write formulas for the Q-values of the reactions shown in Section 4.4. Withthese formulas, evaluate the Q- values.

7. What is the Q-value (in MeV) for each of the following possible nuclear reac-tions? Which are exothermic and which are endothermic?

7

Li +

8. Neutron irradiation of 6Li can produce the following reactions.

' 37Li

+ {p

What is the Q-value (in MeV) for each reaction?

9. What is the net energy released (in MeV) for each of the following fusionreactions? (a) f H + ?H —> iHe + Jn and (b) fH + fH —> ^He + on

10. Calculate the Q-values for the following two beta radioactive decays,(a) ??Na — 38C1 38Ar


04 nuclear energetics.pdf

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