modern physics || nuclear physics

CHAPTER 14

NuclearPhysics

“The people in particle physics have been preoccupied lately with the electroweaktheory. The work of understanding in detail the nature of the strong force has fallen tonuclear physicists.”

MalcolmMacFarlane

14.1 INTRODUCTIONThe histories of elementary particle physics and nuclear physics are intertwined. The rays emittedby radioactive substances were studied by many physicists in the early decades of the twentiethcentury. Using the first three letters of the Greek alphabet, Rutherford classified the differentforms of radiation by the letters, alpha (α), beta (β), and gamma (γ). We now know that α-radiation consists of the nuclei of the most common isotope of helium with two protons andtwo neutrons. β-radiation consists of electrons and positrons, and γ-rays are a very penetratingform of electromagnetic radiation having a very short wavelength.

Scattering experiments designed by Rutherford in 1911 showed that each atom has a smallnucleus at its center containing most of the atomic mass. Rutherford’s discovery lead Bohr, whoworked in Rutherford’s laboratory, to formulate his model of the atom. Important discoveriesmade since the pioneering work of Rutherford and Bohr include the discovery of the neutron byJ. Chadwick, the discovery of the positron by C. Anderson, and the first scattering experimentsusing accelerated beams of protons by J. Cockcoft and E. Watson. These three developments, allof which occurred in 1932, were important milestones in the effort to understand the atomicnucleus.

We begin this chapter by describing the composition of the atomic nucleus and the physicalproperties of nuclei such as their size and binding energy. Ensuing sections will be devoted toradioactive decay processes and to the nuclear shell model.

14.2 PROPERTIESOFNUCLEIAs discussed in the introduction to this book, the number of protons in a nucleus is referredto as the atomic number (Z), and the total number of protons and neutrons is referred to as theatomic mass number (A). We shall use the generic term nucleon to refer to a particle that maybe either a proton or a neutron. For a particular element, the different atomic species havingdiffering numbers of neutrons in their nucleus are referred to as isotopes of the element. Theatomic mass number of an isotope is indicated by a superscript, and the atomic number is

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389

CHAPTER 14Nuclear Physics

390

Table 14.1 Some of the Properties of a Number of Light Isotopes

Name Z A Atomic Mass (u) I Natural Abundance

H 1 1 1.007825 1/2 99.989%2 2.014102 1 0.011%

He 2 3 3.016029 1/2 1.37×10−4%4 4.002603 0 99.99986%

Li 3 6 6.015122 1 7.59%7 7.016004 3/2 92.41%

Be 4 9 9.012182 3/2 100%

B 5 10 10.012937 3 19.9%11 11.009306 3/2 80.1%

C 6 12 12.000000 0 98.93%13 13.003355 1/2 1.07%

N 7 14 14.003074 1 99.632%15 15.000109 1/2 0.368%

O 8 16 15.994915 0 99.757%17 16.999132 5/2 0.038%18 17.999160 0 0.205%

indicated by a subscript. Consider as an example the carbon isotope, 136 C. The atomic number

of this isotope is 6, indicating that it has six protons, and the atomic mass number is 13,indicating that the nucleus has thirteen nucleons and, hence, seven neutrons. In this notation,the α-particle being a helium nucleus with two protons and two neutrons is denoted by 4

2He.Table 14.1 gives some properties of a number of naturally occurring isotopes.

This table gives the atomic number (Z), atomic mass number (A), atomic mass in atomic massunits (u), and nuclear spin (I) of each isotope. The atomic mass of the most common occurringisotope of carbon (12

6 C) is by definition equal to 12.0. The natural abundance of each isotopeis given in the last column of Table 14.1.

14.2.1 Nuclear SizesPrecise information about the structure of nuclei can be obtained from high-energy electronscattering experiments. There is an obvious advantage in using charged leptons to probe nucleisince leptons interact mainly by the electromagnetic force, which is well understood. The wave-length associated with an electron is related to the momentum of the electron by the de Broglierelation, λ = h/p. For electrons to probe the inner structure of nuclei, the wavelength of theincident electrons must be comparable to the size of the nucleus.

Example 14.1

Find the kinetic energy of an electron with a de Broglie wavelength of 10 fm.

Solution

Using eq. (12.14) and the de Broglie relation, the energy and the wavelength of an electron are seen tobe related by the equation

E2 =(

hcλ

)2+ m2c4.

SECTION 14.2Properties of Nuclei

391

As we have discussed in the Introduction, the product of constants hc is equal to 1240 MeV · fm. Substi-tuting this value of hc and λ = 10 fm into the last equation, we obtain

E =√(124 MeV)2 + (0.511 MeV)2 ≈ 124 MeV.

The kinetic energy of an electron is equal to the difference between its energy and its rest energy

K .E. = 124 MeV − 0.511 MeV = 123.5 MeV.

Incident electrons must thus have kinetic energies of a few hundred million electron volts toprovide detailed information about the structure of nuclei.

Point charge

126 MeV

130110 907050� (degrees)

183 MeV10–4

10–3

10–2

10–1

1

10

d�

/dV

(fm

)2 /ra

dia

n2

FIGURE 14.1The differential cross-section

for the elastic scattering of

electrons from gold 19779 Au .

The results of electron scattering experiments usuallyare described by giving the differential cross-sectionas a function of the scattering angle. We recall thatthe differential cross-section is the number of parti-cles scattered in a particular angle divided by the fluxof the incident beam. The differential cross-section forthe elastic scattering of electrons from gold 197

79 Au isshown in Fig. 14.1. In this figure, the scattering to beexpected if the gold nucleus had a point structure isrepresented by a dashed line. As we would expect, thepoint nucleus, which is harder than an extended chargedistribution, produces a larger scattering cross-sectionfor all energies. The differences between the scatteringcross-sections for the extended charge distribution anda point charge becomes larger as the scattering angleincreases.

p2�

p1�

p1

p2

p1�

p2�

p1

p2

q

e

N

(a) (b)

q

e

N

FIGURE 14.2(a) Feynman diagram for the

scattering of an electron

from a point nucleus.

(b) Feynman diagram with

the form factor of the

nucleus represented by a

circle around the nuclear

vertex.

The scattering of an electron from a pointnucleus is described by the Feynman dia-gram shown in Fig. 14.2(a). In this dia-gram, the lower external line corresponds tothe nucleus, and the internal line denotedby q represents the momentum transferredfrom the incident electron to the nucleus.The effect of the charge distribution of thenucleus can be included in the Feynmandiagram by adding a function F(q) of themomentum transfer to the nuclear vertex.This function, which is called the nuclearform factor, is the Fourier transform of the nuclear charge distribution. In Fig. 14.2(b), the formfactor of the nucleus is represented by a circle around the nuclear vertex. Since the scatteringamplitude is proportional to the square of the Feynman amplitude, the scattering cross-sectionfor electrons scattered from an extended nucleus will be equal to |F(q)|2 times the cross-sectionof a point nucleus. The absolute value squared of the nuclear form factor is thus equal to theratio of the scattering cross-sections of the extended and point nuclei. Once the form factor ofthe nucleus is obtained from the scattering data, the charge distribution of the nucleus ρch(r)can be obtained by taking the inverse Fourier transformation.

Nuclei having only one or two nucleons or one or two holes outside closed shells have beenfound to be approximately spherically symmetric. The charge density, which for such nucleidepend only upon the distance from the center of the nucleus, is generally constant for some


392

distance and then falls off very quickly. A charge distribution of this kind can be described bythe function

ρch(r) = ρ0ch

1 + e−(r − R)/a , (14.1)

where R and a may be regarded as adjustable parameters, and ρ0ch is a normalization constant

determined by the condition

∫ρch(r)dV = 4π

∫ρch(r)r

2dr = Z.

FIGURE 14.3The charge distributions of

the light (16O ), medium

(109Ag ), and heavy (208Pb )

nuclei. (Data from

Hofstadter, R., 1963.)1 2 3 4 5 6 7 8 9 10

r (fm)

�ch

(r)2 /

rad

ian2

208Pb

109Ag

16O

0.02

0.04

0.06

By fitting the values of R and a in the empir-ical formula (14.1) to electron scatteringdata, R. Barrett and R. Jackson obtainedthe charge distributions of the light (16O),medium (109Ag), and heavy (208Pb) nucleishown in Fig. 14.3. The charge densities ofthese three nuclei all have the same gen-eral form with an approximately level innerportion and a thin shell region where thecharge falls off exponentially to zero. Theinner regions of the charge distributions ofoxygen and silver are higher than the centralportion of the charge distribution of lead.

Some indication of how protons are distributed within a complex nucleus can be inferredfrom the charge distribution. If the protons within a nucleus were point particles, the densityof protons ρp would be related to the charge density ρch by the formula ρch = eρp. To theextent that the isospin symmetry holds, and protons and neutrons are equivalent, the densityof nucleons within a nucleus would be related to the charge density by the formula

ρ(r) = (A/eZ)ρch(r). (14.2)

FIGURE 14.4The nucleon densities of16O , 109Ag , and 208Pb.

0.04

0.08

0.12

0.16

1 2 3 4 5 6 7 8 9 10r (fm)

� (r

) (fm

)–3

208Pb

109Ag

16O

We note that the factor, A/eZ in this formulahas the effect of lowering the inner regionof the density of light nuclei in relationto the density of heavy nuclei, which haverelatively higher numbers of neutrons. Thenuclear densities of 16O, 109Ag, and 208Pbare shown in Fig. 14.4. These curves showthat at the center of a nucleus the densityof nuclear matter is roughly the same for allnuclei. It increases with A, but appears toapproach a limiting value ρ0 of about 0.17nucleons per fm3 for large A.

The existence of a limiting value of thenuclear density ρ0 for large A is an impor-

tant result. Using this idea, we can obtain an approximate relationship between the atomicmass number A and the nuclear radius R. We set the product of the volume of a sphere ofradius R and the the nuclear density ρ0 equal to the atomic mass number A to obtain

(4π3

)R3ρ0 = A.


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Solving this equation for R and using the fact that ρ0 is equal to 0.17 fm−3, we obtain

R = 1.12A1/3 fm. (14.3)

This formula will be used in later sections to estimate the radius of nuclei having particularvalues of A.

14.2.2 BindingEnergiesThe protons and neutrons in the nucleus can be separated only by working against the strongattractive forces holding them together. The binding energy is the amount of work that wouldbe needed to pull the protons and neutrons in the nucleus entirely apart. We can calculate thebinding energy B(N, Z) of a nucleus A

ZX with Z protons and N neutrons by finding the differencebetween the total rest energy of the constituent protons and neutrons and the nucleus itself:

B(N, Z) = [Zmp + Nmn − mnuc(N, Z)

]c2. (14.4)

The quantity within square brackets in this equation is the mass that would be lost if thenucleus were to be assembled from its constituents. In eq. (14.4), the mass loss is convertedinto a binding energy by multiplying it by c2 according to Einstein’s formula, E = mc2. Theamount of energy corresponding to a single atomic mass unit is 931.5 MeV. Notice that theproton and the neutron, which have masses slightly larger than one atomic mass unit, haverest energies mc2 equal to 938.3 MeV and 939.6 MeV, respectively.

The masses of atoms are measured experimentally rather than the masses of the bare nuclei.Since the binding energy of the electrons in an atom is very much smaller than the bindingenergy of the nucleus, we can find the nuclear binding energy by calculating the difference inmass of the atomic constituents. In place of eq. (14.4), we thus write

B(N, Z) = [Zm(11H)+ Nmn − m(N, Z)

]c2, (14.5)

where m(11H) is the mass of a hydrogen atom, mn is the mass of a neutron, and m(N, Z) isthe mass of the atomic isotope A

ZX. Since the atom AZX has Z electrons, the rest energy of the

electrons in Z hydrogen atoms is equal to the electronic contribution to m(N, Z)c2.

Example 14.2

Using the atomic masses of the 11H and 4

2He isotopes given in Table 14.1 and the mass of the neutrongiven in Appendix A, calculate the binding energy of the 4

2He nucleus.

Solution

The reaction in which two hydrogen atoms combine with two neutrons to form the 42He atom is

211H + 2n → 4

2He.

The terms on each side of this equation contain two protons, two neutrons, and two electrons. To beconcrete, we write the mass under each term in the preceding reaction to obtain

211H + 2n → 4

2He

2(1.007825) u 2(1.008665) u 4.002602 u

The mass lost when two 11H atoms combine with two neutrons to form the 4

2He isotope is

m = 2(1.007825) u + 2(1.008665) u − 4.002602 u = 0.030378 u.


394

The binding energy of the helium nucleus is obtained by multiplying this mass loss by 931.5 MeV. Weobtain

B(2, 2) = 28.297 MeV.

The binding energies of some light nuclei are given in Table 14.2.

The binding energies of all of the nuclei in Table 14.2 can be calculated using the method wehave employed in Example 14.2, and the binding energy per nucleon can then be calculatedby dividing the total binding energy of the nucleus by the number of nucleons. We note thatthe binding energy per nucleon of some nuclei is larger than for the nuclei around them. Thisis true of the 4

2He, 126 C, and 16

8 O nuclei. As we shall find in a later section, the great stability ofthese nuclei can be understood by considering their shell structure.

We can see from Table 14.2 that the binding energy of the 84Be nuclei is 0.1 MeV less than the

binding energy of two 42He nuclei. So, the 8

4Be nuclei is unstable and eventually decays into two42He nuclei.

The binding energy of the last nucleon, which is given in the third column of Table 14.2,is calculated by taking the difference of the binding energy of a particular nucleus with anisotope having one fewer nucleon. We note that the binding energy of the last nucleon ofthe 5

2He nucleus, which is obtained by taking the difference of the binding energies of 52He

Table 14.2 Binding Energies of Some Light Nuclei

Binding Energy Binding EnergyBinding Energy of Last Nucleon per Nucleon

Nucleus (MeV ) (MeV ) (MeV )

21 H 2.22 2.2 1.1

31 H 8.48 6.3 2.8

42 He 28.30 19.8 7.1

52 He 27.34 −1.0 5.5

63 Li 31.99 4.7 5.3

73 Li 39.25 7.3 5.6

84 Be 56.50 17.3 7.1

94 Be 58.16 1.7 6.5

105 B 64.75 6.6 6.5

115 B 76.21 11.5 6.9

126 C 92.16 16.0 7.7

136 C 97.11 5.0 7.5

147 N 104.66 7.6 7.5

157 N 115.49 10.8 7.7

168 O 127.62 12.1 8.0

178 O 131.76 4.1 7.8


395

and 42He, is negative. The unstable 5

2He nucleus decays into a neutron and an α-particle (42He).

Except for 84Be mentioned earlier all the other nuclei in Table 14.2 are stable.

14.2.3 TheSemiempiricalMassFormulaAll nuclei have a shell structure, which contributes to their stability and influences the kindsof decay processes that can occur. The effects of the shell structure are superimposed on aslowly varying binding energy per nucleon. When we studied the spatial distribution of nuclearmatter within the nucleus, we found that nuclei all have an inner region where the density isapproximately uniform and a thin surface region where the distribution of nuclear matter fallsoff exponentially to zero. The same could be said of the drops of a liquid. The validity ofthinking of nuclei as drops of a liquid is made more precise by giving an empirical formulafor the binding energy of nuclei. With a few parameters, this formula fits the binding energyof all but the lightest nuclei to a high degree of accuracy. Following W.N. Cottingham andD.A. Greenwood, whose book is cited at the end of this chapter, we give the following versionof the formula for the binding energy:

B(N, Z) = aA − bA2/3 − dZ2

A1/3 − s(N − Z)2

A− δ

A1/2 , (14.6)

where A is the number of nucleons, N is the number of neutrons, and Z is the number ofprotons. The parameters a, b, d, s, and δ can be obtained by fitting the formula to the measuredbinding energies. The values of these parameters given by Handbuch der Physik, XXXVIII/1 are

a = 15.835 MeVb = 18.33 MeVd = 0.714 MeVs = 23.20 MeV

(14.7)

and

δ =

⎧⎪⎨⎪⎩

+11.2 MeV, odd-odd nuclei (Z odd, N odd)

0, even-odd nuclei (Z even, N odd or N even, Z odd)

−11.2 MeV, for even-even nuclei (Z even, N even).

(14.8)

If nuclear matter were entirely homogeneous, the number of nucleons in a nucleus would beproportional to the volume of the nucleus. In the analogy between nuclei and liquid drops, itis the atomic mass number (A) that is the analogue of the volume of the liquid. The term aA ineq. (14.6) depends upon A in the same way that the cohesive energy of a fluid depends uponthe volume of the fluid. Since the surface area of a sphere depends upon the radius squaredand the volume of a sphere depends upon the radius raised to the third power, the surface areaof a sphere depends upon the volume of the sphere raised to the two-thirds power. Hence, theterm bA2/3 is analogous to the surface energy of a liquid sphere. The surface tension of a liquidkeeps drops spherical when—as for the nucleus—there is no long-range attractive force likegravity. Protons and neutrons on the surface of a nucleus are bound by fewer particles to thecollective than protons and neutrons in the interior of the nucleus. This effect and the fact thatthe surface tension plays an important role in the breakup of a drop when it is distorted helpus understand why the surface term is negative.

The term −dZ2/A1/3 also has a simple explanation. It represents the electrostatic Coulombrepulsion between the positive charged protons in the nucleus. We have already mentionedthat heavy nuclei tend to have many more neutrons than protons. The reason for this is thatthe cumulative effect of the Coulomb repulsion among protons makes heavy nuclei having


396

comparable numbers of protons and neutrons unstable. We can easily estimate the effect ofCoulomb repulsion by using the expression for the nuclear radius obtained previously. Theelectrostatic energy of a uniform sphere of charge eZ and radius R is

ECoul = 35(eZ)2

(4πε0)R.

Substituting into this equation the approximate expression for the nuclear radius R given byeq. (14.3), we find that ECoul is of the form dZ2/A1/3, and we obtain the approximate valueof the constant d = 0.77 MeV, which is fairly close to the empirical value 0.714 MeV givenearlier.

Whereas the Coulomb term −dZ2/A1/3 discourages the formation of states with high numbersof protons, the term −s(N−Z)2/A discourages the formation of states having unequal numbersof protons and neutrons. This can be described as a statistical effect that depends upon theproperties of identical particles. Due to the Pauli exclusion principle, which prohibits twonucleons from being in the same quantum state, the average neutron-proton attraction in anucleus is greater than the average proton-proton and the neutron-neutron attractions. Stateshaving nearly equal values of Z and N have comparatively higher numbers of proton-neutronpairs and are more stable. The factor of A in the denominator of the term −s(N −Z)2/A ensuresthat this term depends linearly upon A for a fixed ratio of neutrons to protons.

The final term in the semiempirical formula (14.6) describes a pairing effect that can be impor-tant for light nuclei. The pairing term makes even-even nuclei more stable than their odd-oddcounterparts with the same A.

FIGURE 14.5The binding energy per

nucleon for stable nuclei.

(From W.N. Cottingham and

D.A. Greenwood, 2001.)

9.0

8.5

8.0

7.5160 2001208040

A

Z 5 82

Z 5 28

Z 5 20

N 5 126

N 5 82

N 5 50

B/A

(MeV

)

Figure 14.5 shows the binding energy pernucleon for stable nuclei. In this figure, thesmooth curve gives the value of B/A obtainedfrom the semiempirical formula (14.6), andthe dots represent experimental data. As weshall discuss in the next section, nuclei candecay in a number of different ways. Theycan emit a compact α-particle, which is a42He nucleus. They can emit a beta particle,which is an electron or a positron, or theycan emit a gamma ray. In drawing Fig. 14.5,we have chosen, for each value of Z, thenucleus that is the most stable with respect toβ-decay. Notice that there are some discrep-ancies between the mass formula and theexperimental data. These differences are dueprimarily to the nuclear shell structure, whichwill be discussed in a later section. More

sophisticated versions of the formula (14.6) include terms describing shell effects; however,these additional terms are relatively unimportant for nuclei heavier than neon with Z = 10.

Of the various contributions to the binding energy, only the bulk term is positive with thenegative surface and Coulomb contributions giving the largest reductions in the bulk value. Theinitial rise of B/A with A may be attributed to the fact the magnitude of the surface contributionto the binding energy decreases relative to the bulk contribution as the size of the nucleusincreases. As A and therefore Z increase further, though, the Coulomb term becomes moreimportant, producing a maximum on the curve.

A formula for the mass of an atom can be obtained by substituting the expression for thebinding energy B(N, Z) given by eq. (14.6) into eq. (14.5) and then solving for m(Z, N). We

SECTION 14.3Decay Processes

397

thus obtain the following empirical formula for the mass m(N, Z) of a neutral atom with atomicmass number A, Z protons, and N neutrons:

m(N, Z)c2 = [Zm(11H)+ Nmn]c2

− aA + bA2/3 + dZ2

A1/3 + s(N − Z)2

A+ δ

A1/2 . (14.9)

This formula is called the semiempirical mass formula.

Example 14.3

Using the semiempirical formulas, calculate the binding energy per nucleon and mass isotopes 5626Fe

and 20882 Pb.

Solution

For 5626Fe, A = 56, Z = 26, and N = 30. Substituting these values into eq. (14.6) with the values of the

parameters given by eq. (14.7), we obtain the binding energy B(N, Z) = 487.1696; MeV. The bindingenergy per nucleon is B(N, Z)/56 = 8.699 MeV.

The mass of 5626Fe can then be calculated using eq. (14.5). We obtain

m(N, Z) = Zm(11H)+ Nmn − B(N, Z)/c2

= 26 × 1.007825 + 30 × 1.008665 − 487.1696 MeV931.5 MeV/u

= 55.9404 u. (14.10)

This value agrees fairly well with the experimental value, 55.9349 MeV.

In exactly the same way, the binding energy of 20882 Pb is found to be 1624.7481 MeV or 7.811 MeV per

nucleon. The mass of the 20882 Pb isotope obtained by using the calculated value of the binding energy

together with eq. (14.5) is 207.9892123, which agrees well with the experimental value of 207.976627.Notice that the binding energy per nucleon we have obtained for 208

82 Pb is less than the binding energy pernucleon of 56

26Fe. The isotope of iron with A = 56 is the most stable isotope.

The importance of the empirical formulas for binding energy and mass is not that they enableus to predict new or exotic nuclear phenomena, but rather that they enable us to under-stand the properties of nuclei in simple physical terms. The empirical models give us someinsight as to which nuclear species should be stable and what decay processes are likely tooccur.

14.3 DECAYPROCESSESIn addition to the stable nuclei existing in nature, there are numerous other unstable nucleithat decay by emitting radiation. Radioactive isotopes can be found in metal ores, and manyother radioactive isotopes have been produced in modern accelerators. Like the cell mutations,which lead to cancer in our own bodies, the process by which the unstable nuclei of radioac-tive isotopes decay is random in nature. The number of nuclei decaying in a time interval isproportional to the time interval dt and to the number of nuclei present. Denoting the numberof nuclei that decay by dN, we may thus write

dN = −λN dt, (14.11)

where the proportionality constant, λ, is called the decay constant. The negative sign that occursin eq. (14.11) is due to the fact that the number of unstable nuclei decreases with time. The


398

significance of the proportionality constant λ can be found by rewriting eq. (14.11) in thefollowing form:

λ = −dN/Ndt

.

We thus see that λ is equal to the fractional number or probability that a nucleus decays pertime.

The solution of eq. (14.11) is

N(t) = N0 e−λt , (14.12)

where N0 is the number of nuclei at time t = 0. The decay rate can be obtained by taking thederivative of this last equation giving

R = −dNdt

= λN0 e−λt .

Notice that both the number of nuclei and the decay rate decay exponentially.

The decay of a radioactive isotope often is described in terms of its half-life (t1/2), which isdefined to be the length of time required for the number of radioactive nuclei to decrease tohalf its original value. Using eq. (14.12), we obtain the following equation for the half-life

12

= NN0

= e−λt1/2 .

This last equation can be written

eλt1/2 = 2.

Taking the natural logarithm of both sides of this equation and solving for t1/2, we obtain

t1/2 = ln 2λ

. (14.13)

Example 14.4

The isotope 23994 Pu has a half-life of 24,100 years. What percentage of an original plutonium sample will

be left after 1000 years?

Solution

Using eq. (14.13), the decay constant λ of 23994 Pu is found to be 2.876 × 10−5 years−1. Equation (14.12)

can be written

NN0

= e−λt .

Substituting the value of λwe have obtained and t = 1000 years into this last equation, we find that N/N0is equal to 0.972 or 97.2%.

One of the reasons that debates concerning the use of nuclear energy lead to controversy isthat experts in different fields have different ideas of the length of time over which they canmake predictions with a reasonable amount of confidence. Nuclear physicists and geologists


399

feel comfortable talking about a thousand years. However, the makeup of the human societysurrounding a waste disposal site will itself change as the buried isotopes decay. No sociologistwith any sense wants to talk about a thousand years!

The conservation laws considered in the previous chapter apply to the decay of unstable nucleijust as they apply to elementary particles. We consider now in turn the most important quantitiesthat are conserved in nuclear reactions, and how the conservation laws help us understandnuclear decay processes.

1. Conservation of energy. The total energy of the nuclei involved in a nuclear reaction mustbe equal to the total energy of the products of the nuclear reaction. For decay processes,this implies that the rest energy of the decaying nucleus must be greater than the totalrest energy of the products of the decay process. The energy liberated in a decay, whichis called the Q-value of the reaction, can be calculated just as we have calculated thebinding energy of nuclei by finding the mass loss in the process and converting the massloss into an energy.

2. Conservation of momentum. The condition that the momentum be conserved implies thatthe total momentum of the decay products must be equal to zero in the rest frame ofthe decaying nucleus.

3. Conservation of electric charge. The total electric charge before a nuclear reaction must beequal to the total charge after the reaction.

4. Conservation of atomic mass number. For nuclear reactions in which the only baryonsinvolved are protons and neutrons, the conservation of baryon number implies that thetotal number of nucleons must remain the same. This in turn means that the atomicmass number is conserved in the reaction.

14.3.1 AlphaDecayIn α-decay, an unstable nucleus disintegrates into a lighter nucleus and an α-particle

(42He

nucleus)

according to the formula

AZ XN → A−4

Z−2X′ + 42He.

The decaying nucleus X is called the parent nucleus. The nucleus X′, which is produced in thedecay, is called the daughter nucleus. In alpha decay, X′ has an atomic mass number that is fourless than the atomic mass number of the nucleus X that decayed, and X′ has an atomic numbertwo less than the atomic number of X.

The fact that α-decay occurs so commonly for the heavy elements can be understood by con-sidering Fig. 14.5, which shows how the binding energy per nucleon varies with atomic num-ber. Since the ratio B/A gradually decreases for heavy nuclei, a heavy nucleus can split intotwo smaller nuclei with a greater total binding energy. The breakup of a nucleus into two ormore smaller nuclei is called fission with α-decay being the most common fission process. Aswe found in Example 14.2, an α-particle, which is a 4

2He nucleus, has a comparatively largebinding energy of 28.2 MeV. The condition for a nucleus (A, Z) to decay by α-emission to anucleus (A − 4, Z − 2) is

B(A, Z) < B(A − 4, Z − 2)+ 28.3 MeV.

This condition is always satisfied for A sufficiently large. Of the nuclei beyond 20983 Bi in the

periodic table only a few isotopes of U and Th are sufficiently stable to have survived sincethe formation of the Earth. All other heavy nuclei are produced either by the decay of theseisotopes, or they have been produced artificially.


400

The energy liberated in α-decay, which is called the Q-value of the reaction, can be calculatedby finding the mass loss in the process and converting the mass loss into an energy. We have

Q = [m(X)− m(X′)− m(4

2He)]c2. (14.14)

The electron masses cancel as they did for calculating binding energies, and atomic masses mayhence be used to calculate the Q-value.

Example 14.5

Calculate the Q-value of the decay of 23892 U into 234

90 Th.

Solution

Using the masses of the 23892 U, 234

90 Th, and 42He nuclei given in Appendix B, we find that the mass loss in the

α-decay is 0.004585 u. The Q-value of the α-decay is found by multiplying the mass loss by 931.5 MeV/ugiving 4.27 MeV.

14.3.2 Theβ-StabilityValleyWe now consider β-decay processes in which a nucleus decays by emitting an electron or apositron. The simplest example of a β-decay process, in which an electron is emitted, is thedecay of the neutron into a proton, an electron, and an antineutrino described by the equation

n → p + e− + ν.

We recall that neutron decay occurs by the weak interaction and that the electron lepton num-ber (Le) is conserved since the electron has Le = +1 and the antineutrino has Le = −1.

Neutron decay can occur within a nucleus with Z protons and N neutrons leading to a nucleuswith Z + 1 protons and N − 1 neutrons:

AZ X →A

Z+1 X′ + e− + νe.

Since the daughter atom X′ has one more electron than the parent atom X and the mass of theneutrino can be neglected, the Q-value for the reaction can be written

Q = [m(AZX)− m(AZ+1X′)]c2, (14.15)

where m(AZX)

and m(AZ+1X′) are the masses of the parent and daughter atoms. The energy

release in the decay (the Q-value) appears as the kinetic energy of the electron, the energy of

FIGURE 14.6The kinetic energy spectrum

of electrons emitted in

β-decay.

N

Ek

Emax

the neutrino, and the recoil kinetic energy of the nucleus X′. Therecoil energy of the daughter nucleus is usually negligible forβ-decay processes, and, hence, the Q-value can be written

Q = Ke + Eν.

The kinetic energy spectrum of electrons emitted in β-decay isshown in Fig. 14.6. The electron has the maximum value ofkinetic energy when the energy of the neutrino is equal to zero.


401

Example 14.6

The 7732Ge isotope decays by two successive β-decay processes to 77

34Se. Calculate the Q-values for each ofthese decay processes. What are the maximum kinetic energies of the emitted electrons?

Solution

The formulas for the two decay processes are

7732Ge → 77

33As + e− + νe,

and

7733As → 77

34Se + e− + νe.

7734Se is the only stable isotope with A = 77.

The Q-value for each of these decay processes can be calculated using eq. (14.15). For the first decayprocess, the Q-value is

Q = [76.923549 u − 76.920648]931.5 MeV/u = 2.70 MeV;

for the second decay process the Q-value is

Q = [76.920648 u − 76.919915]931.5 MeV/u = 0.68 MeV.

The maximum values of the kinetic energy of the electrons emitted in these two decays is slightly less thanthe Q-values due to the small amount of kinetic energy carried off by the daughter nuclei.

Notice that for both of the β-decay processes in Example 14.6, the parent atom is more massivethan the daughter atom. This condition must be satisfied for Q-value to be positive and for theprocess to occur. In a β-decay process, the atomic mass number A remains the same, and thevalue of Z changes by one unit. We can express the mass of an atomic isotope as a function ofZ by replacing N by A − Z in the semiempirical mass formula (14.9) to obtain

m(N, Z))c2 =(Amnc2 − aA + bA2/3 + sA + δA−1/2

)− (

4s + (mn − mp − me)c2)Z +(

4sA−1 + dA−1/3)

Z2. (14.16)

This last equation can be written simply

m(N, Z)c2 = α− βZ + γZ2, (14.17)

where

α= Amnc2 − aA + bA2/3 + sA + δA−1/2

β= 4s + (mn − mp − me)c2

γ = 4sA−1 + dA−1/3.(14.18)

For a particular value of A, eq. (14.17) can be used to plot the atomic mass as a function of Z.In Fig. 14.7, we show the atomic masses of atoms with A = 64 relative to the atomic mass of6428Ni. Open circles indicate odd-odd nuclei and filled circles indicate even-even nuclei. The twocurves, which were drawn using the values of the parameters given by eqs. (14.18), (14.7), and(14.8) have minima near Z = 29. The position of the minima in the curves can be calculatedby taking the derivative of m(N, Z)c2 given by eq. (14.17) and setting the derivative equal to


402

FIGURE 14.7The atomic masses of atoms

with A = 64 relative to the

mass of 6428Ni . Open circles

indicate odd-odd nuclei and

filled circles indicate

even-even nuclei. (From W.N.

Cottingham and D.A.

Greenwood, 2001.)

10

5

0

26 27 28 29 30 31 32z

Rel

ativ

e at

omic

mas

s (M

eV/c

2 )

zero. This gives β/2γ . The displacement of the curve foreven-even nuclei relative to the curve for odd-odd nucleiis due to the pairing interaction, which according toeq. (14.8) is positive for odd-odd nuclei and negative foreven-even nuclei with the displacement of the two curvesbeing equal to 2|δ|A−1/2/c2 for each value of A.

For a beta-decay process in which an electron is emit-ted, the value of Z increases by one unit. We can depictβ-decay processes of this kind by giving the atomic massnumber A and the atomic number Z of the initial andfinal states as follows:

(A, Z) → (A, Z + 1)+ e− + νe.

Since the daughter atom has Z + 1 electrons and the mass of a neutrino can be neglected, thecondition that must be satisfied for this process to occur is

m(A, Z) > m(A, Z + 1),

where, as before, m(A, Z) and m(A, Z+1) are atomic masses. Of the atoms depicted in Fig. 14.7,only 64

27Co, which lies to the left of the minimum, can decay by β− emission. The 6427Co atom,

which has an odd-odd nucleus and is represented by an open circle in Fig. 14.7, decays intothe 64

28Ni atom with an even-even nucleus

6427Co → 64

28Ni + e− + νe.

Beta-decay processes for which a positron is emitted can be depicted

(A, Z) → (A, Z − 1)+ e+ + νe.

Since the daughter atom has Z − 1 electrons, the condition that must be satisfied for positronemission to occur is

m(A, Z) > m(A, Z − 1)+ 2mec2.

The condition that the mass of the daughter atom be less than the mass of the parent atomminus twice the rest mass energy of the electron is somewhat more stringent for positronemission than the condition for electron emission.

Positron emission causes the atomic number Z to decrease and thus causes the atoms in Fig. 14.7to move toward the left. The atom 64

27Co cannot decay by positron emission since that wouldhave the effect of reducing the atomic number Z and transforming 64

27Co into an atom withgreater mass. The atom 64

32Ge with even-even nucleus can decay by positron emission into theatom 64

31 Ga with odd-odd nucleus according to the following formula:

6432Ge → 64

31 Ga + e+ + νe.

We also note that the atom 6429Cu with odd-odd nucleus, which lies near the minimum in

Fig. 14.7, can emit an electron in decaying to 6430Zn or emit a positron in decaying to 64

28Ni. Thetwo atoms, 64

28Ni and 6430Zn, which lie near the minimum of the curve shown in Fig. 14.7 are

stable with respect to β-decay. As can be seen by considering Fig. 14.7, a β-decay of either ofthese two nuclei would result in an atom with greater atomic mass.


403

20

40

60

80

100

120

140

10080604020

50

82

126

N5

Z

Atomic number Z

Neu

tron

num

ber

N

FIGURE 14.8The nuclei (indicated by

black squares), which are

observed to be stable with

respect to β-decay.

The atoms, whose nuclei are observed to bestable with respect to β-decay, are denotedby dark squares in Fig. 14.8. To a very goodapproximation, the bottom of the β-decay val-ley indicated by dark squares in this figurecorresponds to the value Z = β/2γ obtainedpreviously in conjunction with eq. (14.17).Atoms with constant A, which are connectedby β-decay processes, lie on straight lines withN + Z = A. These lines are perpendicu-lar to the line N = Z shown in Fig. 14.8.

14.3.3 GammaDecayThe processes of α- and β-decay usually lead tonuclei that are in excited states. As for excitedatomic states, the excited states of nuclei candecay into lower-lying states by emitting elec-tromagnetic radiation. Since the separation ofthe energy levels of nuclei is much larger thanfor atoms, the radiation emitted by excitednuclei is much more energetic being in theform of γ-rays. Figure 14.9 shows a typicalβ-decay process in which 60

27Co decays into 6028Ni by emitting an electron. As shown in the

figure, the excited 6028Ni nucleus subsequently decays into the ground state by emitting a γ-ray.

6028 Ni

6027 Co

�2

2.506 MeV

FIGURE 14.9A β-decay process in which6027Co decays into 60

28Ni and

the excited 6028Ni nucleus

subsequently decays into

the ground state by emitting

a γ-ray.

Example 14.7

Calculate the mass of the excited 6028Ni nucleus produced in the

β-decay process shown in Fig. 14.9.

Solution

As can be seen in Fig. 14.9, the energy of the excited state of6028Ni produced by the β-decay process lies 2.506 MeV above theground state. The mass of the excited nucleus thus exceeds themass of the unexcited nucleus by

m = 2.506 MeV931.5 MeV/u

= 0.002690 u.

The mass of the excited nucleus is thus

m = 59.930791 u + 0.002690 u = 59.933481 u.

Example 14.8

Calculate the Q-value of the initial β-decay process shown in Fig. 14.9.

Solution

The Q-value of the β-decay can be calculated using eq. (14.15). We obtain

Q = [m(6027Co

)− m(6028Ni

)]c2

= (59.933822 u − 59.933481 u) (931.5 MeV/u)= 0.318 MeV.


404

Apart from a small correction for the kinetic energy of the recoiling nucleus, the Q-value of the reactiongives the total kinetic energy available to the emitted electron and neutrino.

14.3.4 Natural RadioactivityOf the elements in the periodic table, only hydrogen, helium, and a very small amount oflithium can be traced back to the early universe. All other elements were produced in nuclearfusion reactions in stars. The nuclear reactions in stars are the source of the immense amountof energy they radiate and produce all the elements in the periodic table up to iron. As wehave seen, iron has the most stable nucleus and does not fuse with other nuclei to form heav-ier nuclei. All the elements heavier than iron were produced in the supernova explosions ofmassive stars and of binary stars. Supernovas generate a diverse variety of heavy elementsand seed nuclear matter back into space where it can be incorporated into the formation ofnew stars.

The nuclear reactions in stars produce both stable and unstable nuclei. Unstable nuclei decay byα-,β-, and γ-emission into other nuclei, which in turn can also decay. Since most decay processeshave half-lives of a few days or a few years, most radioactive isotopes that were present whenthe Earth formed about 4.5 billion years ago have since decayed into stable elements. However,a few of the radioactive elements produced long ago have half-lives that are comparable to theage of the Earth and still continue to decay.

FIGURE 14.10Members of a natural

sequence of radioactive

decay processes, which lead

ultimately to stable isotopes.

235

231

227

223

219

215

211

207

80 85 90 95z

A

207Ti207Pb

211Po

213Bi

215Po

219Rn

223Ra

227Th

231Pa

235U

231Th

223Fr

227Ac

211Pb

The process of α-decay reduces theatomic mass number A of a radioac-tive isotope by four units and reducesthe atomic number Z by two units;β-decay leaves the atomic mass num-ber unchanged and alters the atomicnumber by one unit. The radioac-tive decay processes that are observedin nature are members of decaysequences, which lead ultimately tostable isotopes. An example of aradioactive sequence is shown inFig. 14.10. In this sequence, theradioactive isotope 235

92 U decays by α-emission to the 231

90 Th isotope, which,in turn, decays by β-emission to23191 Pa isotope and then byα-emission

to 22789 Ac. As can be seen in Fig. 14.10, the entire sequence terminates with the stable iso-

tope 20782 Pb. Three radioactive sequences can be found in nature. Each of these sequences begins

with a relatively long-lived isotope, proceeds through a considerable number ofα- andβ-decays,and ends with a stable isotope.

14.4 THENUCLEARSHELLMODELAlthough the semiempirical theory for the binding energy and the mass of nuclei is very success-ful, experimental atomic masses show deviations from the semiempirical mass formula thatare quantum mechanical in nature. Just as the atom can be described as electrons moving inan average central field due to the nucleus and the electrons, the nucleus itself can be describedas protons and neutrons moving in a field due to both the strong and the electromagneticforces. Because nucleons move in a finite region of space with definite values of the angularmomentum, the table of nuclides shows recurring patterns that are very similar to the pattern

SECTION 14.4The Nuclear Shell Model

405

of atomic elements described by the periodic table. As for atomic systems, the description ofthe nucleus in terms of the angular momentum of individual nucleons is called the nuclear shellmodel.

14.4.1 NuclearPotentialWellsWe begin our description of the shell model by considering the potential energy of nucleons.It is natural to expect that the form of the potential energy in which nucleons move is similarto the distribution of nuclear matter, which is approximately constant near the center of thenucleus and then falls off rapidly to zero. The decline of the nuclear potential energy in thesurface region of the nucleus is actually more rapid than the decline of the nuclear densitysince tunneling of nucleons into the classically forbidden region causes the nuclear density nearthe surface to be more diffuse.

Sn Sp VC(r)

V—

(a) (b)

FIGURE 14.11Potential energy curves for

(a) neutrons and (b) protons.

An illustration of the potential energy ofneutrons and protons is given in Figs.14.11(a) and (b). As shown in Fig. 14.11(a),the potential energy for neutrons is constantnear the center of the nucleus and then risesto a finite value at the boundary. The poten-tial energy for protons shown in Fig. 14.11(b) has a similar contribution due to the strong force,and it has a Coulomb part due to the Coulomb repulsion among the protons. The potentialenergy of protons also has a constant term stemming from the statistical larger effect of theneutron-proton interactions than the neutron-neutron and proton-proton interactions. Sincestable nuclei generally have more neutrons than protons, the average number of neutrons seenby each proton is larger than the average number of protons seen by each neutron.

Each proton in a nucleus with atomic number Z interacts by the Coulomb interaction withZ − 1 other protons. The potential energy due to this interaction can be estimated by findingthe electrostatic potential energy of a positive charge +e due to a sphere of radius R in which acharge of (Z−1)e is uniformly distributed. This gives the following expression for the Coulombenergy of a proton

VC(r) =⎧⎨⎩(Z−1)e2

4πε0R0

[32 − r2

2R20

], r ≤ R0

(Z−1)e2

4πε0r , r > R0,

where R0 is the nuclear radius. In Fig. 14.11(b), the contribution to the potential energy ofa proton due to the Coulomb interaction VC(r) is shown and the statistical effect mentionedpreviously due to proton-neutron interactions is denoted by V . The Coulomb potential VC andthe contribution V both make the potential well of the protons more shallow.

The highest occupied energy levels for neutrons and protons must be the same for the nucleusto be stable with respect to β-decay. If the height of the highest occupied neutron or proton levelwere to exceed the height of the other by more than about mec2, the nucleus could decay byβ-decay. The energy required to detach a neutron from the nucleus is called the neutron separationenergy Sn. Similarly, the energy to detach a proton from the nucleus is called the proton separationenergy Sp. The neutron and proton separation energies are shown in Figs. 14.11(a) and (b). Thelowest energy levels of nucleons are not very sensitive to the precise shape of the potential well,and are similar for finite and infinite wells.

14.4.2 NucleonStatesApproximate wave functions and energies for nucleons can be obtained by solving theSchrödinger equation of a nucleon moving in an infinite potential well. Since nucleons cannot


406

penetrate into the infinite wall of the well, the wave functions must vanish on the boundary.As for the states of the hydrogen atom described in Chapter 4, the wave functions of nucle-ons in spherical coordinates can be written as the product of a radial function and a sphericalharmonic

ψ(r, θ,φ) = R(r)Ylml(θ,φ),

where the orbital angular momentum quantum numbers, l and ml, have the values, l = 0, 1,2, 3, . . . , and ml = l, l − 1, . . . , −l + 1, −l.

The function Ylml(θ,φ) is an eigenfunction of the angular momentum operator l2 corre-

sponding to the eigenvalue l(l + 1)�2 and an eigenfunction of lz corresponding to the eigen-value ml�. The radial part of the wave function is denoted by R(r). The spectroscopic notation isused in nuclear physics as in atomic physics with the letters s, p, d, f , g, . . . being used to denotevalues l = 0, 1, 2, 3, 4, . . . . For each value of the angular momentum, though, the state withthe lowest energy for nuclear physics is denoted by the principal quantum number, n = 1, andthe next energy denoted by n = 2, and so forth. In this respect, the convention used in nuclearphysics differs from the convention in atomic physics where eq. (4.6) serves as the definitionof the principal quantum number.

Since the potential energy for neutrons is equal to zero inside the well, the Schrödinger equationfor neutrons can be written

− �2

2m∇2ψ = Eψ.

The Schrödinger equation may be expressed in terms of the radial and angular coordinatesusing the identity,

∇2ψ = 1r2

ddr

(r2 dR

dr

)Ylml

(θ,φ)− l(l + 1)r2 RYlml

(θ,φ), (14.19)

which is given in Appendix AA (on the web site). The derivation of this equation depends uponthe fact that the spherical harmonic Ylml(θ,φ) is an eigenfunction of the angular momentumoperator l2, corresponding to the eigenvalue l(l + 1)�2.

Using eq. (14.19), the radial function R(r) may be shown to satisfy the equation

− �2

2mr2

ddr

(r2 dR

dr

)+ l(l + 1)�2

2mr2 R = ER,

and this last equation may be simplified by dividing through with �2/2m and by evaluating

the derivatives. We thus obtain the following equation for the radial function of neutrons:

− d2Rdr2 − 2

rdRdr

+ l(l + 1)r2 = k2R, (14.20)

where

k2 = 2mE�2 . (14.21)

The possible energies of neutrons are thus related to the values of k2 for which eq. (14.20)has a solution satisfying the boundary conditions of a particle moving in a potential well. Thefact that the potential well for protons is more shallow has the qualitative effect of shifting theproton levels up slightly above the neutron levels.

SECTION 14.4The Nuclear Shell Model

407

Table 14.3 The Spherical Bessel Functionsjl (r) for l = 0, 1, 2

l jl (r)

0 j0(x) = sin xx

1 j1(x) = sin xx2 − cos x

x

2 j2(x) =(

3x3 − 1

x

)sin x − 3

x2 · cos x

1.0

0.5

0 5 10 15

j2(x)

j1(x)

j0(x)

FIGURE 14.12The spherical Bessel

functions jl (kr ) for

l = 0, 1, and 2.

The solutions of eq. (14.20), whichare finite at the origin, may beexpressed in terms of the sphericalBessel functions denoted jl(kr). Thespherical bessel functions for l = 0, 1,and 2 are given in Table 14.3 and illus-trated in Fig. 14.12.

Knowing the nuclear wave functions, we can now determine the sequence of energy eigenvaluesby imposing the boundary conditions. For l = 0, the radial wave functions can be written

R(r) = sin xx ,

where

x = kr.

The condition that the wave function goes to zero on the surface of the well when r = R0 issatisfied by requiring that x = kR0 = nπ, where n = 1, 2, 3, . . . . The constant k must then beequal to nπ/R0, and eq. (14.21) may be used to obtain the following sequence of eigenvaluesfor s-states

E(n, s) = �2k2

2m= n2h2

8mR20

.

For l = 0, there is thus a sequence of energy eigenvalues, which can be labeled by the integer nor by the corresponding value of kR0.

The procedure we have used to find the energy of s-states can easily be generalized. For anyvalue of l, the values of x = kR0, for which the functions jl(x) are zero, have been tabulated.This leads to well-defined values of k, which determine the energy of higher-lying states in thewell. Table 14.4 gives the lowest energies obtained in this way.

The values of x = kR0, for which the spherical Bessel functions jl(kr) are equal to zero, aregiven in the second column of Table 14.4. The first column of the table gives of values ofl in spectroscopic notation. We note again that the first quantum number associated with eachstate gives the number of the solution for that particular value of l. For instance, the states,1p and 2p, are the first and second states for which l = 1. This convention differs from theconvention used in atomic spectroscopy where the principal quantum number n is definedby eq. (4.6). In atomic physics, the value of n is at least one unit larger than the angularmomentum quantum number l. The first and second p-states for hydrogen are denoted by 2pand 3p, respectively. The third column in Table 14.4 gives the number of nucleons in each shell,and the fourth column gives the total number of nucleons up to that point.


408

Table 14.4 The Lowest Roots of the Spherical Bessel Functions jl (r)

State xnl Number of States Total Number

1s π 2 2

1p 4.49 6 8

1d 5.76 10 18

2s 2π 2 20

1f 6.99 14 34

2p 7.73 6 40

1g 8.18 18 58

2d 9.10 10 68

1h 9.36 22 90

3s 3π 2 92

2f 10.42 14 106

1i 10.51 26 132

3p 10.90 6 138

2g 11.70 18 156

14.4.3 MagicNumbersNuclei with filled shells are important for our understanding of nuclear physics because theyshould enable us to identify values of the atomic number Z and the number of neutrons N forwhich nuclei are particularly stable. There are many ways of studying excited states of nucleiand for determining which nuclear isotopes are stable. As we shall consider in a followingsection, nuclei can be studied by bombarding them with energetic protons or with deuterons.Scattering experiments of this kind enable physicists to study the energies that can be absorbedby nuclei. There is a good deal of experimental evidence indicating that nuclei having certainnumbers of protons and neutrons are especially stable. The values of Z and N for which nucleiare particularly stable are called magic numbers. The magic numbers are 2, 8, 20, 28, 50, 82, and126. Nuclei having Z or N equal to one of these numbers have properties reflecting the existenceof an energy gap between occupied and unoccupied states. For example, tin with Z = 50 hasten stable isotopes, and there are seven stable isotopes with N = 82. The stable 16

8 O isotopehas eight protons and eight neutrons, and the stable 208

82 Pb isotope has 82 protons and 126neutrons.

There is not a close correspondence between the experimentally determined magic numbersand the numbers appearing in the last column of Table 14.4. The numbers 18, 34, and 40,which correspond to filled shells, are not magic numbers, and the magic number 50 correspondsto a partially filled 1g shell. The crucial step in obtaining a correspondence between the nuclearshell structure and the magic numbers is to include the spin-orbit interaction. The effects of thespin-orbit interaction are relatively more important for nuclei than for atoms.

14.4.4 TheSpin-Orbit InteractionThe spin and orbital angular momentum states of any particle with spin s = 1/2 and orbitalangular momentum l > 0 can be combined to form states with the total angular momentum

SECTION 14.5Excited States of Nuclei

409

quantum number j = l ± 1/2. As discussed in Chapter 4, the spin-orbit interaction causes asplitting of these states according to the formula

hs-o|(sl)j >=⎧⎨⎩ξ

2 l�2, j = l + 12

− ξ2 (l + 1)�2, j = l − 12 .

The splitting of levels by the spin-orbit interaction is illustrated in Fig. 4.18.

The spin-orbit interaction thus splits the (4l +2) states with the quantum numbers n and l intoa sublevel corresponding to 2l + 2 states with j = l + 1/2 and another sublevel correspondingto 2l states with j = l − 1/2. For example, the nd levels, which corresponds to ten states, issplit into one sublevel with j = 5/2 corresponding to the six states and another sublevel withj = 3/2 corresponding to four states.

Experiment shows that the spin-orbit function ξ(r) for nuclei is negative so that the states withj = l+1/2 always have lower energy than the states with j = l−1/2. The effect of the spin-orbitinteraction upon the ordering of the energy levels of nuclei is most important for heavy nucleifor which the energy levels are closer together. A revised sequence of energy levels that includesthe effect of the spin-orbit interaction is given in Table 14.5.

With the introduction of the spin-orbit interaction, the magic numbers all correspond to thefilling of energy levels with particular values of the quantum numbers, s, l, and j. The energygaps that occur between shells are represented in Table 14.5 by leaving a row vacant after eachmagic number.

The shell model successfully predicts the angular momentum of nuclei in the ground state.Nuclei with even numbers of protons and even numbers of neutrons (even-even nuclei) haveangular momentum zero and even parity; nuclei with an even number of protons and an oddnumber of neutrons or vice versa (even-odd nuclei) have angular momentum and parity equalto that of the odd nucleon in the shell being filled. As for atomic shells, the parity of a singlenucleon with angular momentum l is even or odd depending upon whether (−1)l is even orodd. It is energetically favorable for pairs of protons and pairs of neutrons to form states havingzero angular momentum and even parity, so that the angular momentum and parity of thenucleus are equal to the angular momentum and parity of the unpaired nucleon. There are veryfew exceptions to this rule.

No simple empirical rule has been found that gives the angular momentum and parity of odd-odd nuclei. Odd-odd nuclei, though, are energetically disfavored and very rare with only fourbeing stable (2

1H, 63Li, 10

5 B, and 147 N). All other odd-odd nuclei undergo β-decay to become

even-even nuclei.

14.5 EXCITEDSTATESOFNUCLEI

pi, Ei

pf, Ef

x

y

�

PFIGURE 14.13Illustration of an experiment

in which a proton of known

momentum is scattered by a

nucleus.

Thus far we have considered the energy levelsof nuclei in their ground state. One method forstudying the excited energy levels of nuclei is toscatter protons of known momentum from thenuclei and observe the momenta of scattered pro-tons for different scattering angles. A scatteringexperiment of this kind is illustrated in Fig. 14.13.In this figure, the momenta of the incident andscattered protons are denoted by pi and pf , respectively, and the scattering angle is denotedby θ. Before the collision, the target nucleus is supposed to be at rest. The momentum of thenucleus after the collision is denoted by P.


410

Table 14.5 A Typical Sequence for the Filling of Energy Levels of Nuclei(From the book by W.N. Cottingham and D.A. Greenwood citedat the end of this chapter)

State xnl Neutron Proton

1s π 1s1/2 2 1s1/2 2

1p 4.49 1p3/2 6 1p3/2 61p1/2 8 1p1/2 8

1d 5.76 1d5/2 14 1d5/2 142s1/2 16 2s1/2 16

2s 2π 1d3/2 20

1d3/2 20

1f 6.99 1f7/2 28

2p3/2 32 1f7/2 28

2p 7.73 1f5/2 38

2p1/2 40 2p3/2 321g 8.18 1g9/2 50 1f5/2 38

2d5/2 56 2p1/2 402d 9.10 1g7/2 64

1h11/2 76 1g9/2 50

1h 3π 3s1/2 782d3/2 82 1g7/2 58

3s 9.42 2f7/2 90 2d5/2 64

1h9/2 100

2f 10.42 3p3/2 104

1i 10.51 1i13/2 118 1h11/2 76

2f5/2 124 2d3/2 80

3p 10.90 3p1/2 126 3s1/2 82

In Fig. 14.13, the direction of the incoming proton is denoted by x, and the transverse directionin the plane of the collision process is denoted by y. The equations describing the conservationof the x- and y-components of the momentum are

pi = pf cos θ + Px

0 = pf sin θ + Py .

Solving these equations for the components of the momentum of the nucleus, we get

Px = pi − pf cos θ

Py = −pf sin θ.

We now express the kinetic energy of the proton and the nucleus in terms of the incom-ing and outgoing momenta of the proton. Since the kinetic energy of the incident protons

SECTION 14.5Excited States of Nuclei

411

in the scattering experiments we will consider are around 10 MeV, nonrelativistic expressionsmay be used for the kinetic energy of the particles. The target nucleus is at rest before thecollision. So, the total kinetic energy before the collision is equal to the kinetic energy ofthe proton

Ei = 12

mpv2i = 1

2mpp2

i .

After the collision, the total kinetic energy is

Ef = 12mp

p2f + 1

2m∗A(pi − pf cos θ)2 + 1

2m∗A(pf sin θ)2,

where m∗A is the mass of the nucleus in its final state.

Since the total energy is conserved in the collision process, the energy absorbed by the nucleusis equal to the loss of kinetic energy, which we denote by E. Collecting together terms, weobtain

E = Ei − Ef = 12mp

p2i − 1

2mpp2

f − 12m∗

A(p2

i + p2f − 2pipf cos θ).

The terms involving the square of the momentum of the proton can now be grouped togetherto give

E = Ei

(1 − mp

m∗A

)− Ef

(1 + mp

m∗A

)+ 2mp

m∗A(EiEf )

1/2 cos θ. (14.22)

Here Ei and Ef are the initial and final kinetic energies of the proton and m∗A is the mass of the

nucleus in its final state. When the kinetic energy of the incident proton is less than 100 MeV,the mass of the nucleus in its final state m∗

A = mA + E/c2 may be replaced by the mass in theground state mA with little error.

3 4 5 6 7 8

8.197.536.616.234.933.853.543.50

Num

ber

of s

catt

ered

pro

tons

(arb

itrar

y sc

ale)

Proton energy Ef (MeV)

FIGURE 14.14Protons scattered within a

small angular range at

θ = 90◦ as a function of the

final energy Ef .

At a fixed scattering angle θ, theenergy of the scattered protons isno longer monoenergetic, but hasseveral well-defined values corre-sponding to the energy levels ofexcited nuclear states. As an exam-ple, we consider a scattering experi-ment performed by B. Armitage andR. Meads. In their experiment, pro-tons with an energy 10.02 MeV werescattered from 10

5 B. Figure 14.14shows the number of protons scat-tered within a small angular rangeat θ = 90◦ as a function of the final energy Ef . Using eq. (14.22), we can show that the highestenergy at 8.19 MeV corresponds to a kinetic energy loss E equal to zero and, hence, correspondsto elastic scattering. The peaks in Fig. 14.14 for lower energies correspond to excited statesof 10

5 B.


412

Example 14.9

Find the energy of the excited states of 105 B corresponding to the peaks in Fig. 14.14.

Solution

Substituting the values, Ei = 10.02 MeV, mp = 1.007276 u, and m(105 B

) = 10.012937 u, into eq. (3.5),the equation becomes

E = 9.0120 MeV − 1.1006 MeV Ef .

We may now substitute the energies Ef given along the bottom of Fig. 14.14 to obtain the correspondingvalues of E given in Table 14.6.

Table 14.6 Excited States of 105 B Calculated by Proton Scattering

Ef (MeV ) 8.19 7.53 6.61 6.23 4.93 3.85 3.54 3.50

E(MeV ) 0.0 0.72 1.74 2.16 3.59 4.77 5.12 5.16

FIGURE 14.15The ground state and the

lowest excited states of 105 B .

E(MeV)

2.16

1.74

0.72

0.0

3.59

4.77

5.125.16

The ground state and the lowest excited states of 105 B are shown in

Fig. 14.15.

Proton scattering experiments provide valuable information about excitednuclear states. Another technique, which can be used to determine theexcited energy levels of nuclei is that of deuteron stripping, in which anucleon of deuterium is taken away by the target nucleus. Consider forexample the following reaction

21H + A

ZX →A+1Z X∗ + p. (14.23)

In this reaction, a neutron is stripped from an incident deuteron by theAZX nucleus. The products of this reaction include the nucleus A+1

Z X∗and a proton. The asterisk on the product nucleus indicates that it maybe in an excited state. As for proton scattering experiments, the conser-vation of momentum and energy can be used to derive an expressionfor the energy of excited nuclear states in terms of the incident kineticenergy of the deuteron and the final kinetic energy of the proton (seeProblem 17).

The number of excited states that nuclei have increases with atomic num-ber. Deuterium, which is the lightest complex nucleus, has only one bound state and henceno excited spectrum. Very light nuclei have only a few excited states. However, the number ofexcited states increases rapidly as A increases. Figure 14.16 shows the lowest energy levels of twomirror nuclei, 11

5 B and 116 C. The nuclei are called mirror nuclei because the number of protons

in one is equal to the number of neutrons in the other and vice versa. The energy levels of thetwo nuclei are very similar. For these two nuclei, the difference in the Coulomb energies is verysmall and the effects of the strong force are almost identical. The similarity of the energy levelsof these two isotopes provides strong evidence for the isospin symmetry.

A qualitative understanding of the energy levels of 115 B and 11

6 C can be obtained from the shellmodel. Consider the 11

6 C nucleus, which has six protons and five neutrons. The six protons fill

SECTION 14.5Suggestions for Further Reading

413

(MeV)

22

22

2

2

1

1

2

1

22

8.67

7Li1 �

8.69

10B1p

7.55

7Be1 �

116 C11

5 B0.0

2.00

4.45

5.02

0.0

2.00

4.32

4.80

6.746.79

6.346.486.91

7.50

8.118.42

7.29

7.98

8.568.92

5 2

1 2

1 2

3 2

23 223 2

5 2

3 2

7 227 2

1 2

5 2

13 2

13 2

13 2

3 223 2

5 225 2

1 2

FIGURE 14.16The lowest energy levels of two mirror nuclei, 11

5 B and 116 C.

MeV

0

2

3

4

Increasingdensity

Increasingdensity

1

FIGURE 14.17The energy levels of 46

20Ca and 10882 Pb.

the 1s1/2 and 1p3/2 shells. Of the five neutrons, two fill the 1s1/2 shell and the other three are inthe 1p3/2 shell. As described earlier, two of the neutrons in the 1p3/2 level are coupled togetherto form a state having zero angular momentum and even parity with the remaining neutrongiving the ground state angular momentum and parity. The parity of a single p-function withl = 1 is odd. This simple way of thinking correctly identifies the angular momentum and parityof the ground states of 11

6 C and 115 B.

For the lowest excited state of 116 C nucleus, the odd neutron from the 1p3/2 shell is excited

into the 1p1/2 shell resulting in a nucleus with angular momentum 1/2 and odd parity. Wemight be inclined to describe the next two excited states, which have angular momenta 5/2and 3/2, as excitations of the 1p3/2 neutron to the 1d5/2 and 1d3/2 states. We note, however,that the next two excited states have odd parity, whereas the parity of a d-state with l = 2is even. The 5/2 and 3/2 states and many other excited states have more than one excitednucleon.

Figure 14.17 shows the energy levels of 4620Ca and 108

82 Pd. Heavier nuclei usually have a greaterdensity of excited states and the density of excited states generally increases as the excitationenergy increases.

SUGGESTIONSFORFURTHERREADING

W.N. Cottingham and D.A. Greenwood, An Introduction to Nuclear Physics, Second Edition(Cambridge, New York, 2001).

Samuel S.M. Wong, Introductory Nuclear Physics, Second Edition (Wiley, New York, 1998).

Kenneth S. Krane, Introductory Nuclear Physics (Wiley, New York, 1988).


414

BASICEQUATIONS

BindingEnergy

B(N, Z) = [Zm

(11H)+ Nmn − m(N, Z)

]c2,

where m(

11H)

is the mass of a hydrogen atom, mn is the mass of a neutron, and m(N, Z) is themass of an atomic isotope.

TheSemiempirical Formula

B(N, Z) = aA − bA2/3 − dZ2

A1/3 − s(N − Z)2

A− δ

A1/2 ,

where A is the number of nucleons, N is the number of neutrons, and Z is the number ofprotons. The parameters a, b, d, s, and δ can be obtained by fitting the formula to the measuredbinding energies.

MagicNumbers

The values of Z and N for which nuclei are particularly stable are

2, 8, 20, 28, 50, 82, and 126

SUMMARY

Information about the structure of nuclei can be obtained from high-energy electron scatteringexperiments. The results of these experiments show that the density of nuclear matter is roughlythe same at the center of all nuclei. The nuclear density increases with A, but appears to approacha limiting value of about 0.17 nucleons per fm−3 for large A. All nuclei appear to have aninner region where the density is approximately uniform and a thin surface region where thedistribution of nuclear matter falls off exponentially to zero.

The binding energy and mass of nuclei can be accurately described by empirical formulas.Superimposed upon the slowly varying properties of the nucleus described by the empiricalformulas are deviations that are quantum mechanical in nature. The protons and neutrons inthe nucleus move in a finite region of space with definite values of the angular momentum. Thedescription of the nucleus in terms of the angular momentum of individual nucleons, whichis known as the shell mode, makes it possible to understand the spin and parity of nuclei andto describe the spectra of excited states.

QUESTIONS

1. What kinds of high-energy scattering experiments are used to determine the structure ofthe nucleus?

2. How is the scattering cross-section related to the nuclear form factor?

3. Make a sketch showing how the density of nuclear matter varies with the distance r fromthe nuclear center.

4. Write down a formula expressing the nuclear binding energy B(N, Z) in terms of the massof the Z protons and N neutrons and the mass of the nucleus mnuc.

5. To how much energy does an atomic mass unit u correspond?

6. To what physical effect does the term −dZ2/A1/3 in the empirical mass formula correspond?

SECTION 14.5Problems

415

7. The formation of which states will be discouraged by the term −s(N−Z)2/A in the empiricalmass formula?

8. The amount remaining of a radiative sample decreases by 50% over a certain period oftime. By how much would the decay rate decline over the same period of time?

9. Is the atomic number of a daughter nucleus produced in the emission of an electron greaterthan or less than the atomic number of the parent nucleus?

10. Which particles are the primary recipients of the energy emitted in a beta-decay process?

11. To which region of the spectrum does the electromagnetic radiation emitted by nucleigenerally belong?

12. Why must the highest occupied energy levels of neutrons and protons in a nucleus be thesame?

13. What condition can we impose on the wave functions of nucleons at the nuclear radius R0?

14. Are the effects of the spin-orbit interaction relatively less important or more important fornuclei than for atoms?

15. Of the two nuclear states, j = l + 1/2 and j = l − 1/2, which has the lower energy?

16. To how many states do the 1d1/2 and 1d3/2 nuclear levels correspond?

17. What are the spin and parity of a nucleus having two protons and two neutrons?

18. What are the spin and parity of a nucleus having two protons and three neutrons?

PROBLEMS

1. How many protons and how many neutrons are there in the nuclei, 73Li, 63

29Cu, and 23892 U?

2. In a nuclear fission reaction, an atomic nucleus breaks up into two or more parts and alsoemits a number of neutrons. How many neutrons would be emitted in the fission reaction

23692 U → 90

36Kr + 14456 Ba?

3. Estimate the sizes of the nuclei 42He, 16

8 O, 5626Fe, 208

82 Pb, and 93237Np.

4. Using the mass of the neutron given in Appendix A and the atomic masses given inAppendix B, calculate the total binding energy and the binding energy per nucleon ofthe nuclei (a) 4

2He, (b) 168 O, (c) 56

26Fe, (d) 20882 Pb, (e) 238

92 U.

5. Find the energy needed to remove a proton from the nuclei (a) 42He, (b) 56

26Fe, (c) 20882 Pb.

6. Find the energy needed to remove a neutron from the nuclei (a) 42He, (b) 56

26Fe, (c) 20882 Pb.

7. Using the mass of the neutron given in Appendix A and the semiempirical formula (14.6),calculate the total binding energy and the binding energy per nucleon of the nuclei givenin Problem 4.

8. Using the semiempirical formula (14.6) without the pairing term, derive an explicit expres-sion for the binding energy per nucleon of a nucleus with atomic mass number A andatomic number Z = N = A/2. Show that the expression for the binding energy pernucleon you obtain has a maximum for Z = A/2 = 26.

9. A radioactive sample emits radiation 1200 times per second at t = 0. Its half-life is 2 min.What will its decay rate be after (a) 4 min, (b) 6 min, (c) 8 min?

10. Tritium (31H) has a half-life of 12.3 years. What fraction of the tritium atoms would remain

after 40 years?

11. The carbon isotope (146 C) is continuously produced in the atmosphere by the reaction

n +147 N → p +14

6 C,


416

where the neutron is due to cosmic rays. 146 C decays back to 14

7 N by the reaction

146 C → 14

7 N + e− + νe,

with a half-life of 5730 years. Since living organisms continually exchange carbon with theatmosphere, they have the same amount of the 14

6 C isotope in a given sample of carbon asdoes the atmosphere.

a. Using the fact that a gram of carbon in the atmosphere or in living organisms on theaverage emits 15.3 beta rays every minute, calculate the proportion of 14

6 C in carbon.b. What rate count would you expect from one gram of carbon extracted from a bone

fragment that was 20,000 years old?

12. Complete the following decay reactions and find out how much energy is released in eachreaction:

a. 20983 Bi → 205

81 Tl +b. 238

92 U → 23490 Th +

c. 7736Kr → 77

35Br +d. 77

35Br → 7734Se +

13. 74Be decays by capturing an orbital electron in the reaction

74Be + e− → 7

3Li + νe.

Find the energy released in the reaction.

14. a. Give possible decay reactions for each of the following unstable nuclei: 62He, 8

4B, 124 B,

158 O, 240

95 Am.

b. How much energy would be released in each process?

15. a. Find how much energy is released in the α-decay

23492 U → 230

90 Th + 42He.

b. How much of the total decay energy is carried off by the α-particle?

16. a. Give the angular momentum and parity of the ground state of the following nuclei:178 O, 17

9 F, 3115P, 31

16S, 3216S, 40

20Ca, 4521 Sc.

b. For each nucleus give a possible single-particle excited state.

17. For the deuteron stripping reaction (14.23), use the conservation of energy and momentumto derive an expression for the energy of excited nuclear states analogous to eq. (14.22) forproton-nuclear scattering.

18. In α-decay, the total kinetic energy of the α-particle and the daughter nucleus is

E = 12mα

pα2 + 12md

pd2.

Use the requirement that the total momentum is conserved in the decay process to showthat the total kinetic energy of α-particle and the daughter nucleus can be written

E = 12m

pα2,

SECTION 14.5Problems

417

where the reduced mass m is defined by the equation

m = mαmd

mα + md.

The total kinetic energy of the α-particle and the daughter nucleus has the form of thekinetic energy of a single particle with the reduced mass m.

19. Calculate the Q-value and mean lifetime of the double α-decay

84Be → 2

(42He

).

The observed mean lifetime of this decay is 2.6 × 10−17 s.

20. Calculate the Q-value and mean lifetime of the α-decay

23894 Pu → 234

92 U + 42He.

The observed mean lifetime of this decay is 128 years.

21. The isotope 19479 Au decays by positron emission

19479 Au → 194

78 Pt + e+ + νe

and by α-decay

19479 Au → 190

77 Ir + 42He.

Both the positron created in the first reaction and the α-particle created in the second reactionmust tunnel through a Coulomb barrier to escape. Calculate the Q-value and the mean lifetimesfor the two processes and explain the difference between the two lifetimes.

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