Chapter 30
`
`Nuclear Physics
`
`
`
`nuclei are stable because of the presence of anotherforce, the nuclear force. This
`force, which is short-range (about 2 fm), is an attractive force that acts betweenall
`nuclear particles. The protons attract cach other via the nuclear force, and at the
`same time they repel cach other through the Coulomb force. The nuclear force
`also acts between pairs of neutrons and between neutrons and protons.
`The nuclear force dominates the Coulomb repulsive force within the nucleus
`(at short ranges). If this were not the case, stable nuclei would not exist. Moreover,
`the strong nuclear force is independent of charge. In other words, the forces as-
`sociated with the proton—proten, proton-neutron, and neutron—neutron inter-
`actions are the same, apart from the additional repulsive Coulomb force for the
`proton—proton interaction.
`There are about 260 stable nuclei; hundreds of others have been observed but
`are unstable. A plot of Nversus Z for a numberof stable nuclei is given in Figure
`30.3, Note that light nuclei are most stable if they contain equal numbersofprotons
`and neutrons—thatis, ifN = Z—but heavy nuclei are more stable if N > Z. This
`can be partially understood by recognizing that as the number ofprotons increases,
`the strength of the Coulomb force increases, which tends to break the nucleus
`apart. As a result, more neutrons are needed to keep the nucleus stable, because
`neutrons experience onlythe attractive nuclear forces. Eventually, when Z = 83,
`the repulsive forces between protons cannot be compensated by the addition of
`more neutrons. In effect, the additional neutrons “dilute” the nuclear charge den-
`sity. Elements that contain more than 83 protons do not havestable nuclei.
`
`A plot of new
`Figure 30.3
`tron number, N, versus atomic
`be
`for thestable mu-
`
`(solid points). The dashed
`clei
`line corresponds to the condi-
`tion N= 2. Theshaded area
`
`shows radioactive nuc
`
` 0
`
`
`50 & 7 80
`40
`oo 2 30
`Proton number 2
`
`of
`
`30.1
`
`Some Properties of Nuclei
`
`895
`
`
`
`The possible orienta-
`Figure 30.4
`tions of the nuclear spin and its
`projections along the z axis for the
`case |
`
`
`It is interesting to note that most stable nuclei have even values of A. In fact,
`certain values of Zand Ncorrespond to nuclei with unusually highstability, These
`values of Nand Z, called magic numbers, are
`Zor N = 2, 8, 20, 28, 50, 82, 126
`
`[30.2]
`
`the helium nucleus (two protons and two neutrons), which has
`For example,
`#=2and N= 2,is verystable.
`
`Nuclear Spin and Magnetic Moment
`In Chapter 29 we discussed the fact that an electron has an intrinsic angular mo-
`mentum, whichis called its spin. Nuclei,
`like electrons, also have an intrinsic an-
`
`gular momentum that arises fromrelativistic properties. The magnitude of the
`nuclear angular momentum is ,/ /(/ + 1)f, where J is a quantum number called
`
`the nuclear spin and maybe anintegerora half-integer. The maximum compo-
`nent of the nuclear angular momentum projected along any direction is JA, Figure
`30.4 illustrates the possible orientations of the nuclear spin and ils projections along
`the z axis for the case where J = 3,
`The nuclear angular momentumhas a nuclear magnetic moment associated
`
`with it. The magnetic moment of a nucleus is
`asured in terms of the nuclear
`magneton, j,,, a unit of magnetic moment defined as
`
`eh
`
`me
`
`= 5,05 X 10°77 J/T
`
`2m,
`
`[30.3]
`
`This definition is analogous to Equation 22.28 for the Bohr magneton, jg. Note
`that yw, is smaller than yy by a factor of about 2000, dueto thelarge difference in
`masses of the proton and electron.
`
`The magnetic moment of a free protonis not y,, but 2.7928 2. Unfortunately,
`
`
`
`thereis no general theory of nuclear magnetismthat
`explains this
`value, Another
`
`surprising pointis the fact that a neutron also has a magnetic moment, which has
`
`
`a value of
`— 1.9135 yw. The minus sign indicates that the neutron’s magnetic mo-
`Ment is opposite its spin angular momentum,
`
`Nuclear Magnetic Resonance and MRI
`It is interesting that nuclear magnetic moments (as well as electronic magnetic
`Moments) precess in an external magnetic field. The frequency at which they pre-
`
`cess, called the Larmorprecessional frequency, «,.
`is directly proportional to the
`magnetic field. This is described schematically in Figure 30.5a, where the magnetic
`field is along the «axis. For example, the Larmor frequency of a proton in a mag-
`
`netic field of | T is equal to 42.577 MHz, Thepotential energy of a magnetic dipole
`moment in an external magnetic field is
`#- B. Whentheprojection of g is along
` is
`— wB;
`that
`is,
`it has its
`the field,
`the potential energy of
`the dipole momer
`minimum value. When the projection of mw is against the field,
`the potential energy
`is «#8 and it has its maximumvalue. These two energy states for a nucleus with a
`spin of 5 are shownin Figure 30.5b.
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`30.2
`
`Binding Energy
`
`897
`
`
`
`Figure $0.8 A computer-
`digitized, color-enhanced MRI of a
`brain with a glioma tumor.
`(0 Soott Comesine/Sctence Smarce
`
`
`
`oo
`Frequency
`
`gradient in the magnetic field, protons in different parts of the body precess at
`different frequencies, so that the resonancesignal can be used to provide infor-
`mation onthe positions of the protons. A computer is used to analyze the position
`information to provide data for constructing a final iniage. An MRI scan taken on
`a human head is shown in Figure 30.8. The main advantage of MRI over other
`imaging techniques in medical diagnostics is that it causes minimal damage to
`cellular structures. Photonsassociated with the radiofrequency signals used in MRI
`have energies of only about 10-7 eV. Because molecular bond strengths are much
`larger (approximately 1] eV), the radio-frequency radiation causes litle cellular
`damage. In comparison, x-rays or ‘y-rays have energies ranging from 107 to 10° eV
`and can cause considerable cellular damage.
`
`30.2 + BINDING ENERGY
`
`The total mass of a nucleus is always less than the sum of the massesofits nucleons.
`Because mass is another manifestation of energy, the total energy of the bound
`system (the nucleus) is less than the combined energy of the separated nucle-
`ons. This difference in energyis called the binding energy of the nucleus and can
`be thought ofas the energythat must be added to a nucleus to break it apart into
`its components. Therefore, in order to separate a nucleus into protons and neu-
`trons, energy must be put into the system.
`
`Conservation of energy
`and the Einstein mass-energy equivalence relationship
`show that the binding energyof any nucleus ?X is
`
`E,(MeV) = [ZM(H) + Nin, — M(3X)] X 931.494 MeV/u
`
`[30.4]
`
`* Binding energy of a nucleus
`
`
`
`q|1
`
`
`
`fay
`
` AE=E, - B= 2H
`
`£, =-uR
`B>o0
`ib)
`
`Nuclear magnetic resonance *
`
`Figure 30.5
`(a) When a nucleus ts placed in an external magnetic field, the nuclear magnetic
`moment precesses about the magnetic field with a frequencythat is proportional to the fieid.
`(b) A proton,the spin of which is 4, can occupy one aftwo energy states whenplaced in an
`external magnetic field, The lower energy state, £, corresponds to the case in which the spin is
`aligned with the field, and the higher energystate, £,, corresponds to the case in which the spin
`is Opposite the held. The reverse is true for electrons.
`
`It is possible to observe transitions between these two spin states by using a
`technique known as nuclear
`resonance. A dc magnetic field is intro-
`duced to align the magnetic moments (Fig, 30.5a), along with a second, weak,
`oscillating magnetic field oriented perpendicular to B. When the frequencyof the
`oscillating field is adjusted to match the Larmor precessional frequency, a torque
`acting onthe precessing moments causes them to “flip” between the two spin states.
`These transitions result in a net absorption ofenergy by the spin system. A diagram
`of the apparatus used to detect a nuclear magnetic resonance signal is illustrated
`in Figure 30.6. The absorbed energy is supplied by the generator producing the
`oscillating magnetic field, Nuclear magnetic resonance and a related technique
`called electron spin resonance are extremely important methods for studying nu-
`clear and atomic systems and how these systems interact with their surroundings.
`A typical NMR spectrumis shown in Figure 30.7.
`Awidely used diagnostic technique called MRI, for magnetic resonance im-
`aging, is based on nuclear magnetic resonance. In MRI, the patient is placed inside
`a large solenoid that supplies a spatially varying magnetic field, Because of the
`
`Tunable
`oscillator
`iJ
`
`
`
`
`Oscilloscope
`
`
`arrangement for nu
`An experimental
`Figure 30.6
`dear magnetic resonance. The radio-frequency mag-
`netic field of the coil, provided by the variable-
`frequencyoscillator, must be perpendicular to the
`dc magnetic field. When the nuciei in the sample
`where M(H) is the atomic mass ofthe neutral hydrogen atom, M(}X) represents
`meet the resonance condition,
`the spins absorb en-
`the atomic mass ofthe element ($X), m, is the mass ofthe neutron, and the masses
`ergy fromthe field of the coil, which changes the ne-
`areall in atomic mass units. Note that the mass of the Z electrons included in the
`sponse of the circuit in which the coil is included.
`Resonance
`Most modern NMR spectrometers use supercomduct-
`first term in Equation 30,4 cancels with the mass of the Zelectrons included in the
`
`signal
`|
`ing magnets at fixed field strengths and operate at
`term M(?X) withinasmall difference associated with the atomic binding energy of
`frequencies of approximately 200 MHz
`theelectrons. Because atomic binding energies aretypically several eV and nuclear
`binding energies are several MeV,this difference is negligible.
`
`
`Figure $0.7) An NMRspect
`
`in a bridged metallic complex cx
`.
`ing platinum. The lines that flank the
`central strong peak are due to the
`inter-
`action between *'P and other
`
` ps is cline
`nuclei, The outermost setofli
`tothe interaction between *'P and
`
`neighboring platinu
`clei. The spec-
`trum was recorded at a fixed field of
`about 4 T, and the meanfrequency was
`200 MHz.
`
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`IPR2023-00783
`
`

`

`Chapter 300
`
`Nuclear Physics
`
`30.2
`
`Binding Energy
`
`899
`
`Using Equation 30.4, we find that the binding energy is
`E, = Amc* = (0.002 388 u) (931.494 MeV/u) = 2.224 MeV
`This result tells us that, to separate a deuteron into its con-
`stituent proton and neutron,
`it
`is necessary to add
`2.224 MeV of energy to the deuteron. One way ofsupplying
`the deuteron with this energy is by bombarding it with en-
`ergetic particles,
`If the binding energy of a nucleus were zero, the nucleus
`would separate into its constituent protons and neutrons
`without the addition ofany energy; that is, it would sponta-
`neously break apart,
`
`
`than the original nucleus. In a similar way, energy can be released when twolight
`
`nuclei with A = 20 combine or fuse to for
`¢ heavier nucleus
`Another important feature of Figure 30.9 is that the binding energy per nu-
`
`es between
`cleon is approximately constant for A > 20. In this case the nuclear fo
`a particular nucleon andall the other nucleonsin the nucleus are said vw be satu
`rated; that is, a particular nucleon forms attractive bonds withonlya limited number
`of other nucleons. Because ofthe short-range character of the nuclear force, these
`other nucleons can be viewed as being the nearest neighbors in the close packed
`structure shownin Figure 30.2.
`
`the nucle
`The general
`features of
`force responsible for
`the binding
`energy of nuclei have been revealed in a wide varicty of experiments and are as
`follows.
`
`Example 30.2 The Binding Energy of the Deuteron
`Calculate the binding energy of the deuteron (the nucleus
`of a deuterium atom), which consists of a proton and a neu-
`tron, given that the atomic mass of deuterium is 2.014 102 u.
`Solution From Table $0.1, we see that masses of a hydrogen
`atom and a neutron are M(H) = 1.007825 u, and m, =
`
`1.008 665 u. Therefore,
`M(H) + m, = 2.016 490 u
`To calculate the mass difference, we subtract the deute-
`rium mass from this value:
`Am = [M(H) + m,] — M(D)
`= 2.016 490 u — 2.014 102 u
`The attractive nuclear forceis the strongest force in nature.
`*
`
`= 0.002 388 u
`* The nuclear forceis a short-range force that rapidlyfalls to zero when the
`
`feeeEeeseparation between nucleons exceeds several fermis. Evidence for the lim-
`ited range of nuclear forces comes fromscattering experiments and from
`the saturation of nuclear forces previously mentioned, The short range of
`the nuclear force is shown in the neutron—proton (n—p) potential energy
`plot of Figure 30,10a obtained byscattering neutrons froma target contain-
`ing hydrogen. The depth of the n—p potential energy curve is 40 MeV to
`50 MeVand contains a strong repulsive component that prevents the nucle-
`ons from approaching muchcloser than0.4 fm. Anotherinteresting feature
`
`of the nuclear forceis thatits size depends onthe relative spin
`orientations
`of the nucleons, as shown byscattering experiments using spin
`polarized
`beams andtargets.
`
`
`
`
` =
`
`¥ (fm)
`
`eo
`=
`=
`= -20
`
`4 f
`
`i)
`
`Hep system
`
`2
`.
`s
`
`(a)
`
`(b)
`
`(a) Potential energyversus separ
`Figure $0.10
`the neutron—proton system.
`(b) Potential energy versus separation for the p
`n=proton system, The difference in t
`
`‘Coulomb repulsion in the case of the proton—proton 1
`
`
`curves is due mainly to thels
`tion.
`
`A plot of binding energy per nucleon, /,/ A, as a function of mass number for
`various stable nuclei
`is shown in Figure 30.9. Except for the lighter nuclei, the
`average binding energy per nucleonis about 8 MeV. For the deuteron,the average
`binding energy per nucleon is E,/A = (2.224 MeV)/2 = 1.112 MeV. Note that the
`curve in Figure 30.9 peaks in the vicinity of A = 60. Thatis, nuclei with mass num-
`bers greater or less than 60 are not as strongly bound as those near the middle of
`the periodic table. The higher values of binding energy near A = 60 imply that
`energyis released when a heavy nucleus with A = 200 splits or fissions into several
`lighter nuclei which lie near A = 60. Energy is released in fission because the final
`state consisting of two lighter fragments is more tightly bound or lower in energy
`
`Region of greatest
`
`| ability
`
`
`
`
`
`a ae
`
`150
`100
`Mass number A
`
`200
`
`950
`
`ingenergyperparticle,MeV
`
`
`
`Figure 30.9 A plot of binding energy per nucleonversus mass number for the stable nuclei in
`Figure 30.3.
`
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`
`

`

`
`
`Chapter 30
`
`Nuclear Physies
`
`30.3 - RADIOACTIVITY
`
`30.3
`
`Radioactivity
`
`* From n=n, n-p, and p—p scattering experiments and other indirect evi-
`dence, it is found that the nuclear force is independentofthe electric charge
`of the interacting nucleons, As might be expected from this “charge-blind”
`character of the nuclear force, the nuclear force does notaffect electrons,
`enabling energetic electronstoserve as point-like probes of the charge den-
`sity of nuclei, The charge independence of the nuclear force also means
`that the main difference benween the n—p and p—pinteractionsis that the
`P-p potential energy consists of a superposition of nuclear and Coulomb
`interactions as shown in Figure 30.10b. At distances less than 2 fm, both
`p-p and n-p potential energies are nearly identical, but for distances
`greater than this, the p-p potential has a positive energy barrier with a
`maximumof about 1 MeVat 4 fm,
`
`
`Thinking Physics 1
`
`Figure 30.9 shows a graph of the amount of energy necessary to remove a nucleon,
`from the nucleus, Figure 29.15 in the previous chapter shows the energy necessary to
`
`remove an electron
`from an atom. Why does Figure 30.9 showan approximately constant
`amount of energy necessary to remove a nucleon (above about A = 56), whereas Figure
`29.15 shows widely varying amounts of energy necessary to remove an electron from the
`atom?
`
`
`
`
`
`In 1896 Henri Beequerel accidentally discovered that uranyl potassium sulfate crys-
`tals emitted an invisible radiation that could darken a photographic plate when the
`plate was coveredto exclude light. After a series of experiments, he concluded that
`the radiation emitted by thecrystals was of a new type, onethat required noexternal
`stimulation and was so penetrating that it could darken protected photographic
` #
`plates and ionize gases.
`This spontaneous emission ofradiation was soon to be
`called radioactivity. Subsequent experiments by otherscientists showedthat other
`substances were alsoradioactive.
`
`
`Marie Curie
`The most significant investigations of this type were conducted by
`
`(1867-1954) and Pierre Curie (1859-1906). After several years of careful and. la-
`
`
` active ore,
`borious chemical separation processes on tons of pitchblende,a r
`the Cures reported the discovery of two previously unknown radioactive elements.
`
`These were 1
`d polonium and radium. Subsequent experiments,
`including
`
`
`Rutherford's famous work on alpha-particle scattering, suggested that radioactivity
`was the result ofthe decay, or disintegration, of unstable nuclei.
`
`Threetypesofradiation can be emitted by a radioactive
`substance: alpha (a)
`
`rays, where the emitted particles are *He nuclei; beta (f) rays, in which the emitted
`
`
`
`cles are either electrons or positrons;
`and gamma(+) rays,
`in which the emitted
`
`are high-energy photons. A positron is a
`particle similar to the electron inall
`
`Marie Curie (1867-1934)
`
`respects except that it has a charge of + ¢
`(The positron is said to be the antipar-
`ticle of the electron.) The symbol ec”
`is used to designate an electron and e€* des
`
`Reasoning In the case of Figure 30.9, the approximately constant value of the nuclear
`A Polish scientis!, Marie Curie shared
`binding energyis a result of the short-range nature of the nuclear strong force. A given
`ignates a positron.
`the Nobelprize in 1903 with her hus-
`
`jon using the scheme
`nucleon interacts only withits few nearest neighbors, rather than withall of the nucle-
`band, Pierre, and with Becquerelfor
`It
`is possible to distinguish these three forms of radi
`ons
`the nucleus. Thus, no matter how many tucleoms are present in the nucheus,
`their work om spontaneous radioactiv-
`illustrated in Figure 30.11. Theradiation froma radioactive sample is directed into
`ity and the radiation emitted by radio:
`pulling one nucleon out involves separating it only fromits nearest neighbors. The
`a region in which there is a magnetic field. The beam splits into three components,
`active substances.
`“I persist in believ-
`
`eC
`energy to do this, therefore, is approximately independent of how many nucleons are
`
`
`
`two bending in opposite directions and thethird experiencing no change in c
`present.
`ing thot the ideos thotthan guided us
`tion. From this simple observation, one can conclude that the radiation of
`the
`are the only ones which can lead to
`However, the electrical force holding the electrons to the nucleus in an atom is a
`
`the true social progress. We connol
`long-range fonce,
`An electron in the atom interacts with all of the protons in the
`undeflected beamcarries no charge (the gamma ray), the component deflected
`
`hope te build o better world without
`
`nucleus. When the nuclear charge increases, there isastronger attraction between the
`upward corresponds topositively charged particles (alpha particles), and the com-
`
`improving the individual. Toward this
`nucleus and the electrons. Asa result, as the nuclear charge increases, more energyis
`ponent deflected downward correspondsto negatively charged particles (e~). If the
`end, each of us must work toward his
`
`necessary bo remove an electron. This is demonstrated by the upward tendencyof the
`beam includes a positron (e7), it is deflected upward.
`own highest development, acceptin
`
`
`at the sometime his share of responsi
`ionization energy in Figure 29.15 for each period.
`bility in the generallife of humanity,"
`
`
`Thinking Physics2
`
`When
`avy nucleus undergoes fission intojust two lighter nuclei, the product maclei
`are invariably very unstable, Why is this?
`Reasoning According to Figure 30.3, the ratio of the numberof neutrons to the num-
`
`ber of protons increases with ZAsa result, when a heavy nucleus splits in a fission
`reaction to two lighter nuclei, the lighter nuclei tend to have too many neutrons. This
`leads to instability
`as the nucleus returns to the curve in Figure 30.5 by decay processes
`that reduce the numberof neutrons
`
`
`
`Photographic
`plate
`
`
`Figure 30.11
`Theradiation from a radioac-
`tive source
`
`be separated into three compo-
`am:
`nents by
`etic field, The photographic
`plate at the right records the events. The
`gammaray is not deflected by the magnetic
`
`field.
`
`CONCEPTUAL PROBLEM 1SE
`Isotopes of a given clement can have different masses, different magnetic moments, but
`same chemical behavior, Whyis this?
`
`
`
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`IPR2023-00783
`
`

`

`
`
` 30.3
`
`Nuclear Physies
`
`Chapter 30
`
`Radioactivity
`
`903
`
`The three types ofradiation have quite different penetrating powers. Alpha
`particles barely penetrate a sheet of paper, beta particles can penetrate a few mil-
`limeters of aluminum, and gamma rays can penetrate several centimeters of lead.
`The rate at which a particular decay process occurs in a radioactive sample is
`proportional to the number of radioactive nuclei present (that is, those nuclei that
`have mot yet decayed). If Nis the number ofradioactive nuclei present at some
`instant, the rate of change of Nis
`
`dN
`Soret
`150.5]
`dt
`where A is called cither the decay constant or the disintegration constant. The
`minus sign indicates that dN/ dt is negative; that is, Nis decreasing in time.
`If we write Equation 30.5 in the form
`
`we canintegrate
`
`feerhea ae
`
`nN dN
`i
`; Ww =—-A\ f, dt
`N
`In (=) =-M
`N= Ne
`
`[30.6]
`
`0.693
`
`In2
`ha=
`
`(30.8)
`
`* Halfstife equation
`
`This is a convenient expression relating the halflife to the decay constant. Note
`
`that after an elapsed timeof one halflife, Nj
`/2 radioactive nuclei remain (by def-
`inition); after two half-lives, half of these will have decayed and N,//4 radioactive
`
`nuclei will be left; after three half-lives,
`N,/8 will be left; and so on. In general
`after » half-lives, the numberofradioactive nuclei remaining is N,/2". Thus, we
`see that nuclear decay is independent ofthe past history of the sample.
`Activity can be measured with a unit called the curie (Ci), defined as
`1 Ci =3.7 X 10!decays/s
`
`is the approximate
`This unit was selected as the original activity unit because it
`activity of 1 g of radium. The SI unit ofactivity is called the beequerel (Bq):
`1 Bq = 1 decay/'s
`1 Ci = 3.7 * 10° Bg. The most commonly used units of activity are
`Therefore,
`millicuries and microcuries.
`
`* The cunie
`
`* The becquerel
`
`Ne Nger*!
`
`
`
`
`} i
`
`Eaae
`Ty 27
`Figure 30.12 A plot of the expo-
`nential decay law for radioactive
`nuclei. The vertical axis represents
`the number ofradioactive nuclei
`present at any time 4, and the hori-
`zontal axis is time, The time 7) -s is
`the halflife of the sample.
`
`Thinking Physics 3
`[f youstart with a sample
`Theisotope '#Cis radioactive and hasa half-tife of 5730 years.
`of 1000 carbon-14 nuclei, how manywill still be around in 17 190 years?
`Reasoning In 5730 years, half the samplewill have decayed, leaving 500 radioactive
` 460 years), the number
`*4C nuclei. In another 5730years (for a total elapsed timeof
`
`will be reduced to 250 nuclei. After another 5730 years (total time 17 190 years), 12!
`The constant N, represents the number ofradioactive nuclei at ¢ = 0 and¢is the
`remain.
`base of the natural logarithm.
`These numbers represent ideal circumstances. Radioactive decay is an averaging
`The decay rate, R( = ldN/di!), can be obtained by differentiating Equation
`process over a verylarge number ofatoms, and the actual outcome depends onstatis-
`
`30.6 with respect to time:
`ocrtainly not a
`tics. Our original samplein this example contained only 1000 nuch
`
`verylarge number. Thus, if we were actually to count the nu
`ber remaining after one
`half-life for this small sample,
`it probably would not be exactly 500
`
`'
`aN
`[30.7]
`= Nar= Rye™
`c= at
`
`where R= NA and &, = N)A is the decay rate at t= 0. The decay rate of a sample
`
`
`Example 30.3 The Activity of Radium
`is often referred to as its activity. Note that both N and FR decrease exponentially
`Wecan calculate the activity of
`the sam
`0 using
`with time. The plot of N versus tin Figure 30.12 illustrates the exponential decay
`
`The half-life of the
`radioactive nucleus "Z}Ra is 1.6% 10"
`law.
`Ny
`is the
`here A, is the decay rate at
`f=
`()
`
`
`
`years. If a sample contains 3 * 10'* such nuclei, determine Ry=A
`theactivity at this initial time ¢=0.
`Another parameterthat is useful for characterizing radioactive decayis the half-
`number of radioactive nuclei present air=0
`life, 1». The half-life of a radioactive substance is the time it takes half of a
`
`= AN, = (14 X 10-5
`Solution
`First,
`Jet us convert the half-life wo seconels
`R
`given number of radioactive nuclei to decay. Setting N = N,/2 and f= 7, in
`10° decays/'s
`Equation 30.6 gives
`
`
`
`Tig=(1.6 % 107 s/year)=5.0 * 10! 510" years) (3.16 * 11
`*
` Noe0)
`ATi
`CCALISE
`a
`10"
`decays,’
`s
`tivity,
`or decay
`1Ci
`we
`B
`15
`rate, at f= 0 is
`
`Writing this in the form e4!* = 2 and taking the natural logarithmof both sides,
`we get
`
`constant:
`Now we can use this value in Equation 30.8 to get
`0.693
`0.693
`A=S = apa 14 xs
`Ta
`5.0 x 10's
`
`the decay
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`The quantitygiven by Equation 30.12 is sometimes referred to as the Q-value of the
`nuclear reaction. Note that Q will be in joules if the massesare in kilograms, andc
`is the usual 3.00 X 10° m/s. We make this expression more convenient forcalcu-
`Example 30.4 The Activity of Carbon
`lations by including in all terms the masses of enough electrons to make neutral
`A radioactive sample contains 3.50 ag ofpure "LC. which has
`atoms, Also, we convert atomic mass units for mass to MeVunits for energy:
`a half-tife of 20.4 min. (a) Determine the number of nuclei
`present initially.
`Q= [M(X) — M(Y)
`— MCHe)] = 931,494 MeV/u
`[30,13]
`Solution The atomic mass of || is approximately11.0, and
`
`
`
`Thedisintegration energy Qappearsinthe formof kinetic energyof the daugh-
`therefore 11,0 g contains Avogadro's number of nuclei,
`Hence, 3.50 ag contains Nnuclei, where
`ter nucleus and the alphaparticle. In the case of the ““Ra decay described in Figure
`30.13, if the parent nucleus decaysat rest, the residual kinetic energyof the prod-
`N
`_ 3.50 x 10 “2
`ucts is 4.87 MeV. Most of the kinetic energyis associated with the alpha particle
` nuclei/mol
`11.0 g/mol
`becausethis particle is much less massive than the recoiling daughter nucleus. That
`N= 1.92 x 107 nuclei
`is, because momentum must be conserved, the lighter alpha particle recoils with a
`R= Roe= (1.09 * 104 decays/s)e
`much higher speed than the daughter nucleus, Generally, light particles carry off
`= 8.96 * 10° decays/s
`is the activity of
`the sample initially andafter
`(b) What
`BOM) he
`most of the energy in nuclear decays
`
`Finally,itis interesting to note that if one assumed that **U (or other alpha
`Solution
`EXERCISE 1 Calculate the number of radioactive nuclei re-
`Because 7). = 20.4 min = 1224s, the decay con-
` Ar wer
`start is
`emitters) decayed by emitting protons and neutrons,the mass of the decay products
`maining after 8.00 h.
`:
`N= 1.58 x 10" nuclei
`
`would exceed that of the parent nucleus, corresponding to negative Q values
`Therefore, such spontaneous decays do not occur.
`
`= 5.66 * 10-4 s7!
`
`
`0.693
`0.695
`1294s
`Tig
`Therefore,the initial activity of the sample is
`Ro = ANg = (5.66 * 10-' s-')(1.92 % 10!)
`= 1.08 x 10" decays/s
`We can use Equation 30.7 to find the activity at any time t
`For t= 8.00 h = 2.88 X 10*s, we see that At = 16.3, and so
`
`
`
` 30.4
`
`Chapter 30
`
`Nuclear Physics
`
`TheDecay Processes
`
`905
`
`
`
`The hands and numbers of this luminous watch contain minute
`amounts of radium salt. The radioactive decay of radium causes the
`(© Rickard Magna 1990, Fundamental!Photo
`watch to glow in the dark,
`wraphs)
`
`
`[30.10]
`U0 — “iTh + jHe
`[30.11]
`8 Ra —~ *GRn + JHe
`The half-life for **U decay is 4.47 x 10° years, and the half-life for “"°Ra decayis
`1.60 * 10° years. In both cases, note that the mass number A of the daughter nu-
`cleus is 4 less than that of the parent nucleus. Likewise, Z is reduced by 2. The
`differences are accounted for in the emitted alpha particle (the ‘He nucleus)
`
`The decayof "Rais shown in Figure 30.13. When one element changesinto
`another, as happensin alpha decay, the processis called spontaneous decay. As a
`general rule, (1) the sumof the mass numbers A must be the same on both sides
`of the equation, and (2) the sum of the atomic numbers 7 must be the same on
`both sides of the equation. In addition, the total energy must be conserved. If we
`
`call My the mass ofthe parent nucleus, M; the
`mass of the daughter nucleus, and
`M, the mass ofthe alpha particle, we can define the disintegration energy, Q:
`x
`Y
`eh.
`2 = (My — My— M,)c2
`[30.12]
`
`
`
`Kes
`Ka
`/
`a —
`=
`Pm
`\
`Po
`After decay
`
`Figure 30.13 Alpha decay ofra-
`dium. The
`radium nucleus is ini-
`tially at rest. After the decay, the
`radon nucleus has kinetic energy
`
`Ky, and mom
`HM Pep, and the
`alpha particle has kinetic energy
`K, and momentump,
`
`A=
`
`30.4 - THE DECAY PROCESSES
`
`|
`
`As we stated in the preceding section, a radioactive nucleus spontaneously decays
`via alpha, beta, or gamma decay. Let us discuss these three processes in more detail.
`
`Example 30.5 The Energy Liberated When Radium Decays
`
`Th
`Ra nucleus undergoes alpha decay according to Equa-
` mass of
`tion M.11. Calculate the Q value for this process. Take the
`
`Ra to be
`226.025 406 u, the mass of "Rn to be
`222.017 5740, and the mass of {He to be 4.002 603 u, as
`foundin Table A.5
`
`MeV
`
`the
`that
`ow
`4 MeV, whereas
`is only
`about
`
`(931 44
`*
`(0.005 279 u)
`
`i
`to end-ofchapter Problem
`is left
`It
`
`
`kinetic energy of the alphaparticle is al
`Alpha Decay
`that
`of
`the
`recoiling
`daughter
`nucleus
`O.] Mey
`If a
`nucleus emits an alpha particle ({He), it loses nwo protons and two neutrons,
`Solution Using Equation 30.13, wesee that
`
`rfore, N decreases by 2, Z decreases by 2, and A decreases by 4. The decay can
`MC2Rn) — MC@He)]*931.494 MeV/u
`O=
`(MCZ*Ra)
`—
`be written symbolically as
`(226.025 406 uw
`=
`222.017 574 uw
`— 4.002 603 u)
`[30.9]
`IN pV+ dHe
`Alpha decay
`~ 931.494 MeV/u
`whereXis called the parent nucleus and ¥the daughter nucleus. As examples,
`“8U) and
`Ra are both alpha emitters and decay according to the schemes
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

` 30.4
`
`Chapter 30
`
`NuclearPheysies
`
`The Decay Processes
`
`907
`
`
`
`Figure 36.14 Potential energy
`versus separation for the alpha par-
`ticle—nucleus system, Classically,
`the energy ofthe alpha particle is
`not sufficient to overcome the bar-
`rier, and so the particle should not
`be able to escape the nucleus.
`
`We now turn to the mechanism of alpha decay. Figure 30.14 is a plot of the
`potential energy versus distance rfrom the nucleus for the alpha particle—nucleus
`system, where Ris the range of the nuclear force. The curve represents the com-
`bined effects of (1) the Coulomb repulsive energy, which gives the positive peak
`for r > R, and (2) the nuclear attractive force, which causes the curve to be negative
`for r< KR. As we saw in Example 30.5, the disintegration energy is about 5 MeV,
`whichis the approximate kinetic energy of the alpha particle, represented by the
`lower dotted line in Figure 30.14. According to classical physics, the alpha particle
`is trapped in the potential well. How, then, doesit ever escape from the nucleus?
`The answer to this question was provided by Gamow and, independently, Gur-
`ney and Condon in 1928, using quantum mechanics. In brief, the view of quantum
`mechanics is that there is always some probability that the particle can penetrate
`(or tunnel) through the barrier (Section 28.11). Recall that the probability of lo-
`cating the particle depends on its wave function, #, and that the probability of
`tunneling is measured by lyl*. Figure 30.15 is a sketch of the wave function for a
`particle of energy £, meeting a square barrier offinite height, which approximates
`the nuclear barrier. Note that the wave function is oscillating both inside and out-
`side the barrier but is greatly reduced in amplitude because of the barrier. As the
`energy £ increases, the probability of escaping also increases, Furthermore, the
`probability increases as the width of the barrier is decreased.
`
`Inside the
`uy nucleus
`
`Outside
`the nucleus
`
`
`
`aP
`
`R
`
`R, "i
`
`Figure 30.15 The nuclear poten-
`tial energyis modeled as a square
`barner. The energy of the alpha
`particle is £, whichis less than the
`height of the barner. According 10
`quantum mechanics, the alpha par-
`
`ticle has somec
`ce of tunneling
`through the barrier, ax indicated by
`th