Supplement 1: Beta Decay

Transcription

1 Phys. 69: Nuclear Instrumentation Physics Department Yarmouk University Supplement : Beta Decay Introduction Energy Release in β Decay Dr. Nidal M. Ershaidat Introduction Electron emission, positron emission (9, I. & F. Joliot-Curie) and (orbital) electron capture (98, lvarez) are all known as beta decay processes n p + e + ν + p n + e + ν p + e n + ν Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay e β β ε The electron resulting from a β - decay process is created thanks to the energy available. (Electrons do not preexist inside a nucleus). The processes involving protons occur only for bound protons in nuclei (The presence of the nucleus field is a sine qua non condition) + What really happens! We know now that the weak interaction is responsible for beta decay. n p + e + ν e Fig : β - decay In β -, The W - mediates the interaction. One of the d quarks of the neutron transform into a u quark. The neutron, thus, becomes a proton. n electron and its anti-neutrino are emitted in the process. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay

2 Typical Beta Decay Processes Decay Type Q(MeV) t / Ne Na + e - + ν e β Tc 99 Ru + e - + ν β - e 5 l 5 Mg + e + + ν e β s. 0 5 y 7. s 5 Energy Continuous Spectrum in β Decay We expect to have mono-energetic electrons as we observe the mono-energetic alpha s in a decay. But instead we have for the electrons resulting from a β decay a continuous spectrum starting at 0 and ending at E max (Endpoint energy) which is the energy an electron should have in this decay! 6 I Te + e + + ν e β +.. d 5 O + e - 5 N + ν e ε.75. s Ca + e - K + ν e ε y Exercise: Check the Q-value for all these reactions. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay Fig : β - decay spectrum, i.e. Energy Distribution of the electrons. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay Energy Release in β Decay Beta decay of 0 Bi 0 8 Energy Release in β Decay 0 Bi7 8Po6 + e Q = ( ) = MeV Neglecting the recoil energy of the daughter, the maximum energy an electron can have is : E max = =.6 MeV 8 Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay

3 The Neutrino Experiments showed that the shape of the spectrum of electrons emitted in β decay is characteristic of the electron themselves. In 9 Pauli, suggested the presence of a second particle emitted in the decay which can carry a part of the available energy and linear momentum. This particle should have a zero mass, be neutral and interacts so weakly with matter that detectors do not see it! Pauli called it The ghost particle and Fermi gave it the name neutrino (small neutron in Italian). The neutrino was discovered in 957! 9 Kinematics Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay Kinematics Consider the β - decay of a free neutron (t / = 0 min) n p + e + ν e β Q = (m n - m p - m e - m ν ) c In a frame attached to the decaying neutron, the available energy will be shared by the resulting particles: Q = T + T p e + T νe Neglecting the proton s recoil energy T p, which is measured to be 0. kev, Q is essentially shared by the electron and the antineutrino. The β - Decay Electron is Relativistic The energy carried by the electron in β - is of the order of its rest mass energy T e / m e c > 0., while the recoil energy is low and can be taken non relativistically. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay

4 Q The electron-(anti)neutrino is massless = mn c m p c m e = m = 0.78 MeV m The measured value is Q = 0.78 ± 0.0 MeV. This suggests that the mass of the antineutrino = 0 within the experimental error ( kev). New experiments give very much lower limits (few ev's) Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay νe c c m Measurements of the linear momentum of the electron and the proton indicate that a rd particle should be present. νe c νe c Q β - Decay Kinematics Z X N Z + Y N + e + ν e ( X ) m ( Y ) m [ c = mn Z N Z ] β + e where N indicates the nuclear mass and energy and considering a massless antineutrino. Neglecting the electrons binding energies we have : Q = [ m( Z X ) m( Z + Y )] c β Masses here are neutral atomic masses (tables) Q = Te + T β νe Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay Q is shared between e - and ν e 5 β + Decay Kinematics 6 This is the energy shared between the electron and the antineutrino.we saw that in the case of β - Decay of 0 Bi, Q =.6 MeV is in good agreement with the measured value. This measurement is used to calculate the mass of the 0 Po isotope. Q + Z X N Z Y N + e + ν e + = m Z Z ] β e ( X ) m( Y ) m [ c Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay

5 Z Q ε X Electron Capture N + e Z Y N + [ + ν The calculation of Q should take into account the fact that the daughter nucleus is in an excited state. The resulting X-ray (or X-rays) should have the binding energy of the captured electron ( Z X ) m( Z Y )] c Bn = m Where B n is the binding energy of the captured electron from shell n (K, L, M, ) Note that here the neutrino is mono-energetic. nd if we neglect the recoil energy of Y, E ν = Q. e 7 Fermi Theory of β Decay Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay Fermi Theory of β Decay The theory to explain β Decay should include the following information : ) The electron and the neutrino do not preexist in the nucleus ) The electron and the neutrino are relativistic. ) The continuous distribution of electron energies Enrico Fermi proposed, in 9, a theory of β decay based on Pauli s neutrino hypothesis. The major idea is that β Decay is the result of a weak interaction (compared to that responsible for the quasi-stationary states). The characteristic times in β decay is of the order of seconds or longer where the nuclear characteristic time is of the order of 0-0 s. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay 9 Fermi Theory - Transition probability The barrier potential used in alpha decay (See Suppl-lpha decay) does not exist in the case of β Decay. nd even though if it exists, the transmission probability is nearly. Fermi s idea was to consider β decay as a perturbation forcing the quantum system (The parent nucleus) in a transition. The transition probability is given by : = π V h ( ) fi ρ E f ρ(e f ) is the final density of final states = dn/de dn is the number of final sates per energy interval de and V fi is the matrix element: * V fi = ψ f Vψi dv 0 5

6 Fermi Theory - Perturbation Potential Fermi had no idea about the weak interaction potential so he tried all possible forms consistent with special relativity, and showed that V can be replaced by an operator O X, where X gives the form of the operator O. X = V (vector potential), X = (xial vector) X = S (Scalar) X = Pseudoscalar X = Tensor Only experiment can help deciding which transformation is the appropriate one. Now we know that X is the so-called V- (Vector-xial transformation). Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Supplement Beta Decay 6

7 Phys. 69: Nuclear Instrumentation Physics Department Yarmouk University Supplement : lpha Decay Basic α Decay Processes (Introduction and Kinematics) Dr. Nidal M. Ershaidat Seven years after Becquerel s discovery, Rutherford (and Mme Curie) identified the naturally emitted α particles as being less penetrating comparatively to the other emitted ones (β & γ). β Radioactive source Introduction By simply using a deflecting magnetic field Marie Curie demonstrated that a particles are doubly positively charged. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Suppl. lpha Decay γ α Magnetic Field α Particles are Helium Nuclei Rutherford, using an evacuated closed chamber with a thin wall accumulated α particles emitted by radium for several days, proved by atomic spectroscopy that helium gas is formed and thus that an α particle is in fact a He nucleus (or a doubly ionized helium atom). This means that an α source ejects a cluster of nucleons (p & n). Why? One can imagine that it is easier and simpler for an unstable nucleus to eject a single nucleon or nucleons. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Suppl. lpha Decay

8 In a nuclear reaction Q-Value a + B + b The net energy release called the Q-value of the reaction is given by: Q = ( m c + m c ) ( m c m c ) a b + The Q-value is simply the difference between the available initial rest energy and the resulting final rest energy. B 5 Q-Value - Generalization More generally, in a nuclear reaction involving s nuclei or nucleons and the result of which is the creation of t nuclei or nucleons the (general) definition of the Q-value is: Q = s i = m i c t f = m f c 6 The reaction is energetically possible if and only if Q > 0. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Suppl. lpha Decay Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Suppl. lpha Decay α Decay Kinematics In an α decay, an unstable nucleus X emits an α ( He nucleus) and transforms to nucleus Y. Z X N Z Y N + α Example 6 Eα =. 8 MeV 88 Ra6 86Rn + α t / ( 6 Ra)=600y If we consider that the nucleus X decays while at rest (or equivalently we consider a referential attached to this nucleus) then conservation of energy gives: m c = m c + T + m c + T α X Y Y α In these conditions the only available energy is the rest mass energy of X. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Suppl. lpha Decay 7 Momentum and Kinetic Energy of α Q = m X c m Y c m c = T + α Y Tα In the cm referential, X at rest (P X = 0), the resulting nucleus Y and the α particle are emitted in opposite directions and we have : T P m α α α my PY = Pα = = > T P m m Y The available energy Q is shared by Y and the α particle inversely proportionally to their respective masses, i.e. the α particle kinetic energy is much bigger than the nucleus Y s kinetic energy. Q T = α Q + m m α Y Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Suppl. lpha Decay Y >> >> Y α 8

9 Emitted particle (mass in MeV/c ) α Decay is energetically favored Let s look at the possible emission of a decaying U nucleus (mass =.066 u). Decay Q (MeV) n (99.57) 9U 0 9U9 + 0n p (98.80) 9U 0 9Pa0 + H 0-6. H (875.68) 0 9U 0 9Pa9 + H He (78.) U Th + He Q = m m 9U0 9U9 m 0 c n Q=( ) = MeV Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Suppl. lpha Decay 9 Computing Q for possible decays of U Decay 9U 9U0 + 0n Q = ( ) = MeV Decay 9U 0 9Pa0 + p0 Q = ( ) = - 6. MeV Decay 0 9U 0 9Pa9 + H Q = ( ) = MeV Decay 8 9U 0 90 Th8 + He Q = ( ) = + 5. MeV Decay 6 6 9U 0 89c7 + Li Q=( ) 9.50 = -.79 MeV Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Suppl. lpha Decay 0 Is Q > 0 the unique condition for an α Decay to occur? The previous calculations show that only alpha decay can occur spontaneously (Q > 0). But this is not the only criterion. For example the following decays are energetically possible(q > 0) 8 9U 0 88Pb6 + Be Exercise : Check that! 0 9U 0 86Rn + 6 C6 However, nuclear spectroscopy shows that such decays have vanishingly small partial decay constants compared to a decay, and thus they are not seen. In some cases, beta decay is intense enough to mask possible α decays. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Suppl. lpha Decay Criteria for an α Decay to occur. Q-value should be > 0.The partial decay constant should be large enough This corresponds, according to our technological limits to half-lives of the order of 0 6 years.. α decay should not be masked by β decay. Half of the unstable nuclei against alpha decay ( > 90 and many nuclei in the range 50 < < 90) verify this criterion. Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter : Radiation Sources - Suppl. lpha Decay

10 Phys. 69: Nuclear Instrumentation Physics Department Yarmouk University Supplement : Part - Systematics and Theory of α emission Systematics of α emission Dr. Nidal M. Ershaidat Systematics of α Decay Geiger & Nuttall ) Emitters with large disintegration energies are short-lived and vice versa (Geiger & Nuttall, 9) Fig. 6. : Q α vs. t / Nuclide t / Q α (MeV) Th. x 0 0 y.08 8 Th.0 x 0-7 s = 00 ns Q α ( Th) t ( Th) =. Qα ( Th) t ( = 0! For a factor in energy, the half-lives are 0 - times different!! Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay

11 α Decay & Stability ) dding a neutron to an unstable nucleus reduces the disintegration energy which means, according to Geiger-Nuttall findings, a longer lifetime or more stability. (See Fig for > ). There is a distinguished discontinuity at (N=6, =). This is conform to the shell-model and the existence of magic numbers. 5 α Decay & The Semiempirical Mass Formula Q α = m(z,) c m(z-, -) c - m α c m(z,) c = (Z m p c + N m n c B(Z,)) + Z m e c + i= Q α = B( He) + B(Z-, -) B(Z,) Z 6 e B i Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay B B ( Z, ) = a V α Decay & The Semiempirical Mass Formula ( Z, ) = av ( ) as ( ) ac ( Z ) ( Z ) ( ) (( ) ( Z ) ) asym ( ) + a ( ) p a a S Sym a Z ( Z ) ( Z ) + a Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay C p 7 - The volume term: a V ( - ) - a V = - a V - The surface term: as = B(Z-,-) - B(Z,) >> + as 8 as = + as + + Z Z - The Coulomb term : a C Z >> Z >> a C ( Z )( Z ) ( Z ) + ac Z ( Z( Z ) ) Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay 8

12 Q α = B ( He) + B(Z-,-) - B(Z,) Q α = B( He) + B(Z-, -) B(Z,) 8 C Z Qα = 8. a V + as + a 7 + Z asym ap Z 0 Explaining the Difference The small difference (0. MeV) between the values which comes from the fact that the semiempirical mass formula s parameters are chosen so as to reproduce the maximum number of nuclei masses, is not significant. Q α = = 6.75 MeV For 6 Th this gives Q = 6.75 MeV, while the measured value is 6.5 MeV. Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay Theory of α Decay Gamow and Condon & Gurney proposed independently in 98 a theory to explain the alpha decay and its systematics. Theory of α Emission The theory is based on the idea that alpha particles exist in the parent nucleus (or considered to behave as if they do exist). The interaction potential between an alpha particle and the residual (daughter) nucleus is represented by the following: Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay

13 Nuclear Potential an α suffers The Coulomb potential extends inward to a and then suffers a cut off V(r) Q α V 0 V ( r ) r a b r ) Spherical Potential Well Three regions of interest appear : Region I (r < a) : This is the spherical part (potential well of depth V 0 ) where the alpha particle, with a kinetic energy E - V = Q + V α 0 moves. The radius a is the sum of the alpha particle and the residual nucleus. V(r) Q α V 0 I a V ( r) r Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay b r 5 ) Barrier potential Region II (a < r < b) : This is the annular shell region which plays the role of a barrier potential because the potential energy here is greater than the particle s total energy. Classically the particle cannot enter this region (Kinetic energy = Q - V < 0). 6 ) Classical ccessible Region Region III (r > b) : This is a classically permitted region. if the particle succeeds to escape! then it can move freely in this rd region 7 V(r) V(r) V ( r) r V ( r) r Q α I a II b r Q α I a II b III r V 0 V 0 Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay

14 But alpha can escape! Classically the alpha particle rebounds each time it hits the well at r = a. In quantum mechanics there is a small probability that it leaks through the wall. The wave function digs a tunnel and the particle can escape the potential well!! The escape (tunneling) probability which is related to the decay constant depends on the penetrability (P) of the barrier region. The decay constant is given by: = f P where f represents the frequency with which the alpha particle presents itself at the barrier and P is the penetrability, i.e. probability of transmission from one side to another of the barrier. Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay 8 with D Barrier Potential - Semi Classical Treatment The transmission probability of a flux of incident particles on a barrier potential (depth V 0 and Length L) can be easily calculated in quantum mechanics. In a D problem with spherical symmetry (V depends only on r) and zero angular momentum P the probability to penetrate the complete barrier, also called penetrability, is given by : γ P e r γ = h r = π ε [ m ( V ( r) E) ] dr.88 Z r Here: V ( r) = ( MeV ) 0 z Z e r Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay in F 9 Height of the (Coulomb) Barrier Potential The height of the barrier part is :.88 Z B( r = a) = ( MeV ) & B(r = b) = 0 ainf The height of the barrier varies from (B - Q) above the particle s energy at r = a to 0 at r = b. We shall take as a representative height (the energy difference (E - V 0 )) the average height : ( B Q) The average width which we shall take as L is : ( b a) Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay 0 Parameter b t a first approximation using the average height and width the penetrability is : P e k L = e k (( ba) ) k ( b a = e ) Typical values for a heavy parent (Z = 90) are : Q = 6 MeV, a = 7.5 F b can be calculated as being the radius at which the alpha particle can leave the barrier. t this point Q = V(b) or z Z e Q = π ε b 0 Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay b.88 Z = Q (F) (MeV) 5

15 Penetrability Which gives for Q = 6 MeV : b = F 6 mc ( BQ) k = hc For a = 7.5 F we have : B = B r = a = = ( 6) k = =.6 F 97 Thus we have : P e P e k L.6 ( ) MeV = e k Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay (( b a ) ) k ( b a = e ) ( 7.5) 5 =.5 0 Estimating f! = f P f is roughly of the order of v/a, where v is the relative velocity of the alpha particle. for a = 7.5 F & Q = 6 MeV and V 0 = (B) = MeV: ( Q V ) + f = v = a a 0 / m c 0 / f = c = s 7.5 = = s -! Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay c Gamow s theory ln t = 700 s If we Change Q from 6 MeV to 5 MeV then P becomes 0-0 and t / = 0 8 s The theory, although using a semi-classical treatment explain remarkably the major observation by Geiger and Nuttall. Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay Quantum Theory of α Decay Refined calculation is achieved using quantum mechanics principles. First we consider the system (Daughter, alpha) in the a cm (center of mass) where the problem reduces to considering a particle with the reduced mass : mα MY = + µ = µ M Y m mα + M α Y Then one divides the Coulomb barrier potential into spherical shells of radii (r,r+dr). The transmission probability between r and r + dr is µ z Z e dp = exp Q α dr h π ε0 r Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay 5 6

16 The Gamow factor G The penetrability is: P = e -G Where G (the Gamow factor) is: b z Z e G = a µ Q R α h 0 r πε Taking x = a/b = Q/B, the integral gives : z Z e G = π ε0 hv Exercise : Check that! Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay dr [( cos x ) x ( x) ] 6 Thus we have : P in Quantum Mechanics For x << ( Q << B or a << b), which is the case in the case for most decays of interest, G becomes: z Z e π G = x π ε h 0 v z Z e µ c π P = exp π ε c Q 0 h. Z π P = exp 97 Q Q B Q B Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay 7 with f given by: t f = a = 0.69 c Half-life in Quantum Mechanics t v a µ c ln = = ln f P ( Q V ) + = a /m c π exp Z ( V0 + Q) Q B Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay 0 c 8 Half-life in Quantum Mechanics For the even-even isotopes of Th (Z =90), the previous calculations give the following table : t / (s) Q α (MeV) Measured Calculated x x 0-6. x x x x 0 7. x x 0.0 x x x

17 The Real Situation The rough and approximate calculations give good results (discrepancies are within to orders of magnitude over a range of more than 0 orders of magnitude). To improve the calculations we should have taken into consideration the following arguments : Fermi s Golden Rule : The initial and final wave functions of the transition, The fact that an alpha particle carries an angular momentum, The non spherical shape of the majority of nuclei. For the highly deformed nuclei ( > 0), the differences become very significant. We use this fact in the reverse order! Life-times are used to have an approximation of the nuclei radii. pplication of Gamow s Theory Measurements of the radii of highly deformed nuclei. Calculation and Prediction of heavier nuclei ( C ) emission Calculation and Prediction of single-proton decay processes. See Introductory Nuclear Physics (Kenneth. S. Krane) pages Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter : Radiation Sources - Supplement lpha Decay 8

18 Phys. 69: Nuclear Instrumentation Physics Department Yarmouk University Supplement : Gamma Decay Introduction Dr. Nidal M. Ershaidat Introduction The gamma decay is the emission of an energetic photon when an excited state of a nucleus decays to a lower energy state. In general a series of γ decays is necessary to reach the stable ground state. These decays are encountered each time a nucleus is excited. This excitation could be the result of a or b decays or a nuclear reaction. The resulting photons are just like the atomic X- rays in nature, i.e. they are electromagnetic radiations, but are more energetic (roughly 0. to 0 MeV, wavelengths are between 0 F and 00 F * ) * (F) = hc/e = 0/E(MeV) Importance of γ spectroscopy The Study of excited states is very rich in information about the nuclei properties. γ photons are relatively easy to detect which makes them a very popular tool for spectroscopists. The study of the competitive process to gamma decay namely, the internal conversion, is an excellent tool to obtain the spins and parities of nuclear states!

19 Energetics of γ Decay Consider a nucleus, at rest, in an excited state E i decaying to a lower energy state E f, a γ photon (E γ = p γ c) is emitted: * + γ Conservation of energy: E i = E f + E γ + T R () The symbol R stands for "recoil" of the parent. Conservation of momenta: in the cm we have: r r r r E γ 0 = p + p p = p p R = pγ = () R γ R γ c Equation can be written as: Eγ E = Ei E f = Eγ () M c 5 E γ E E γ is the solution of the quadratic equation, i.e. E Eγ = M c ± + () M c E is typically of the order of MeV. The rest energy Mc is of the order of 0 MeV, i.e. E << Mc Expanding the square root we have: E E E γ = M c (5) M c M c From (or 5), if we neglect the term E/Mc, we have E = E. γ 6 E γ E The correction to E γ due to the recoil energy [( E) /Mc ] is negligible (0-5 ) and except a special case in nuclear spectroscopy *, E γ is simply taken as equal to E E γ = E ( E ) M c (6) 7 Lifetimes for γ Emission, Selection Rules * Mössbauer Spectroscopy which is dedicated to the use of this correction

20 Example: Fig. shows the energy levels of the (even-even) 7 Se (Z=) isotope Fig. Energies and γ transition energies are given in kev Weisskopf Estimates vs. Experiment ) Evaluation of the partial decay rate for γ emission ll details are shown in an energy level scheme: spinparity, energies and/or γ transition energies and halflives of excited states. 9 Example: 7 Se rd excited state We ll have a closer look on the rd excited state of 7 Se. Spin-parity = + Energy = 7 kev t / = 8.7x0 - s Transition rd - nd = 80 kev Transition rd - st = 55 kev The measured relative intensities (or branching ratios) are γ,7 : γ,5 : γ,80 = 5:9:0 The decay constant t = ln/t / = s - Neglecting the (internal) conversion factors, t is simply the sum of the decay rates of the three transition that depopulate this excited state!, i.e. t = g,7 + g,5 + g,80 0 Comparison with Weisskopf Estimates The partial decay rates of the transitions (7,55,80) are: γ,7 = 0.5 * t =. 0 0 s - γ,55 = 0.9 * t =. 0 0 s - γ,80 = 0.0 * t = s - The following tables give the calculations of (EL) and (ML) for the energies involved in this example (=7) Weisskopf Estimates for E = 7 kev L (EL) (s - ) (ML) (s - ) (E) = (M) =.8 0 (E) = (M) = = γ,7. 00 s - (E) = (M) =.0 0 Weisskopf Estimates for E = 55 kev L (EL) (s - ) (ML) (s - ) (E) =.6 0 (M) = = γ, s - (E) = (M) = (E) = 7. 0 (M) =.9 0

21 Weisskopf Estimates for E = 80 kev The M Transitions (Systematics) Case L (EL) (s - ) (E) = = γ, s - (E) = (E) =.0 0 (ML) (s - ) (M) =.07 0 (M) = (M) = 5.8 Fig. represents the experimental data for different nuclei. The straight line represents Weisskopf estimate The previous calculations indicate that the favored transitions are the E ones. But they also show that the measured values are one order of magnitude greater than Weisskopf Estimate. There is a strong evidence for the collective structure of the nucleus!, since Weisskopf used the shell (a single individual particle) model to make his estimations. τ(m) = E -9 Fig. : log(τ ) vs. E(in kev) This figure shows, in particular, the good agreement with the expected E -9 dependence. : Selection Rules Multipoles and ngular Momenta n em field produced by oscillations of charges and currents produces also angular momentum. In QM this angular momentum carried by the quanta of energy (photons of energy E = h ν) is quantized. The rate at which this angular momentum is radiated, is proportional to the rate at which energy is radiated. The proportionality is preserved if each emitted photon carries a definite angular momentum. multipole operator of order L includes a spherical harmonic Y lm (θ,φ), which is associated with an angular momentum L. Conclusion: a multipole of order L transfers an angular momentum of L per photon. 6

22 ngular Momentum and Parity Selection Rules Consider a γ transition from an initial excited state of angular momentum I i and parity p i to a final state (I f, π f ). ssume I i I π f. (Spin-parity for these states are i π I i and I f f respectively) Conservation of angular momentum is expressed by: r r r Ii = I f + L 7 Thanks to the rules of addition of angular momenta we know that L could only have restricted values. I i I f L I i + I f Example: For I i = / and I f = 5/, the possible values for L are:,, and and the radiated field would be a mixture of dipole, quadrupole, octupole and hexadecapole radiation! The relative parity of the initial and final levels determine the type of the emitted radiation (electric or magnetic). The following table resumes the parities related to em radiations 8 7 Parity and EM Transitions The following table resumes what we know about the parity associated to electric and magnetic transition. Parity L Electric Transition Magnetic Transition Even + - Odd - + The following notations are used when studying parity changes see Table 0 I i I f π L π = no + + Even π = yes - Odd The two tables are used to determine the type of the emitted radiation (electric or magnetic) Example Let s take again the previous example I i = / and I f = 5/. For L = and π = no, the transition cannot be electric because the associated parity is (-) negative and this would give a final parity different from the initial one. In this case the transition is the magnetic dipole M transition. In the case π = yes, the transition cannot be magnetic; the associated parity is (-) + positive. In this case the transition is the electric dipole E transition. See Table 8 for the other cases. π = no π = yes L Transition M E M E L Transition E M E M 9 Selection Rules The precedent example suggests restrictions on the possible transitions. These restrictions are called selection rules π = no π = yes I i I f L I i + I f ; I i I f. even electric, odd magnetic odd electric, even magnetic 0 5

23 The Case I i = I f. This exception to the previous selection rules occurs because it would mean that L = 0 is a possible value, and there are no monopole transitions in which a single photon is emitted! * The lowest multipole order allowed in the case where I i I f. is L = Pure Multipole Transition nother interesting case is when (I i 0, I f. =0) or (I i =0, I f 0) L is equal to I i for the first type of transition (or I f for the second one). For an even Z even N nucleus (like 7 Se) the first excited state + decays to the ground state 0 + through the emission of a pure E (quadrupole) transition. * The magnetic monopole does not exist. For an electric monopole, a spherical distribution of charge (like a single point charge) the Coulomb field is not affected by radial oscillations and thus no corresponding radiation is produced. Exercise: Pure Multipole Transition The I i = I f = 0 Case Internal Conversion Find the type of transitions for the decay from the fourth state to the ground state in the case of 7 Se Here the only possible value for L is zero. This case is not permitted for radiative transitions. The transition between (excited) states with spin = 0 * occurs through the competitive process we mentioned before, namely the internal conversion. Internal Conversion In this process, the available energy is transmitted to an orbital electron which in turn is ejected. Orbital electrons with wave functions penetrating the nucleus field are concerned. * few even Z even N nuclei have 0 + first excited states. Those are forbidden to decay to the 0 + ground state by γ emission. 6

24 Selection Rules and Weisskopf Estimates 5 Selection Rules and Weisskopf Estimates 6 In table 8 we showed which transitions are possible between the states I i = / and I f = 5/. Several multipoles are permitted. For example in the π = no case, M, E, M and E are allowed. To decide which are the ones we observe, we shall make a simple calculation using Weisskopf estimates. Let s assume a medium-weight nucleus (=5) * and E = MeV. ccording to Weisskopf, the transition probabilities are as follows: (σl, =5, E= MeV) (M) = s - (E) = s - (M) =.00 0 s - (E) =.7 s - (σl)/(m) π = no L Transition M E M E * / = 5, / = 65 Selection Rules and Weisskopf Estimates 7 Expectations based on the single-particle Estimates 8 The previous results show that the lower orders are dominant. They also indicate that this transition could be composed of M radiation with possibly a small mixture of E. For the π = yes case, the calculations show that the E is dominant and the other modes are most likely not to occur! (σl, =5,E= MeV) (σl)/(e) π = yes (E) = s - L Transition (M) = s E (E) = 5. 0 s M E (M) = 0.70 s M ) The lowest permitted multipoles usually dominate ) Electric multipole emission is orders of magnitude more probable than the same magnetic multipole emission. ) Emission of multipole L+ is less probable than emission of multipole L by a factor of the order of 0-5 ) Points and combined give the following relations (L = L +) ( EL' ) ( ML) ( ML' ) ( EL) = = ( EL' ) ( EL) ( ML' ) ( ML) ( EL) ( ML) ( ML) ( EL) = 0 5 = = 0 = 0 7 7

25 Important Remark 9 The precedent calculations are based on a single particle model with simple approximations. We observe in the lab. transitions in which (E) > (M) especially in transitions between vibrational and rotational collective states. 8