Radioactive Decay, Part II

7. Radioactive decay 7.1 Gamma decay 7.1.1 Classical theory of radiation 7.1.2 Quantum mechanical theory 7.1.3 Extension to Multipoles 7.1.4 Selection Rules 7.2 Beta decay 7.2.1 Reactions and phenomenology 7.2.2 Conservation laws 7.2.3 Fermi’s Theory of Beta Decay Radioactive decay is the process in which an unstable nucleus spontaneously loses energy by emitting ionizing particles and radiation. This decay, or loss of energy, results in an atom of one type, called the parent nuclide, transforming to an atom of a different type, named the daughter nuclide. The three principal modes of decay are called the alpha, beta and gamma decays. We already introduced the general principles of radioactive decay in Section 1.3 and we studied more in depth alpha decay in Section 3.3. In this chapter we consider the other two type of radioactive decay, beta and gamma decay, making use of our knowledge of quantum mechanics and nuclear structure. 7.1 Gamma decay Gamma decay is the third type of radioactive decay. Unlike the two other types of decay, it does not involve a change in the element. It is just a simple decay from an excited to a lower (ground) state. In the process of course some energy is released that is carried away by a photon. Similar processes occur in atomic physics, however there the energy changes are usually much smaller, and photons that emerge are in the visible spectrum or x-rays. The nuclear reaction describing gamma decay can be written as A ∗ ZX →A Z X +γ where ∗ indicates an excited state. We have said that the photon carries aways some energy. It also carries away momentum, angular momentum and parity (but no mass or charge) and all these quantities need to be conserved. We can thus write an equation for the energy and momentum carried away by the gamma-photon. From special relativity we know that the energy of the photon (a massless particle) is � E = m2 c4 + p2 c2 → E = pc 2 p (while for massive particles in the non-relativistic limit v ≪ c we have E ≈ mc2 + 2m .) In quantum mechanics we have seen that the momentum of a wave (and a photon is well described by a wave) is p = hk with k the wave number. Then we have E = hkc = hωk This is the energy for photons which also defines the frequency ωk = kc (compare this to the energy for massive 2 2 k particles, E = n2m ). Gamma photons are particularly energetic because they derive from nuclear transitions (that have much higher energies than e.g. atomic transitions involving electronic levels). The energies involved range from E ∼ .1 ÷ 10MeV, 4 giving k ∼ 10−1 ÷ 10−3 fm−1 . Than the wavelengths are λ = 2π k ∼ 100 ÷ 10 fm, much longer than the typical nuclear dimensions. Gamma ray spectroscopy is a basic tool of nuclear physics, for its ease of observation (since it’s not absorbed in air), accurate energy determination and information on the spin and parity of the excited states. Also, it is the most important radiation used in nuclear medicine. Ei, Ii, Πi Eγ=ħω=Ei-Ef Πγ=ΠiΠf Lγ 7.1.1 Classical theory of radiation From the theory of electrodynamics it is known that an accelerating charge radiates. The power radiated is given by the integral of the energy flux (as given by the Poynting vector) over all solid angles. This gives the radiated power as: P = Ef, If, Πf Fig. 42: Schematics of gamma decay 2 e2 |a|2 3 c3 where a is the acceleration. This is the so-called Larmor formula for a non-relativistic accelerated charge. Example. As an important example we consider an electric dipole. An electric dipole can be considered as an oscillating charge, over a range r0 , such that the electric dipole is given by d(t) = qr(t). Then the equation of motion is r(t) = r0 cos(ωt) and the acceleration a = r̈ = −r0 ω 2 cos(ωt) Averaged over a period T = 2π/ω, this is � 2� ω a = 2π T dta(t) = 0 1 2 4 r ω 2 0 Finally we obtain the radiative power for an electric dipole: PE1 = 1 e2 ω 4 |'r0 |2 3 c3 A. Electromagnetic multipoles In order to determine the classical e.m. radiation we need to evaluate the charge distribution that gives rise to it. The electrostatic potential of a charge distribution ρe (r) is given by the integral: V ('r) = 1 4πǫ0 V ol′ ρe (r'′ ) |'r − r'′ | When treating radiation we are only interested in the potential outside the charge and we can assume the charge 1 (e.g. a particle!) to be well localized (r ′ ≪ r). Then we can expand |!r−r !′ | in power series. First, we express explicitly J √ ( r ′ /2 ′ ′ the norm |'r − r'′ | = r2 + r′2 − 2rr′ cos ϑ = r 1 + r − 2 rr cos ϑ. We set R = rr and ǫ = R2 − 2R cos ϑ: this is a small quantity, given the assumption r′ ≪ r. Then we can expand: ( ) 1 1 1 1 1 3 5 = √ = 1 − ǫ + ǫ2 − ǫ3 + . . . r 1+ǫ r 2 8 16 |'r − r'′ | Replacing ǫ with its expression we have: ( ) 1 1 1 1 3 5 √ = 1 − (R2 − 2R cos ϑ) + (R2 − 2R cos ϑ)2 − (R2 − 2R cos ϑ)3 + . . . r 1+ǫ r 2 8 16 ) ( 15 15 5 1 1 3 3 3 5R6 + R5 cos(ϑ) − R4 cos2 (ϑ) + R3 cos3 (ϑ)] + . . . = 1 + [− R2 + R cos ϑ] + [ R4 − R3 cos ϑ + R2 cos2 ϑ] + [− r 2 8 2 2 16 8 4 2 ( ) ) ) ( ( 2 3 5 cos (ϑ) 3 cos(ϑ) 3 cos ϑ 1 1 = 1 + R cos ϑ + R2 − + R3 − + ... r 2 2 2 2 We recognized in the coefficients to the powers of R the Legendre Polynomials Pl (cos ϑ) (with l the power of Rl , and note that for powers > 3 we should have included higher terms in the original ǫ expansion): ∞ ∞ l=0 l=0 1 1 10 10 l √ = R Pl (cos ϑ) = r 1+ǫ r r ( ′ )l r Pl (cos ϑ) r With this result we can as well calculate the potential: 1 1 V ('r) = 4πǫ0 r Z ∞ 10 ρ('r ) r V ol′ ′ l=0 ( ′ )l r Pl (cos ϑ)dr'′ r The various terms in the expansion are the multipoles. The few lowest ones are : Z 1 1 Q ρ('r′ ) dr'′ = 4πǫ0 r 4πǫ0 r V ol′ Z Z ' 1 1 1 1 ′ ′ ′ ′ ′ ' '′ = r̂ · d ρ(' r )r P (cos ϑ) dr = ρ(' r )r cos ϑd r 1 4πǫ0 r2 ZV ol′ 4πǫ0 r)2 4πǫ0 r2 Z V ol′ ( 1 1 3 1 1 1 ρ('r′ )r′2 P2 (cos ϑ) dr'′ = ρ('r′ )r′2 cos2 ϑ − dr'′ 4π ǫ0 r3 V ol′ 2 2 4πǫ0 r3 V ol′ Monopole Dipole Quadrupole This type of expansion can be carried out as well for the magnetostatic potential and for the electromagnetic, time-dependent field. At large distances, the lowest orders in this expansion are the only important ones. Thus, instead of considering the total radiation from a charge distribution, we can approximate it by considering the radiation arising from the first few multipoles: i.e. radiation from the electric dipole, the magnetic dipole, the electric quadrupole etc. Each of these radiation terms have a peculiar angular dependence. This will be reflected in the quantum mechanical treatment by a specific angular momentum value of the radiation field associated with the multipole. In turns, this will give rise to selection rules determined by the addition rules of angular momentum of the particles and radiation involved in the radiative process. 7.1.2 Quantum mechanical theory In quantum mechanics, gamma decay is expressed as a transition from an excited to a ground state of a nucleus. Then we can study the transition rate of such a decay via Fermi’s Golden rule W = 2π |�ψf | V̂ |ψi � |2 ρ(Ef ) h There are two important ingredients in this formula, the density of states ρ(Ef ) and the interaction potential Vˆ . A. Density of states ny The density of states is defined as the number of available states per Ns , where Ns is the number of states. We have seen energy: ρ(Ef ) = dd E f at various time the concept of degeneracy: as eigenvalues of an oper­ ator can be degenerate, there might be more than one eigenfunction sharing the same eigenvalues. In the case of the Hamiltonian, when there are degeneracies it means that more than one state share the same energy. By considering the nucleus+radiation to be enclosed in a cavity of volume L3 , we have for the emitted photon a wavefunction represented by the solution of a particle in a 3D box that we saw in a Problem Set. As for the 1D case, we have a quantization of the momentum (and hence of the wave-number k) in order to fit the wavefunction in the box. Here we just have a quantization in all 3 directions: kx = 2π 2π 2π nx , ky = ny , kz = nz , L L L dn n nx Fig. 43: Density of states: counting the states (2D) (with n integers). Then, going to spherical coordinates, we can count the number of states in a spherical shell between L3 n and n + dn to be dNs = 4πn2 dn. Expressing this in terms of k, we have dNs = 4πk 2 dk (2π) 3 . If we consider just a small solid angle dΩ instead of 4π we have then the number of state dNs = finally obtain the density of states: ρ(E) = L3 2 (2π)3 k dkdΩ. Since E = hkc = hω, we L3 2 dk L3 k 2 ω 2 L3 dNs = k dΩ = dΩ = dΩ (2π)3 hc hc3 (2π)3 dE (2π)3 dE B. The vector potential Next we consider the potential causing the transition. The interaction of a particle with the e.m. field can be expressed 'ˆ of the e.m. field as: in terms of the vector potential A V̂ = e 'ˆ ˆ A · p' mc 'ˆ in QM is an operator that can create or annihilate where p'ˆ is the particle’s momentum. The vector potential A photons, 0 2πhc2 ! ! 'ˆ = A (âk eik·!r + â†k e−ik·!r )'ǫk V ωk k annihilates (creates) one photon of momentum 'k. Also, 'ǫk is the polarization of the e.m. field. Since where gamma decay (and many other atomic and nuclear processes) is able to create photons (or absorb them) it makes sense that the operator describing the e.m. field would be able to describe the creation and annihilation of photons. ! The second characteristic of this operator are the terms ∝ e−ik·!r which describe a plane wave, as expected for e.m. waves, with momentum hk and frequency ck. âk (âk† ) C. Dipole transition for gamma decay To calculate the transition rate from the Fermi’s Golden rule, W = 2π |�ψf | Vˆ |ψi �i |2 ρ(Ef ), h we are really only interested in the matrix element h�ψf | Vˆ |ψi �i, where the initial state does not have any photon, and the final has one photon of momentum hk and energy hω = hkc. Then, the only element in the sum above for the ˆ†k , with the appropriate 'k momentum: vector potential that gives a non-zero contribution will be the term ∝ a Vif = e mc ) ( 2πhc2 ! 'ǫk · pe 'ˆ −ik·!r V ωk i ˆ2 ˆ This can be simplified as follow. Remember that [p'ˆ2 , 'rˆ] = −2ihp'ˆ. Thus we can write, p'ˆ = 2n [p' , 'r] = 2 2 ˆ ˆ p ! ! im p ˆ ˆ ˆ r), 'r]. We introduced the nuclear Hamiltonian Hnuc = 2m + Vnuc ('r): thus we have p'ˆ = n [ 2m + Vnuc (' Taking the expectation value � im � h�ψf | Hnuc'rˆ |ψi �i − h�ψf | 'rˆHnuc |ψi �i h�ψf | p'ˆ |ψi �i = h and remembering that |ψi,f �i are eigenstates of the Hamiltonian, we have im (Ef − Ei ) h�ψf | 'rˆ |ψi �i = imωk h�ψf | 'rˆ |ψi �i , h�ψf | p'ˆ |ψi �i = h where we used the fact that (Ef − Ei ) = hωk by conservation of energy. Thus we obtain Vif e = mc � ( ) ( ) 2πhc2 2πhe2 ωk ! ! ik ! r − · imω'ǫk · 'rˆe =i 'ǫk · 'rˆe−ik·!r V V ωk ˆ2 ! im p rˆ] = n [ 2m , ' im rˆ]. n [Hnuc , ' We have seen that the wavelengths of gamma photons are much larger than the nuclear size. Then 'k · 'r ≪ 1 and we L ! ' · 'r)l = L 1 (−ikr cos ϑ)l . This series is very similar in meaning can make an expansion in series : e−k·!r ∼ l l!1 (−ik l l! to the multipole series we saw for the classical case. For example, for l = 0 we obtain: � r 2πhe2 ωk ( ˆ) 'r · 'ǫk Vif = V ˆ which is the dipolar approximation, since it can be written also using the (electr ) ic dipole ( ) operator e'r. ˆ ˆ ˆ The angle between the polarization of the e.m. field and the position 'r is 'r · 'ǫ = 'r sin ϑ The transition rate for the dipole radiation, W ≡ λ(E1) is then: ω 3 ( ˆ) 2 2 2π | 'r | sin ϑ dΩ |�ψf | Vˆ |ψi �i |2 ρ(Ef ) = h 2πc3 h Jπ J 2π and integrating over all possible direction of emission ( 0 dϕ 0 (sin2 ϑ) sin ϑdϑ = 2π 43 ): 4 e2 ω 3 ( ˆ) 2 λ(E1) = | 'r | 3 hc3 Multiplying the transition rate (or photons emitted per unit time) by the energy of the photons emitted we obtain the radiated power, P = W hω: 4 e2 ω 4 ( ˆ) 2 | 'r | P = 3 c3 Notice the similarity of this formula with the classical case: λ(E1) = 1 e2 ω 4 |'r0 |2 3 c3 We can estimate the transition rate by using a typical energy E = hω for the photon emitted (equal to a (typical ) energy difference between excited and ground state nuclear levels) and the expectation value for the dipole ( | 'rˆ | ∼ PE1 = Rnuc ≈ r0 A1/3 ). Then, the transition rate is evaluated to be λ(E1) = e2 E 3 2 2/3 r A = 1.0 × 1014 A2/3 E 3 hc (hc)3 0 (with E in MeV). For example, for A = 64 and E = 1MeV the rate is λ ≈ 1.6 × 1015 s−1 or τ = 10−15 (femtoseconds!) for E = 0.1MeV τ is on the order of picoseconds. Obs. Because of the large energies involved, very fast processes are expected in the nuclear decay from excited states, in accordance with Fermi’s Golden rule and the energy/time uncertainty relation. 7.1.3 Extension to Multipoles We obtained above the transition rate for the electric dipole, i.e. when the interaction between the nucleus and the e.m. field is described by an electric dipole and the emitted radiation has the character of electric dipole radiation. This type of radiation can only carry out of the nucleus one quantum of angular momentum (i.e. Δl = ±1, between excited and ground state). In general, excited levels differ by more than 1 l, thus the radiation emitted need to be a higher multipole radiation in order to conserve angular momentum. A. Electric Multipoles We can go back to the expansion of the radiation interaction in multipoles: 01 ˆ ' · 'rˆ)l Vˆ ∼ (ik l! l Then the transition rate becomes: 8π(l + 1) e2 λ(El) = l[(2l + 1)!!]2 hc ( E hc )2l+1 ( 3 l+3 )2 ( ) 2l c |'rˆ| ( ) Notice the strong dependence on the l quantum number. Setting again | 'rˆ | ∼ r0 A1/3 we also have a strong dependence on the mass number. Thus, we have the following estimates for the rates of different electric multipoles: - λ(E1) = 1.0 × 1014 A2/3 E 3 - λ(E2) = 7.3 × 107 A4/3 E 5 - λ(E3) = 34A2 E 7 - λ(E4) = 1.1 × 10−5 A8/3 E 9 B. Magnetic Multipoles The e.m. potential can also contain magnetic interactions, leading to magnetic transitions. The transition rates can be calculated from a similar formula: )� ( )2 ( ) � ( 2l−2 1 8π(l + 1) e2 E 2l+1 3 h ˆ| λ(M l) = µ − c |' r p l+3 mp c l+1 l[(2l + 1)!!]2 hc hc where µp is the magnetic moment of the proton (and mp its mass). 1 Estimates for the transition rates can be found by setting µp − l+1 ≈ 10: - λ(M 1) = 5.6 × 1013 E 3 - λ(M 2) = 3.5 × 107 A2/3 E 5 - λ(M 3) = 16A4/3 E 7 - λ(M 4) = 4.5 × 10−6 A2 E 9 7.1.4 Selection Rules The angular momentum must be conserved during the decay. Thus the difference in angular momentum between the initial (excited) state and the final state is carried away by the photon emitted. Another conserved quantity is the total parity of the system. A. Parity change The parity of the gamma photon is determined by its character, either magnetic or electric multipole. We have Πγ (El) = (−1)l Πγ (M l) = (−1)l−1 Electric multipole Magnetic multipole Then if we have a parity change from the initial to the final state Πi → Πf this is accounted for by the emitted photon as: Π γ = Πi Πf This of course limits the type of multipole transitions that are allowed given an initial and final state. ΔΠ = no → Even Electric, Odd Magnetic ΔΠ = yes → Odd Electric, Even Magnetic B. Angular momentum From the conservation of the angular momentum: ˆ ˆ 'ˆ γ I'i = I'f + L the allowed values for the angular momentum quantum number of the photon, l, are restricted to lγ = |Ii − If |, . . . , Ii + If Once the allowed l have been found from the above relationship, the character (magnetic or electric) of the multipole is found by looking at the parity. In general then, the most important transition will be the one with the lowest allowed l, Π. Higher multipoles are also possible, but they are going to lead to much slower processes. Multipolarity M1 M2 M3 M4 M5 Angular Momentum l 1 2 3 4 5 Parity Π + + + Multipolarity E1 E2 E3 E4 E5 Angular Momentum l 1 2 3 4 5 Parity Π + + - Table 3: Angular momentum and parity of the gamma multipoles C. Dominant Decay Modes In general we have the following predictions of which transitions will happen: 1. The lowest permitted multipole dominates 2. Electric multipoles are more probable than the same magnetic multipole by a factor ∼ 102 (however, which one is going to happen depends on the parity) λ(El) ≈ 102 λ(M l) 3. Emission from the multipole l + 1 is 10−5 times less probable than the l-multipole emission. λ(E, l + 1) ≈ 10−5 , λ(El) λ(M, l + 1) ≈ 10−5 λ(M l) λ(E, l + 1) ≈ 10−3 , λ(M l) λ(M, l + 1) ≈ 10−7 λ(El) 4. Combining 2 and 3, we have: Thus E2 competes with M 1 while that’s not the case for M 2 vs. E1 D. Internal conversion What happen if no allowed transitions can be found? This is the case for even-even nuclides, where the decay from the 0+ excited state must happen without a change in angular momentum. However, the photon always carries some angular momentum, thus gamma emission is impossible. Then another process happens, called internal conversion: A ∗ ZX − →A Z X +e − where A Z X is a ionized state and e is one of the atomic electrons. Besides the case of even-even nuclei, internal conversion is in general a competing process of gamma decay (see Krane for more details). 7.2 Beta decay The beta decay is a radioactive decay in which a proton in a nucleus is converted into a neutron (or vice-versa). In the process the nucleus emits a beta particle (either an electron or a positron) and quasi-massless particle, the neutrino. Courtesy of Thomas Jefferson National Accelerator Facility - Office of Science Education. Used with permission. Fig. 44: Beta decay schematics Recall the mass chain and Beta decay plots of Fig. 7. When studying the binding energy from the SEMF we saw that at fixed A there was a minimum in the nuclear mass for a particular value of Z. In order to reach that minimum, unstable nuclides undergo beta decay to transform excess protons in neutrons (and vice-versa). 7.2.1 Reactions and phenomenology The beta-decay reaction is written as: A Z XN → A ′ Z+1 XN −1 + e− + ν̄ This is the β − decay. (or negative beta decay) The underlying reaction is: n → p + e− + ν̄ which converts a proton into a neutron with the emission of an electron and an anti-neutrino. There are two other types of reactions, the β + reaction, A Z XN → A ′ Z−1 XN +1 + e+ + ν ⇐⇒ p → n + e+ + ν which sees the emission of a positron (the electron anti-particle) and a neutrino; and the electron capture: A Z XN + e− → A ′ Z−1 XN +1 +ν ⇐⇒ p + e− → n + ν a process that competes with, or substitutes, the positron emission. Examples 64 29 Cu ր ց 64 − ¯ 30 Zn + e + ν, 64 + Ni + e + ν, 28 The neutrino and beta particle (β ± ) share the energy. Qβ = 0.57M eV Qβ = 0.66M eV Since the neutrinos are very difficult to de­ tect (as we will see they are almost massless and interact very weakly with matter), the elec­ trons/positrons are the particles detected in beta-decay and they present a characteristic en­ ergy spectrum (see Fig. 45). The difference be­ tween the spectrum of the β ± particles is due to the Coulomb repulsion or attraction from the nucleus. Notice that the neutrinos also carry away an­ gular momentum. They are spin-1/2 particles, with no charge (hence the name) and very small mass. For many years it was actually believed to have zero mass. However it has been confirmed that it does have a mass in 1998. © Neil Spooner. All rights reserved. Fig. 45: Beta decay spectra: Distribution of momentum (top plots) and kinetic energy (bottom) for β − (left) and β + (right) decay. 7.2.2 Conservation laws As the neutrino is hard to detect, initially the beta decay seemed to violate energy conservation. Introducing an extra particle in the process allows one to respect conservation of energy. Besides energy, there are other conserved quantities: - Energy: The Q value of a beta decay is given by the usual formula: ′ 2 Qβ − = [mN (A X) − mN (A Z+1 X ) − me ]c . Using the atomic masses and neglecting the electron’s binding energies as usual we have ′ 2 A A ′ 2 Qβ − = {[mA (A X) − Zme ] − [mA (A Z+1 X ) − (Z + 1)me ] − me }c = [mA ( X) − mA (Z+1 X )]c . The kinetic energy (equal to the Q) is shared by the neutrino and the electron (we neglect any recoil of the massive nucleus). Then, the emerging electron (remember, the only particle that we can really observe) does not have a fixed energy, as it was for example for the gamma photon. But it will exhibit a spectrum of energy (which is the number of electron at a given energy) as well as a distribution of momenta. We will see how we can reproduce these plots by analyzing the QM theory of beta decay. - Momentum: The momentum is also shared between the electron and the neutrino. Thus the observed electron momentum ranges from zero to a maximum possible momentum transfer. - Angular momentum (both the electron and the neutrino have spin 1/2) - Parity? It turns out that parity is not conserved in this decay. This hints to the fact that the interaction responsible violates parity conservation (so it cannot be the same interactions we already studies, e.m. and strong interactions) - Charge (thus the creation of a proton is for example always accompanied by the creation of an electron) - Lepton number: we do not conserve the total number of particles (we create beta and neutrinos). However the number of massive, heavy particles (or baryons, composed of 3 quarks) is conserved. Also the lepton number is conserved. Leptons are fundamental particles (including the electron, muon and tau, as well as the three types of neutrinos associated with these 3). The lepton number is +1 for these particles and -1 for their antiparticles. Then an electron is always accompanied by the creation of an antineutrino, e.g., to conserve the lepton number (initially zero). 7.2.3 Fermi’s Theory of Beta Decay The properties of beta decay can be understood by studying its quantum-mechanical description via Fermi’s Golden rule, as done for gamma decay. W = 2π |�ψf | Vˆ |ψi �i |2 ρ(Ef ) h In gamma decay process we have seen how the e.m. field is described as an operator that can create (or destroy) photons. Nobody objected to the fact that we can create this massless particles. After all, we are familiar with charged particles that produce (create) an e.m. field. However in QM photons are also particles, and by analogy we can have also creation of other types of particles, such as the electron and the neutrino. For the beta decay we need another type of interaction that is able to create massive particles (the electron and neutrino). The interaction cannot be given by the e.m. field; moreover, in the light of the possibilities of creating and annihilating particles, we also need to find a new description for the particles themselves that allows these processes. All of this is obtained by quantum field theory and the second quantization. Quantum field theory gives a unification of e.m. and weak force (electro-weak interaction) with one coupling constant e. The interaction responsible for the creation of the electron and neutrino in the beta decay is called the weak interaction and its one of the four fundamental interactions (together with gravitation, electromagnetism and the strong interaction that keeps nucleons and quarks together). One characteristic of this interaction is parity violation. A. Matrix element The weak interaction can be written in terms of the particle field wavefunctions: Vint = gΨe† Ψν̄† where Ψa (Ψa† ) annihilates (creates) the particle a, and g is the coupling constant that determines how strong the interaction is. Remember that the analogous operator for the e.m. field was ∝ a†k (creating one photon of momentum k). Then the matrix element Vif = h�ψf | Hint |ψi �i can be written as: Vif = g Z d3 'x Ψp∗ ('x) [Ψe∗ ('x)Ψν̄∗ ('x)] Ψn ('x) (Here † → ∗ since we have scalar operators). To first approximation the electron and neutrino can be taken as plane waves: Vif = g Z ! ! eike ·!x eikν ·!x √ Ψn ('x) d3 'x Ψp∗ ('x) √ V V and since kR ≪ 1 we can approximate this with Vif = g V Z d3 'x Ψp∗ ('x)Ψn ('x) We then write this matrix element as g Mnp V where Mnp is a very complicated function of the nuclear spin and angular momentum states. In addition, we will use in the Fermi’s Golden Rule the expression Vif = |Mnp |2 → |Mnp |2 F (Z0 , Qβ ) where the Fermi function F (Z0 , Qβ ) accounts for the Coulomb interaction between the nucleus and the electron that we had neglected in the previous expression (where we only considered the weak interaction). B. Density of states In studying the gamma decay we calculated the density of states, as required by the Fermi’s Golden Rule. Here we need to do the same, but the problem is complicated by the fact that there are two types of particles (electron and neutrino) as products of the reaction and both can be in a continuum of possible states. Then the number of states in a small energy volume is the product of the electron and neutrino’s states: d2 Ns = dNe dNν . The two particles share the Q energy: Qβ = Te + Tν . For simplicity we assume that the mass of the neutrino is zero (it’s much smaller than the electron mass and of the kinetic mass of the neutrino itself). Then we can take the relativistic expression Tν = cpν , while for the electron E 2 = p2 c2 + m2 c4 → E = T e + me c 2 with Te = � p p2e c2 + m2e c4 − me c2 and we then write the kinetic energy of the neutrino as a function of the electron’s, T ν = Qβ − Te . The number of states for the electron can be calculated from the quantized momentum, under the assumption that ! the electron state is a free particle (ψ ∼ eik·!r ) in a region of volume V = L3 : dNe = ( L 2π )3 4πke2 dke = 4πV 2 p dpe (2πh)3 e and the same for the neutrino, dNν = 4πV 2 p dpν , (2πh)3 ν where we used the relationship between momentum and wavenumber: p' = h'k. At a given momentum/energy value for the electron, we can write the density of states as ρ(pe )dpe = dNe d Nν V2 2 d pν V2 = 16π 2 pe dpe pν2 = [Q − Te ]2 p2e dpe 6 d Tν (2π h) d Tν 4π 4 h6 c3 where we used : dd Tpνν = c and pν = (Qβ − Te )/c. The density of states is then ρ(pe )dpe = � p V2 V2 2 2 [Q − T ] p dp = [Q − ( p2e c2 + m2e c4 − me c2 )]2 p2e dpe e e e 4 6 3 4 6 3 4π h c 4π h c or rewriting this expression in terms of the electron kinetic energy: ρ(Te ) = V2 V2 2 2 dpe [Q − T ] p = [Q − Te ]2 e e 4π 4 h6 c3 dTe 4c6 π 4 h6 J ( / Te 2 + 2Te me c2 Te + me c2 (as pe dpe = (Te + me c2 )/c2 dTe ) Knowing the density of states, we can calculate how many electrons are emitted in the beta decay with a given energy. This will be proportional to the rate of emission calculated from the Fermi Golden Rule, times the density of states: �2 � p p2 p2 � 2 N (p) = C F (Z, Q)|Vf i |2 2 [Q − T ] = CF (Z, Q)|Vf i |2 2 Q − ( p2e c2 + m2e c4 − me c2 ) c c and N (Te ) = � p C 2 2 [Q − T ] Te2 + 2Te me c2 (Te + me c2 ) F (Z, Q)|V | e f i c5 These distributions are nothing else than the spectrum of the emitted beta particles (electron or positron). In these expression we collected in the constant C various parameters deriving from the Fermi Golden Rule and density of states calculations, since we want to highlight only the dependence on the energy and momentum. Also, we introduced a new function, F (Z, Q), called the Fermi function, that takes into account the shape of the nuclear wavefunction and in particular it describes the Coulomb attraction or repulsion of the electron or positron from the nucleus. Thus, F (Z, Q) is different, depending on the type of decay. These distributions were plotted in Fig. 45. Notice that these distributions (as well as the decay rate below) are the product of three terms: - the Statistical factor (arising from the density of states calculation), p2 c2 - the Fermi function (accounting for the Coulomb interaction), F (Z, Q) - and the Transition amplitude from the Fermi Golden Rule, |Vf i |2 2 [Q − T ] These three terms reflect the three ingredients that determine the spectrum and decay rate of in beta decay processes. C. Decay rate The decay rate is obtained from Fermi’s Golden rule: W = 2π |Vif |2 ρ(E) h where ρ(E) is the total density of states. ρ(E) (and thus the decay rate) is obtained by summing over all possible states of the beta particle, as counted by the density of states. Thus, in practice, we need to integrate the density of states over all possible momentum of the outgoing electron/positron. Upon integration over pe we obtain: ρ(E) = V2 4π 4 h6 c3 Z pemax 0 dpe [Q − Te ]2 p2e ≈ V 2 (Q − mc2 )5 4π 4 h6 c3 30c3 (where we took Te ≈ pc in the relativistic limit for high electron speed). We can finally write the decay rate as: W = V 2 (Q − mc2 )5 2π 2π � g �2 |Vif |2 ρ(E) = |Mnp |2 F (Z, Qβ ) 4 6 3 30c3 h h V 4π h c = G2F |Mnp |2 F (Z, Qβ ) where we introduced the constant (Q − mc2 )5 60π 3 h(hc)6 gm2e c 2π 3 h3 GF = √ 1 which gives the strength of the weak interaction. Comparing to the strength of the electromagnetic interaction, as 2 1 given by the fine constant α = ne c ∼ 137 , the weak is interaction is much smaller, with a constant ∼ 10−6 . We can also write the differential decay rate dd W pe : dW 2π = |Vif |2 ρ(pe ) ∝ F (Z, Q)[Q − Te ]2 p2e h d pe The square root of this quantity is then a linear function in the neutrino kinetic energy, Q − Te : s dW 1 ∝ Q − Te d pe p2e F (Z, Q) This is the Fermi-Kurie relation. Usually, the Fermi-Kurie plot is used to infer by linear regression the maximum electron energy (or Q) by finding the straight line intercept. © Neil Spooner. All rights reserved. . Fig. 46: Example of Fermi-Kurie plot (see also Krane, Fig. 9.4, 9.5)