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Beta decay Examine the stability against beta decay by plotting the rest mass

Beta decay Examine the stability against beta decay by plotting the rest mass energy M of nuclear isobars (same value of A ) along a third axis perpendicular to the N/Z plane. Recalling the semi-emperical mass formula. where the binding energy was modeled by.

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Beta decay Examine the stability against beta decay by plotting the rest mass

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  1. Beta decay Examine the stability against beta decay by plotting the rest mass energy M of nuclear isobars (same value of A) along a third axis perpendicular to the N/Z plane.

  2. Recalling the semi-emperical mass formula where the binding energy was modeled by Note that for constant A this gives a parabola in MvsZ which implies the mass takes on some minimum value!

  3. A = 127 isobars -80 -80 -84 -84 Mass defect  (MeV) -86 -86 -88 -88 50 52 54 56 Atomic number, Z

  4. For odd A the isobars lie on a single parabola as a function of Z. In this case there is only a single stable isobar to which the other members on the parabola decay by electron or positron emission depending on whether their Z value is lower or higher than that of the stable isobar.

  5. For even A the pairing term in the Mass formula splits the parabola into two, one for even Z and one for odd Z. The beta decay transition switches from one parabola to the other and in this case there may be two or even three stable isobars. Note also this implies some even A nuclei (eg 29Cu64) can decay by either electron or positron emission.

  6. 42Mo A = 104 isobars  43Tc 103.912 A Z A Z+1 ? ? -decay: X  Y + - N N-1 ? A Z A Z-1 e-capture: X + e  Y 48Cd 103.910 N N+1 Odd Z 47Ag Mass, u 103.908 45Rh Even Z 103.906 44Ru  46Pd 103.904 42 44 46 48 Atomic Number, Z

  7. Electron emission If Mc2(A,Z)is the rest mass energy of the parent atom and Q is the energy released in the decay then where M+(A,Z+1)c2 is the rest mass energy of the positive ion produced. Now since the ionization potential energy is relatively small we can write - energy will bereleasedin the decay(ie Q > 0) providedM(A,Z)c2is greater thanM(A,Z+1)c2.

  8. What is the maximum energy of the electron emitted in the b- decay of 3 1 H ? The reaction is: Neglecting the kinetic energy of the nucleus, and mass of the neutrino the Q is shared between e and n. Keat maximum whenKn0, so

  9. Positron emission The charge on the nucleus is reduced and the daughter atom has an excess electron. Thus The2mec2account for the emitted positron and the excess atomic electron in the final state. In this case the decay process can only proceed (Q > 0) if M(A,Z)c2 > M(A,Z-1)c2 + 2mec2. i.e, the difference in rest mass energies must be at least 1 MeV.

  10. Electron capture Instead of pn + e+ + occuring inside the nucleus it is possible for an atomic electron to be captured in the process p + e-    n +  and the daughter atom produced has the correct compliment of electrons. butin an excited state (with a vacancy in its electron configuration). This process is often called K capture because it is an inner shell atomic electron which is captured. E*is the excitation energy of the daughter atom (sufficiently small to be ignored). Electron capture can proceed (Q > 0) if M(A,Z)c2 < M(A,Z-1)c2. It strongly competes with positron decay but is the only decay route available to nuclei with a proton excess but a mass difference between the atoms of less than1 MeV/c2.

  11. -decay -decay

  12. Some Alpha Decay Energies and Half-lives Isotope KEa(MeV) t1/2l(sec-1) 232Th 4.01 1.41010 y 1.610-18 238U 4.19 4.5109 y 4.910-18 230Th 4.69 8.0104y 2.810-13 238Pu 5.50 88years 2.510-10 230U 5.89 20.8 days 3.910-7 220Rn 6.29 56 seconds 1.210-2 222Ac 7.01 5 seconds 0.14 216Rn 8.05 45.0msec 1.5104 212Po 8.78 0.30msec 2.3106 216Rn 8.78 0.10msec 6.9106

  13. 1930 Series of studiesofnuclear beta decay, e.g., Potassiumgoestocalcium19K40 20Ca40 Coppergoestozinc29Cu64 30Zn64 Borongoestocarbon5B12  6C12 Tritiumgoes tohelium1H3  2He3 Potassium nucleus Before decay: After decay: A B A) B) C) both the same Which fragment has a greater momentum? energy?

  14. 1932 Once the neutron was discovered, included the more fundamental n  p + e For simple 2-bodydecay, conservation of energy and momentum demand both the recoil of the nucleus and energy of the emitted electron be fixed (by the energy released through the loss of mass) to a single precise value. Ee = (mA2 - mB2 + me2)c2/2mA but this only seems to match the maximum value observed on a spectrum of beta ray energies!

  15. No. of counts per unit energy range 0 5 10 15 20 Electron kinetic energy in KeV The beta decay spectrum of tritium ( H  He). Source: G.M.Lewis, Neutrinos (London: Wykeham, 1970), p.30)

  16. 1932n  p + e- + neutrino charge0 +1 -1 ? mass939.56563938.272310.51099906? MeV MeV MeV neutrino mass < 5.1 eV < me /100000  0

  17. 1936 Millikan’s group shows at earth’s surface cosmic ray showers are dominated by electrons, gammas, and X-particles capable of penetrating deep underground (to lake bottom and deep tunnel experiments) and yielding isolated single cloud chamber tracks

  18. 1937 Street and Stevenson • 1938 Andersonand Neddermeyer • determine X-particles • are charged • have 206× the electron’s mass • decay to electrons with • a mean lifetime of 2msec 0.000002 sec

  19. 1947 Lattes, Muirhead, Occhialini and Powell observe pion decay Cecil Powell (1947) Bristol University 

  20. C.F.Powell, P.H. Fowler, D.H.Perkins Nature 159, 694 (1947) Nature 163, 82 (1949)

  21. Consistently ~600 microns (0.6 mm) 

  22. Under the influence of a magnetic field m+ p+ m+ energy always predictably fixed by Ep simple 2-body decay! p+m+ + neutrino? charge +1 +1 ?

  23. n p + e- + neutrino? p+m+ + neutrino? Then m-e- + neutrino? p ??? m e As in the case of decaying radioactive isotopes, the electrons’s energy varied, with a maximum cutoff (whose value was the 2-body prediction) 3bodydecay! e m 2 neutrinos

  24. 1953, 1956, 1959 Savannah River (1000-MWatt) Nuclear Reactor in South Carolina looked for the inverse of the process n p + e- + neutrino p + neutrino n + e+ with estimate flux of 51013 neutrinos/cm2-sec observed 2-3 p + neutrino events/hour also looked for n + neutrino p + e- but never observed! Cowan & Reines

  25. Fermi Theory starting from Fermi’s Golden Rule where M is the “matrix” element for the transition and Hint is the interaction responsible for it and as we’ve talked about before is the “density of states” Recall: The Golden Rule’s Wif is the rate of transition from initial state i to final state f , i.e. Wif = 

  26. For a single particle we’re dealing with a 6-dimensional phase space – three position coordinates and three momentum components. The smallest element of volume a state can occupy in this phase space is h3. (ignoring for now particle spin) recall the uncertainty relation dxdpx ~ h Thus (dx)3(dpx)3 ~ h3 and the number of states in a the “volume” element d3x d3p is given by d3xdx dy dzandd3pdpxdpydpz.

  27. For a coordinate space volume V and a momentum volume 4 p2dp for both electron and neutrino sharing the same volume: already integrating overd3xeandd3x

  28. With three particles in the final state peand pare independent. Within the overall energy constraint the momenta is balanced by that of the daughter nucleus. This means that any electron momentum can be expressed in terms of the neutrino state. Its the electron which can be directly detected, so often the neutrino quantities are written in terms of

  29. To calculate the matrix element we replace Hintby a constant g (coupling constant) and assume e and are ~constant inside the volumeV (and ~zero elsewhere). Actually: For 1 MeV electron p/ħ=0.007 fm-1 and within the nucleus r<6 fm

  30. dNf/dE = 162V2(Tmax - Te)2(pe)2dpe/(h6c3) Energy spectrum of beta decay electrons from 210Bi Intensity Kinetic energy, MeV

  31. Then, under this approximation: thus and where P(pe) is the probability that a beta particle with momentum pe will be emitted in unit time.

  32. The individual contributions, dl , per specific pe value: This quadratic equation has zeroes at pe = 0 and Te = Tmax Q

  33. This phase space factor determines the decay electron momentum spectrum. (shown below with the kinetic energy spectrum for the nuclide).

  34. This does not take into account the effect of the nucleus’ electric charge which accelerates the positrons and decelerates the electrons. Adding the Fermi function F(Z,pe) , a special factor (generally in powers of Z and pe), is introduced to account for this.

  35. P(pe) is  the measured beta momentum spectrum

  36. Including the measured probability Pmeas(pe) in which when plotted against Te yields a straight line the Fermi-Kurie plot.

  37. The overall decay probability   is obtained by integratingP(pe)dpe from 0 topmax: with energy and momenta expressed in units of mc2 and mc respectively. The integral is often tabulated when it cannot be solved analytically.

  38.  = G2 [MN]2f(Z,Tmax) Since the decay probability= ln(2)/t1/2 This ‘ft' value provides a measurement of the nuclear matrixelement MN of the decay transition. Note: the nuclear matrix element depends on how alikeA,ZandA,Z±1are.

  39. the shortest half-lives (most common) b-decays “super-allowed” 0+ 0+ 10C 10B* 14O 14N* 1p1/2 1p3/2 1s1/2 The space parts of the initial and final wavefunctions are identical! What differs? The iso-spin space part (Chapter 11 and 18) |MN|2 =

  40. Table 9.2 ft values for “Superallowed” 0+0+ Decays Decay ft (seconds) 10C10B 3100  31 14O14N 3092  4 18Ne18F 3084  76 22Mg22Na 3014  78 26Al26Mg 3081  4 26Si26Al 3052  51 30S30P 3120  82 34Cl34S 3087  9 34Ar34Cl 3101  20 38K38Ar 3102  8 38Ca38K 3145  138 42Sc42Ca 3091  7 46V46Ti 3082  13 50Mn50Cr 3086  8 54Co54Fe 3091  5

  41. Note: the nuclear matrix element depends on how alikeA,ZandA,Z±1are. When A,ZA,Z±1|MN|2~ 1 otherwise |MN|2 < 1. If the wavefunctions correspond to states of different J or different parities then |MN|2= 0. Thus the Fermi selection rules for beta decay DJ = 0 and 'the nuclear parity must not change'.

  42. Mirror Nuclei two nuclei with the same closed shell core of nucleons but one with a single odd proton outside this core, the other a single odd neutron. Such nuclear wavefunctions are identical except for small coulomb effects. This implies|MN|2 ~ 1 thus any beta decay transition between the ground states of mirror nuclei can be used to measure the coupling constantg. It is found that log10(ft1/2) ~ 3.4 so g ~ 1.4 10-62 J m3 0.8810-4 MeV·fm3 39 20 13 7 13 6 39 19 N6 C7 Ca19 K20

  43. 11.13 width 0.26 MeV 3/2 9.7 3/2 7.48 5/2 6.56 width 1.0 MeV 5/2 4.63 7/2 0.478 1/2 0 3/2 10.79 width 0.29 MeV 3/2 9.2 7.19 6.51 width 1.2 MeV 5/2 4.55 7/2 0.431 1/2 0 3/2 Excited states of Li7 Excited states of Be7 The charge symmetryof nuclear forces is illustrated by the existence of mirror nuclei like Li7andBe7 C14andO14

  44. Mirror Nuclei two nuclei with the same closed shell core of nucleons but one with a single odd proton outside this core, the other a single odd neutron. Such nuclear wavefunctions are identical except for small coulomb effects. This implies|MN|2 ~ 1 thus any beta decay transition between the ground states of mirror nuclei can be used to measure the coupling constantg. It is found that log10(f1/2) ~ 3.4 so g ~ 1.4 10-62 J m3 0.8810-4 MeV·fm3 39 20 13 7 13 6 39 19 N6 C7 Ca19 K20

  45. j total angular momentum of any single nucleon I total angular momentum of the entire nucleus Often a single valence nucleon determines a nuclei’s properties I = j When necessary to consider 2 valence particles:I = j1 + j2 And when there’s an odd particle atop a filled core: I = jnucleon + jcore All (hundreds) of known (stable and radio-active) even-Z, even-N nuclei have spin-0 ground states. Powerful evidence for the nucleon-pairing we’ve described as a fundamental part of nuclear structure.

  46. Obviouslythe ground state of odd-A nuclei have I = j of the odd proton or neutron. Since individual p,n are fermions (with spin = 1/2 ) the individual j= s + ℓ = ½ + ℓ 1 2 3 2 5 2 i.e. must be half-integral: , , , … 1 2 5 2 3 2 with z-components: ħ, ħ, ħ, … Then for even-A nuclei: Iz = integral values Iis aninteger!

  47. odd-A nuclei: I = half-integer even-Anuclei: I = integer For nuclei, Parity is only in principal calculable, In practice it is inferred by studying the reactions nuclei participate in or analyzing scattering distributions.

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