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 M vs Z which implies the mass takes on some minimum value!

3. A = 127 isobars -88 -86 -84 -80 -88 -86 -84 -80 52545056 Atomic number, Z Mass defect  (MeV)

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 29 Cu 64 ) can decay by either electron or positron emission.

6. 103.912 103.910 103.908 103.906 103.904 Mass, u 42444648 Atomic Number, Z A = 104 isobars 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd Odd Z Even Z  -decay: X  Y +   AZAZ N ? ???? N-1 A Z+1   e -capture: X + e  Y N N+1 A Z-1 AZAZ

7. If Mc 2 (A,Z) is the rest mass energy of the parent atom and Q is the energy released in the decay then M(A,Z)c 2 = M + (A,Z+1)c 2 + m e c 2 + Q where M + (A,Z+1)c 2 is the rest mass energy of the positive ion produced. Now since the ionization potential energy is relatively small we can write - M + (A,Z+1)c 2 + m e c 2 ~ M(A,Z+1)c 2 Thus M(A,Z)c 2 = M(A,Z+1)c 2 + Q energy will be released in the decay (ie Q > 0) provided M(A,Z)c 2 is greater than M(A,Z+1)c 2. Electron emission

8. What is the maximum energy of the electron emitted in the   decay of 3131 H ? The reaction is: Neglecting the kinetic energy of the nucleus, and mass of the neutrino the Q is shared between e and K e at maximum when K  0, so

9. The charge on the nucleus is reduced and the daughter atom has an excess electron. Thus M(A,Z)c 2 = M(A,Z-1)c 2 + 2m e c 2 + Q The 2m e c 2 account 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)c 2 > M(A,Z-1)c 2 + 2m e c 2. i.e, the difference in rest mass energies must be at least 1 MeV. Positron emission

10. 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. but in 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. M(A,Z)c 2 = M(A,Z-1)c 2 + E * + Q E * is the excitation energy of the daughter atom (sufficiently small to be ignored). Electron capture can proceed (Q > 0) if M(A,Z)c 2 < M(A,Z-1)c 2. 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 than 1 MeV/c 2. Electron capture

11.  - decay  - decay

12. Some Alpha Decay Energies and Half-lives Isotope KE  (MeV)  1/2 (sec -1 ) 232 Th4.011.4  10 10 y 1.6  10  18 238 U4.194.5  10 9 y 4.9  10  18 230 Th4.698.0  10 4 y 2.8  10  13 238 Pu5.5088 years 2.5  10  10 230 U5.8920.8 days 3.9  10  7 220 Rn6.2956 seconds 1.2  10  2 222 Ac7.01 5 seconds 0.14 216 Rn8.0545.0  sec 1.5  10  212 Po8.780.30  sec 2.3  10  216 Rn8.780.10  sec 6.9  10 

13. B Before decay: After decay: Which fragment has a greater momentum? Potassium nucleus energy? A) B) C) both the same A 1930 Series of studies of nuclear beta decay, e.g., Potassium goes to calcium 19 K 40  20 Ca 40 Copper goes to zinc 29 Cu 64  30 Zn 64 Boron goes to carbon 5 B 12  6 C 12 Tritium goes to helium 1 H 3  2 He 3

14. 1932 Once the neutron was discovered, included the more fundamental n  p + e For simple 2-body decay, 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. but this only seems to match the maximum value observed on a spectrum of beta ray energies! E e = (m A 2 - m B 2 + m e 2 )c 2 /2m A

15. No. of counts per unit energy range Electron kinetic energy in KeV 51015200 The beta decay spectrum of tritium ( H  He). Source: G.M.Lewis, Neutrinos (London: Wykeham, 1970), p.30)

16. 1932 n  p + e  + neutrino charge0 +1  1 ? mass 939.56563 938.27231 0.51099906 ? MeV MeV MeV neutrino mass < 5.1 eV < m e /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 Anderson and Neddermeyer determine X-particles are charged have 206× the electron’s mass decay to electrons with a mean lifetime of 2  sec 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) 

23.    + energy always predictably fixed by E  Under the influence of a magnetic field simple 2-body decay!  +   + + neutrino? charge +1 +1 ?

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

26. 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  10 13 neutrinos/cm 2 -sec observed 2-3 p + neutrino events/hour also looked for n + neutrino  p + e  but never observed! Cowan & Reines

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

28. For a single particle we’re dealing with a 6-dimensional phase space – three position coordinates and three momentum components. d 3 x  dx dy dz and d 3 p  dp x dp y dp z. Thus (dx) 3 (dp x ) 3 ~ h 3 and the number of states in a the “volume” element d 3 x d 3 p is given by recall the uncertainty relation dx  dp x ~ h The smallest element of volume a state can occupy in this phase space is h 3. (ignoring for now particle spin)

29. For a coordinate space volume V and a momentum volume 4  p 2 dp for both electron and neutrino sharing the same volume: dN f = 16  2 V 2 (p e ) 2 dp e (p ) 2 dp /(h 6 ) already integrating over d 3 x e and d 3 x

30. With three particles in the final state p e and 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 p = (E max - E e )/c = (T max -T e )/c dp /dE = 1/c Hence dN f /dE = 16  2 V 2 (T max - T e ) 2 (p e ) 2 dp e /(h 6 c 3 )

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

32. Energy spectrum of beta decay electrons from 210 Bi dN f /dE = 16  2 V 2 (T max - T e ) 2 (p e ) 2 dp e /(h 6 c 3 ) Kinetic energy, MeV Intensity

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

34. The individual contributions, d , per specific p e value: This quadratic equation has zeroes at p e = 0 and T e = T max  Q

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

36. 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,p e ), a special factor (generally in powers of Z and p e ), is introduced to account for this.

37. P (p e ) is  the measured beta momentum spectrum

38. Including the measured probability P meas (p e ) in which when plotted against T e yields a straight line the Fermi-Kurie plot.

40. The overall decay probability is obtained by integrating P(p e )dp e from 0 to p max : = {64  4 g 2 [ M N ] 2 /(h 7 c 3 )} ∫ F(Z,p e ) (T max - T e ) 2 (p e ) 2 dp e = {64  4 g 2 m 5 c 4 [ M N ] 2 /(h 7 )}f(Z,T max ) = G 2 [ M N ] 2 f(Z,T max ) where f(Z,T max ) = ∫ F(Z,p e ) {(T max -T e )/(mc 2 )} 2 (p e /(mc)) 2 dp e /(mc) with energy and momenta expressed in units of mc 2 and mc respectively. The integral is often tabulated when it cannot be solved analytically.

41. Since the decay probability = ln(2)/  1/2 f (Z,T max )  1/2 = ln(2)/(G 2 | M N | 2 ) This ‘ft' value provides a measurement of the nuclear matrix element  M N  of the decay transition. Note: the nuclear matrix element depends on how alike  A,Z and  A,Z±1 are. = G 2 [ M N ] 2 f(Z,T max )

42. the shortest half-lives (most common)  -decays “super-allowed” 0 +  0 + 10 C  10 B* 14 O  14 N* The space parts of the initial and final wavefunctions are identical! What differs? The iso-spin space part (Chapter 11 and 18) | M N | 2 = 1s 1/2 1p 3/2 1p 1/2

43. Table 9.2 ft values for “Superallowed” 0+  0+ Decays Decayft (seconds) 10 C  10 B3100  31 14 O  14 N3092  4 18 Ne  18 F3084  76 22 Mg  22 Na3014  78 26 Al  26 Mg3081  4 26 Si  26 Al3052  51 30 S  30 P3120  82 34 Cl  34 S3087  9 34 Ar  34 Cl3101  20 38 K  38 Ar3102  8 38 Ca  38 K3145  138 42 Sc  42 Ca3091  7 46 V  46 Ti3082  13 50 Mn  50 Cr3086  8 54 Co  54 Fe3091  5

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

45. 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 | M N | 2 ~ 1 thus any beta decay transition between the ground states of mirror nuclei can be used to measure the coupling constant g. It is found that log 10 (ft 1/2 ) ~ 3.4 so g ~ 1.4 10 -62 J m 3 0.88  10 -4 MeV·fm 3 13 7 N6N6 13 6 C7C7 39 20 Ca 19 39 19 K 20

46. 11.13width 0.26 MeV3/2 9.73/2 7.485/2 6.56width 1.0 MeV5/2 4.637/2 0.4781/2 03/2 Excited states of Li 7 Excited states of Be 7 10.79width 0.29 MeV3/2 9.2 7.19 6.51width 1.2 MeV5/2 4.557/2 0.4311/2 03/2 The charge symmetry of nuclear forces is illustrated by the existence of mirror nuclei like Li 7 and Be 7 C 14 and O 14

47. 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 | M N | 2 ~ 1 thus any beta decay transition between the ground states of mirror nuclei can be used to measure the coupling constant g. It is found that log 10 (f  1/2 ) ~ 3.4 so g ~ 1.4 10 -62 J m 3 0.88  10 -4 MeV·fm 3 13 7 N6N6 13 6 C7C7 39 20 Ca 19 39 19 K 20

48. 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 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. When necessary to consider 2 valence particles: I = j 1 + j 2 And when there’s an odd particle atop a filled core: I = j nucleon + j core

49. Obviously the 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 i.e. must be half-integral: 1212 3232 5252,,, … with z-components: 1212 3232 5252  ħ,  ħ,  ħ, … Then for even-A nuclei: I z = integral values I is an integer!

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

51. An approximation based on the assumption that over the nuclear volume: While fine for estimating energy differences,  spectra over-simplifies picture w.r.t angular momentum considerations Assumes, effectively,, e created at r=0 with ℓ=0 and j=s=1/2 for each.

52. 1932 n  p + e  + neutrino charge0 +1  1 ? mass 939.56563 938.27231 0.51099906 ? MeV MeV MeV neutrino mass < 5.1 eV < m e /100000  0 0 spin½ ½ ½  ? but what else? ½ together carry net momentum from the nucleus

53. Total S = 0 (anti-parallel spins)Total S = 1 parallel spins) Fermi DecaysGamow-Teller Decays Nuclear  I = 0 I i = I f + 1  I = 0 or 1 With P e, = (  1) ℓ = +1 P A,Z = P A,Z  1  I = 0,1 with no P change If the electron and neutrino carry NO orbital (relative) angular momentum

54. 10 C  10 B * 14 O  14 N * 0 +  0 + Fermi Decays 6 He  6 Li 13 B  13 C n  p 3 H  3 He 13 N  13 C 0 +  1 + 3/2   1/2  Gamow-Teller Decays e, pair account for  I = 1 change carried off by their parallel spins 1/2 +  1/2  1/2   1/2 

55. Forbidden Decays ℓ=1 “first forbidden” With either Fermi decays s = 0 Gamow-Teller decays s = 1 with P arity change!

56. Forbidden Decays ℓ=2 “second forbidden” With either Fermi decays s = 0 Gamow-Teller decays s = 1 With no P arity change! even rarer! Fermi and Gamow-Teller already allow (account for)  I= 0, 1 with no parity change and contributions from these deacys will dominate