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For the alpha particle D m= 0.0304 u which gives 28.3 MeV binding energy!

For the alpha particle D m= 0.0304 u which gives 28.3 MeV binding energy!. 236 94. Is Pu unstable to -decay?. 236 94. 232 92. 4 2. Pu  U + . + Q. Q = ( M Pu – M U - M  ) c 2. = ( 236.046071 u – 232.037168 u – 4.002603 u ) 931.5MeV/ u. = 5.87 MeV > 0.

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For the alpha particle D m= 0.0304 u which gives 28.3 MeV binding energy!

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  1. For the alpha particle Dm= 0.0304 u which gives 28.3 MeV binding energy!

  2. 236 94 Is Puunstable to -decay? 236 94 232 92 4 2 Pu U +  + Q Q = (MPu – MU - M)c2 = (236.046071u – 232.037168u – 4.002603u)931.5MeV/u = 5.87 MeV > 0

  3. 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.

  4. 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

  5. Fourier Transforms Generalization of ordinary “Fourier expansion” or “Fourier series” Note how this pairs canonically conjugate variables  and t.

  6. Breit-Wigner Resonance Curve 1.0  MAX 0.5 = FWHM E Eo

  7. Incompressible Nucleus R=roA1/3 ro1.2 fm

  8. scattered particles Incident mono-energetic beam v Dt A dW N = number density in beam (particles per unit volume) Solid angle dWrepresents detector counting the dN particles per unit time that scatter through qinto dW Nnumber of scattering centers in target intercepted by beamspot FLUX = # of particles crossing through unit cross section per sec = NvDt A / Dt A = Nv Notice: qNv we call current, I, measured in Coulombs. dN NF dW dN = s(q)NF dW dN =NFds -

  9. Cross section incident particle velocity, v Nscattered= NFdsTOTAL The scattering rate per unit time Particles IN (per unit time) = FArea(ofbeamspot) Particles scattered OUT (per unit time) = F NsTOTAL

  10. D. R. Nygren, J. N. Marx, Physics Today 31 (1978) 46 a p p d m dE/dx(keV/cm) e Momentum [GeV/c]

  11. Notice the total transition probability t and the transitionrate

  12. vz Classically, for free particles E = ½ mv2 = ½ m(vx2 + vy2 + vz2 ) vy Notice for any fixed E, m this defines a sphere of velocity points all which give the same kinetic energy. vx The number of “states” accessible by that energy are within the infinitesimal volume (a shell a thickness dv on that sphere). dV = 4v2dv

  13. Classically, for free particles E = ½ mv2 = ½ m(vx2 + vy2 + vz2 ) We just argued the number of accessible states (the “density of states”) is proportional to 4v2dv dN dE  E1/2

  14. What if there were initially some daughter products already there when the rock was formed? Which we can rewrite as: y = x  m + b

  15. Rb-Sr dating method Allows for the presence of initial 87Sr

  16. Calculation of the kinetic energy of an alpha particle emitted by the nucleus 238U. The model for this calculation is illustrated on the potential energy diagram at right.

  17. In simple 1-dimensional case V E x = r1 x = r2 III II I probability of tunneling to here

  18. Where E r2 R So let’s just write as

  19. When the result is substituted into the exponential the expression for the transmission becomes

  20. (2l + 1)( l- m)! 4p( l + m)! ml(q,f ) = Pml (cosq)eimf Pml (cosq) = (-1)m(1-cos2q)m[()mPl (cosq)] d d (cosq) 1 2l l! d d (cosq) Pl(cosq) = [()l(-sin2q)l ] So under the parity transformation: P:ml(q,f ) =ml(p-q,p+f)=(-1)l(-1)m(-1)m ml(q,f ) = (-1)l(-1)2m ml(q,f ) )=(-1)l ml(q,f ) An atomic state’s parity is determined by its angular momentum l=0 (s-state)  constant parity = +1 l=1 (p-state)  cos parity = -1 l=2 (d-state)  (3cos2-1) parity = +1 Spherical harmonics have (-1)l parity.

  21. In its rest frame, the initial momentum of the parent nuclei is just its spin: Iinitial = sX and: Ifinal = sX'+ sa + ℓa 1p1/2 1p3/2 1s1/2 4He So |sX'– sX| < ℓa< sX'+ sX Sa = 0

  22. Since the emitted a is described by a wavefunction: the parity of the emitted a particle is(-1)ℓ Which defines a selection rule: restricting us to conservation of angular momentum and parity. If P X' = P X then ℓ= even If P X' = -P X then ℓ= odd

  23. 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.

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

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

  26. 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'.

  27. Total S = 0 (anti-parallel spins) Total S = 1 parallel spins) Fermi Decays Gamow-Teller Decays Nuclear I = 0 Ii = If + 1 I = 0 or1 With Pe, = (-1)ℓ = +1 PA,Z = PA,Z1 I = 0,1 with no P change

  28. 10C10B* 14O14N* 0+ 0+ Fermi Decays 0+ 1+ 6He6Li 13B13C Gamow-Teller Decays 3/2- 1/2- e, pair account for I = 1 change carried off by their parallel spins n  p 3H3He 13N13C 1/2+ 1/2+ 1/2+ 1/2+ 1/2- 1/2-

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

  30. Forbidden Decays even rarer! ℓ=2 “second forbidden” With either Fermi decays s = 0 Gamow-Teller decays s = 1 With no Parity change! Fermi and Gamow-Teller already allow (account for) I= 0, 1 with no parity change

  31. Mössbauer Effect If this change is large enough, the  will not be absorbed by an identical nucleus. In fact, for absorption, actually need to exceed the step between energy levels by enough to provide the nucleus with the needed recoil: pN2 2mN p2 2mN TN = = p=Eg/c The photon energy is mismatched by

  32. As an example consider the distinctive 14.4 keV gfrom 57Fe. ~90% of the 57Fe* decays are through this intermediate level produce 14.4 keV s. t=270d 7/2 57Co The recoil energy of the iron-57 nucleus is EC 136keV 5/2 t=10-7s 3/2 14.4keV 1/2 57Fe With  = 10-7 s,  =10-8eV this is 5 orders of magnitude greater than the natural linewidth of the iron transition which produced the photon!

  33. The Compound Nucleus [ Ne]* 20 10

  34. The Optical Model To quantum mechanically describe a particle being absorbed, we resort to the use of a complex potential in what is called the optical model. Consider a traveling wave moving in a potential V then this plane wavefunction is written where If the potential V is replaced by V + iW then k also becomes complex and the wavefunction can be written and now here

  35. A possible (and observed) spontaneous fission reaction 8.5 MeV/A 7.5 MeV/A Gains ~1 MeVper nucleon! 2119 MeV = 238 MeV released by splitting 238U 119Pd

  36. Z2/A=36 such unstable states decay in characteristic nuclear times ~10-22sec Z2/A=49 Tunneling does allow spontaneous fission, but it must compete with other decay mechanisms (-decay) The potential energy V(r) = constant-B as a function of the separation, r, between fragments.

  37. At smaller values of x, fission by barrier penetration can occur, However recall that the transmission factor (e.g., for -decay) is where m while for  particles (m~4u) this gave reasonable, observable probabilities for tunneling/decay for the masses of the nuclear fragments we’re talking about,  can become huge and X negligible.

  38. only the Natural uranium (0.7% 235U, 99.3% 238U) undergoes thermal fission Fission produces mostly fast neutrons Mev but is most efficiently induced by slow neutrons E (eV)

  39. The proton-proton cycle The sun 1st makes deuterium through the weak (slow) process: Q=0.42 MeV then Q=5.49 MeV 2 passes through both of the above steps then can allow Q=12.86 MeV This last step won’t happen until the first two steps have built up sufficient quantities of tritium that the last step even becomes possible. 2(Q1+Q2)+Q3=24.68 MeV plus two positrons whose annihilation brings an extra 4mec2 = 40.511 MeV

  40. The CNO cycle Q=1.95 MeV Q=1.20 MeV Q=7.55 MeV Q=7.34 MeV Q=1.68 MeV Q=4.96 MeV carbon, nitrogen and oxygen are only catalysts

  41. The 1st generation of stars (following the big bang) have no C or N. The only route for hydrogen burning was through the p-p chain. In later generations the relative importance of the two processes depends upon temperature. Rate of energy production Shown are curves for solar densities 105 kg m-3 for protons and 103kg m-3 for 12C.

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