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

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

5. 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.
                                                                                       

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

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

8. Breit-Wigner Resonance CurveEo E
1.0
0.5 
MAX
 = FWHM

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

10. N number of scattering scattered particles
Incident mono-energetic beam
v D t A
d W
N = number density in beam
(particles per unit volume)
Solid angle d W represents
detector counting the dN
particles per unit time that
scatter through q into d W
N number of scattering
centers in target
intercepted by beamspot
FLUX = # of particles crossing through unit cross section per sec
= Nv Dt A / Dt A = Nv
Notice: qNv we call current, I, measured in Coulombs.
dN N F d W dN = s(q)N F d W dN = N F d s -

11. Nscattered = N F dsTOTAL
The scattering rate
per unit time
Particles IN (per unit time) = FArea(of beam spot)
Particles scattered OUT (per unit time) = F N sTOTAL
Cross section 
incident particle velocity, v

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

13. Notice the total transition probability  t
and the transition rate

14. dV = 4v2dv vz vy vx 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

15. dN  E1/2 dE 4v2dv 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

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

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

19. 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.

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

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

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

23. P:ml (q,f ) =ml (p-q,p+f )=(-1)l(-1)m(-1)m ml (q,f )
(2l + 1)( l- m)!
4p( l + m)!
ml (q,f ) = Pml (cosq)eimf
Pml (cosq) = (-1)m(1-cos2q)m [( )m Pl (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.

24. 4He So |sX' – sX| < ℓa< sX' + sX Sa = 0
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
Sa = 0
So |sX' – sX| < ℓa< sX' + sX

25. Which defines a selection rule:
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

26. which accelerates the positrons and decelerates the electrons.
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.

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

28. 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 =

29. If the wavefunctions correspond to states of
Note:
the nuclear matrix element depends on how alike A,Z and A,Z±1 are.
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'.

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

31. 0+  0+ 0+  1+ 10C10B* 14O14N* Fermi Decays 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-

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

33. Forbidden Decays ℓ=2 “second forbidden” even rarer!
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

34. levels by enough to provide the nucleus with the needed recoil:
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

35. As an example consider the distinctive 14.4 keV g from 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
5/2
136keV
t=10-7s
3/2
14.4keV
1/2
57Fe
With  = 10-7 s,  =10-8 eV
this is 5 orders of magnitude greater than the natural linewidth
of the iron transition which produced the photon!

36. The Compound Nucleus
[ Ne]* 20 10

37. The Optical Model where and now here
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

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

39. Z2/A=36 Z2/A=49 The potential energy V(r) = constant-B
such unstable states
decay in characteristic
nuclear times ~10-22 sec
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.

40. m for the masses of the nuclear fragments we’re talking about,
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.

41. but is most efficiently induced by slow neutrons
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)

42. 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mec 2 = 40.511 MeV

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

44. 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 103 kg m-3 for 12C.