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 Dm= 0.0304 u which gives
28.3 MeV binding energy!
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
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.
42Mo
A = 104 isobars
43Tc
103.912
A
Z
A
Z+1
?
?
decay: X Y + 
N
N1
?
A
Z
A
Z1
ecapture: 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
Generalization of ordinary “Fourier expansion” or “Fourier series”
Note how this pairs canonically conjugate variables and t.
Incident monoenergetic 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

incident particle velocity, v
Nscattered= NFdsTOTAL
The scattering rate
per unit time
Particles IN (per unit time) = FArea(ofbeamspot)
Particles scattered OUT (per unit time) = F NsTOTAL
Notice the total transition probability t
and the transitionrate
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 = 4v2dv
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
4v2dv
dN
dE
E1/2
What if there were initially some daughter products
already there when the rock was formed?
Which we can rewrite as:
y = x m + b
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.
When the result is substituted into the exponential the expression for the transmission becomes
(2 expression for the transmission becomes l + 1)( l m)!
4p( l + m)!
ml(q,f ) = Pml (cosq)eimf
Pml (cosq) = (1)m(1cos2q)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(pq,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 (sstate) constant parity = +1
l=1 (pstate) cos parity = 1
l=2 (dstate) (3cos21) parity = +1
Spherical harmonics have (1)l parity.
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
Since the emitted is just itsa 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
This does is just itsnot 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.
This phase space factor determines the decay electron momentum spectrum.
(shown below with the kinetic energy spectrum for the nuclide).
the shortest halflifes ( momentum spectrum. most common) bdecays “superallowed”
0+ 0+
10C 10B*
14O 14N*
The space parts of the initial and final wavefunctions are idenitical!
What differs?
The isospin space part (Chapter 11 and 18)
MN2 =
Note: momentum spectrum.
the nuclear matrix element depends on how alikeA,ZandA,Z±1are.
When A,ZA,Z±1MN2~ 1
otherwise MN2 < 1.
If the wavefunctions correspond to states of
different J or different parities
then MN2= 0.
Thus the Fermi selection rules for beta decay
DJ = 0 and
'the nuclear parity must not change'.
Total momentum spectrum. S = 0 (antiparallel spins)
Total S = 1 parallel spins)
Fermi Decays
GamowTeller Decays
Nuclear I = 0
Ii = If + 1
I = 0 or1
With Pe, = (1)ℓ = +1
PA,Z = PA,Z1
I = 0,1 with no P change
10 momentum spectrum. C10B*
14O14N*
0+ 0+
Fermi Decays
0+ 1+
6He6Li
13B13C
GamowTeller Decays
3/2 1/2
e, pair account for I = 1 change
carried off by their parallel spins
n p
3H3He
13N13C
1/2+ 1/2+
1/2+ 1/2+
1/2 1/2
Forbidden Decays momentum spectrum.
ℓ=1 “first forbidden”
With either Fermi decays s = 0
GamowTeller decays s = 1
with Parity change!
Forbidden Decays momentum spectrum.
even rarer!
ℓ=2 “second forbidden”
With either Fermi decays s = 0
GamowTeller decays s = 1
With no Parity change!
Fermi and GamowTeller already allow (account for)
I= 0, 1 with no parity change
M momentum spectrum. össbauer 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
p2
2mN
TN =
=
p=Eg/c
The photon energy is mismatched by
As an example consider the distinctive 14.4 keV momentum spectrum. 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 iron57 nucleus is
EC
136keV
5/2
t=107s
3/2
14.4keV
1/2
57Fe
With = 107 s, =108eV
this is 5 orders of magnitude greater than the natural linewidth
of the iron transition which produced the photon!
The Optical Model momentum spectrum.
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
A possible ( momentum spectrum. and observed) spontaneous fission reaction
8.5 MeV/A
7.5 MeV/A
Gains ~1 MeVper nucleon!
2119 MeV = 238 MeV
released by splitting
238U
119Pd
Z momentum spectrum. 2/A=36
such unstable states
decay in characteristic
nuclear times ~1022sec
Z2/A=49
Tunneling does allow spontaneous fission, but it must compete with other decay mechanisms (decay)
The potential energy V(r) = constantB
as a function of the separation, r, between fragments.
At smaller values of momentum spectrum. 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.
only momentum spectrum.
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)
The protonproton cycle momentum spectrum.
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
4mec2 = 40.511 MeV
The CNO cycle momentum spectrum.
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
The 1 momentum spectrum. st generation of stars (following the big bang) have no C or N.
The only route for hydrogen burning was through the pp 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 m3 for protons
and 103kg m3 for 12C.