For the alpha particle Dm= 0.0304 u which gives

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For the alpha particle Dm 0.0304 u which gives
28.3 MeV binding energy!
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Is Pu unstable to ?-decay?
236 94
Pu ? U ?
236 94
232 92
4 2
Q
Q (MPu MU - M?)c2
(236.046071u 232.037168u 4.002603u)?931.5MeV
/u
5.87 MeV gt 0
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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.
                                                                                       
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42Mo
A 104 isobars
??
43Tc
103.912
?-decay X ? Y ? -
A Z
? ?
A Z1
N
?
N-1
e-capture X e ? Y
A Z-1
A Z
48Cd
103.910
N
N1
Odd Z
47Ag
Mass, u
103.908
45Rh
Even Z
103.906
44Ru
??
46Pd
103.904
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44
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Atomic Number, Z
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Fourier Transforms
Generalization of ordinary Fourier expansion or
Fourier series
Note how this pairs canonically conjugate
variables ? and t.
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Breit-Wigner Resonance Curve
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Incompressible Nucleus
RroA1/3 ro?1.2 fm
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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
-
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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
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D. R. Nygren, J. N. Marx, Physics Today 31 (1978)
46
a
p
p
d
m
dE/dx(keV/cm)
e
Momentum GeV/c
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Notice the total transition probability ? t
and the transition rate
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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
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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
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What if there were initially some daughter
products already there when the rock was formed?
Which we can rewrite as
y x ? m
b
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Rb-Sr dating method Allows for the presence of
initial 87Sr
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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.
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In simple 1-dimensional case
V
E
x r1
x r2
III
II
I
probability of tunneling to here
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Where
E
r2
R
So lets just write
as
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When the result is substituted into the
exponential the expression for the transmission
becomes
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(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)
d d (cosq)
1 2l l!
Pl (cosq) ( )l (-sin2q)l

So under the parity transformation
P?ml (q,f ) ?ml (p-q,pf )(-1)l(-1)m(-1)m ?ml
(q,f ) (-1)l(-1)2m ?ml (q,f ) )(-1)l
?ml (q,f )
An atomic states parity is determined by its
angular momentum l0 (s-state) ??
constant parity 1 l1 (p-state) ??
cos? parity -1 l2 (d-state) ??
(3cos2?-1) parity 1 Spherical
harmonics have (-1)l parity.
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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 lt lalt sX' sX
Sa 0
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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
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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.
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This phase space factor determines the decay
electron momentum spectrum. (shown below with
the kinetic energy spectrum for the nuclide).
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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)
MN2
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Note the nuclear matrix element depends on how
alike ?A,Z and ?A,Z1 are. When ?A,Z ? ?A,Z1
MN2 1 otherwise MN2 lt 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'.
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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
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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-
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Forbidden Decays
l1 first forbidden
With either Fermi decays s 0 Gamow-Teller
decays s 1
with Parity change!
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Forbidden Decays
even rarer!
l2 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
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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

pEg /c
The photon energy is mismatched by
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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.
t270d
7/2
57Co
The recoil energy of the iron-57 nucleus is
EC
136keV
5/2
t10-7s
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!
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The Compound Nucleus
Ne
20 10
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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
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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
238U
119Pd
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Z2/A36
such unstable states decay in characteristic nucle
ar times 10-22 sec
Z2/A49
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.
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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 (m4u) this gave
reasonable, observable probabilities for
tunneling/decay
for the masses of the nuclear fragments were
talking about, ? can become huge and X
negligible.
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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)
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The proton-proton cycle
The sun 1st makes deuterium through the weak
(slow) process
Q0.42 MeV
then
Q5.49 MeV
2 passes through both of the above steps then can
allow
Q12.86 MeV
This last step wont happen until the first two
steps have built up sufficient quantities of
tritium that the last step even becomes possible.
2(Q1Q2)Q324.68 MeV
plus two positrons whose annihilation brings an
extra
4?mec 2 4?0.511 MeV
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The CNO cycle
Q1.95 MeV
Q1.20 MeV
Q7.55 MeV
Q7.34 MeV
Q1.68 MeV
Q4.96 MeV
carbon, nitrogen and oxygen are only catalysts
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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.