Transcript beta and gamma decays

Beta Decays
• Beta decays are proton  neutrons or neutron 
proton transitions
• involve W exchange and are weak interaction
M
M
Z ,A
 M
Z ,A
 M

e M
Z ,A

( p  ne  )

Z 1 , A
 e  e
Z 1, A
 e  e
( n  pe  )
 e
( ep  n  )
 M
Z 1 , A
• the last reaction is electron capture where one of
the atomic electrons overlaps the nuclei. Same
matrix element (essentially) bit different kinematics
• the semi-empirical mass formula gives a minimum
for any A. If mass difference between neighbors is
large enough, decay will occur
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Beta Decays - Q Values
• Determine Q of reactions by looking at mass
difference (careful about electron mass)


 : X Z , A  Y Z  1, A  e   e
m X (  Zm e )  m Y  ( Zm e )  K Y  m e  K e  K 
Q  AtomicMass
X

 AM
Y
 K Y  K e  K

 : X Z , A  Y Z 1, A  e   e
m X (  Zm e )  m Y  ( Zm e )  K Y  m e  K e  K 
Q  AM
X
 AM
Y
 2me

EC : e  X Z , A  Y Z 1 , A   e
m e  m X (  Zm e )  m Y  ( Zm e )  K Y  K e  K 
Q  AM
X
 AM
Y
• 1 MeV more Q in EC than beta+ emission. More
phase space BUT need electron wavefunction
overlap with nucleus..
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Beta+ vs Electron Capture
• Fewer beta+ emitters than beta- in “natural” nuclei
(but many in “artificial” important in Positron
Emission Tomography - PET)
• sometimes both beta+ and EC for same nuclei.
Different widths
• sometimes only EC allowed
3
Li
4
Be
7
7
M  7 . 01600 u
M  7 . 01693 u
 M  . 00093 u  2  m e  2  . 00055 u
4

Be  e  Li  
7
3
7
• monoenergetic neutrino. E=.87 MeV. Important
reaction in the Sun. Note EC rate different in Sun
as it is a plasma and not atoms
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Beta+ vs Electron Capture
• from Particle Data Group

p  p H  e  
2
8
7

B  Be  e  
8
Be  e  Li  
7
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Beta Decay - 3 Body
• The neutrino is needed to conserve angular
momentum
• (Z,A)  (Z+1,A) for A=even have either
Z,N even-even  odd-odd or odd-oddeven-even
• p,n both spin 1/2 and so for even-even or odd-odd
nuclei I=0,1,2,3…….
• But electron has spin 1/2
I(integer)  I(integer) + 1/2(electron) doesn’t
conserve J
• need spin 1/2 neutrino. Also observed that electron
spectrum is continuous indicative of >2 body decay
• Pauli/Fermi understood this in 1930s
electron neutrino discovered 1953 (Reines and
Cowan)
muon neutrino discovered 1962 (Schwartz
+Lederman/Steinberger)
tau neutrino discovered 2000 at Fermilab
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3 Body Kinematics
• While 3 body the nuclei are very heavy and easy
approximation is that electron and neutrino split
available Q (nuclei has similar momentum)
• maximum electron energy when E(nu)=0
X  Y  e 
p y  pe
let E   0
conserve
momentum
( m x  E e  m )  E y
2
2
27
Mg
12
energy
2
(  m )
2m x
K e  Ee  me 
• example
conserve
m x  m y  me
2
E e max 
2
( m x  m y  m e )( m x  m y  m e )
Q
2m x
 Al
27
13

 e 
m 27 ,12  26 . 9843 , m 27 ,13  26 . 981
m e  . 00055  Q  2 . 8 MeV
pe 
E e  m e  2 . 75 MeV ,  e 
K Al 
p
2
2
E
 5 .5
m
2
 0 . 2 keV  small
2m
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Beta decay rate
• Start from Fermi Golden Rule
Rates 
2

| M |  Final
2
M    F  d 
*
• first approximation (Fermi).
Beta=constant=strength of weak force
*
M     Z  1 Z d 
M  M 
• Rule 1: parity of nucleus can’t change (integral of
odd*even=0)
• Rule 2: as antineutrino and electron are spin 1/2
they add to either 0 or 1. Gives either
Fermi :  i  i ZA  i Z  1 A  0
42
Sc
21
 Ca
42
20


 e 
0  0
Gamow  Teller :  i   1
32
P  S
15
22
16

 e 

( not 0  1  0 )
1  0
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

7
Beta decay rate II
• Orbital angular momentum suppression of 0.001
for each value of (in matrix element calculation)
36
Sc
17
 Ca
36

 e 
18

2  0

i  1  L  1
• look at density of states factor. Want # quantum
states per energy interval
Rates 
2

| M |  Final
n 
2
dN n
dE n
• we know from quantum statistics that each particle
(actually each spin state) has
dN  4
p
2
h
3
dp
• 3 body decay but recoil nucleus is so heavy it
doesn’t contribute
2
2
dN  4
pe
h
3
dp e 4
p
h
3
dp 
p  ( Q  K e ) / c
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Beta decay rate III
• Conservation of energy allows one to integrate over
the neutrino (there is a delta function)
dN
Rates 
2


dp e
2

| M |  Final 
2
4 p e 4 ( Q  K e )
2
|M |
2
h
3
( hc )
K e  ( pe  me )
2
2 1/ 2
2
3
 me
• this gives a distribution in electron
momentum/energy which one then integrates over.
(end point depends on neutrino mass)
Rate 
1
T
5

me c
4
2 
3
| M |  F ( E e max )
2
7
• F is a function which depends on Q. It is almost
loqrithmic
log F  A log K e max
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actual. not “linear” due to
electron mass
log F  A log K e max  4 . 4 log K  . 5
 F  3K
4 .4
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Beta decay rate IV
• FT is “just kinematics”
• measuring FT can study nuclear wavefunctions M’
and strength of the weak force at low energies
• lower values of FT are when M’ approaches 1
• beta decays also occur for particles

K





 e  e
0
0

 e  e
• electron is now relativistic and E=pc. The integral
is now easy to do. For massive particles (with
decay masses small), Emax = M/2 and so rate goes
as fifth power of mass
p max
 ( Q  K e ) p e dp e  E max / 30
2
2
5
0
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Beta decay rate V
• M=M’   is strength of weak interaction. Can
measure from lifetimes of different decays
  10
 62
joule  m  100 eV  F
3
3
• characteristic energy


vol
100 eV * F
(10 F )
3
3
 0 . 1eV
• strong energy levels ~ 1 MeV
 weak
 strong
 10
7
 relative strength  10
 14
• for similar Q, lifetimes are about
 strong  10
 EM  10
 23
 16
 weak  10
s
s
 10
s
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Parity Violation in Beta Decays
• The Parity operator is the mirror image and is NOT
conserved in Weak decays (is conserved in EM and
strong) P ( x , y , z )  (  x ,  y ,  z )
P ( r , ,  )  ( r ,    ,    )
• non-conservation is on the lepton side, not the
nuclear wave function side
• spin 1/2 electrons and neutrinos are (nominally)
either right-handed (spin and momentum in same
direction) or left-handed (opposite)
• Parity changes LH to RH
•
RH


P( p)   p
  

P(L  r  p)  L
LH
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“Handedness” of Neutrinos
• “handedness” is call chirality. If the mass of a
neutrino = 0 then:
• all neutrinos are left-handed
all antineutrinos are right-handed
• Parity is maximally violated
• As the mass of an electron is > 0 can have both LH
and RH. But RH is suppressed for large energy (as
electron speed approaches c)
• fraction RH vs LH can be determined by solving
the Dirac equation which naturally incorporates
spin
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Polarized Beta Decays
• Some nuclei have non-zero spin and can be
polarized by placing in a magnetic field
• magnetic moments of nuclei are small (1/M factor)
and so need low temperature to have a high
polarization (see Eq 14-4 and 14-5)
60

Co  Ni  e  
i5
60
i4 s
1
2
, 12
• Gamow-Teller transition with S(e-nu) = 1
• if Co polarized, look at angular distribution of
electrons. Find preferential hemisphere (down)
Spin antinu-RH
Pnu
pe
Spin e - LH
Co
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Discovery of Parity Violation in
Beta Decay by C.S. Wu et al.
• Test parity conservation by observing a
dependence of a decay rate (or cross section) on
a term that changes sign under the parity
operation. If decay rate or cross section changes
under parity operation, then the parity is not
conserved.
• Parity reverses momenta and positions but not
angular momenta (or spins). Spin is an axial
vector and does not change sign under parity
operation.
180o
Pe
neutron

Beta decay of a neutron in a
real and
mirror worlds:
If parity is conserved, then the
probability of electron
emission at  is equal to that at
180o-.
Selected orientation of neutron
spins - polarisation.
Pe
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Wu’s experiment
• Beta-decay of 60Co to 60Ni*. The
excited 60Ni* decays to the
ground state through two
successive  emissions.
• Nuclei polarised through spin
alignment in a large magnetic
field at 0.01oK. At low
temperature thermal motion does
not destroy the alignment.
Polarisation was transferred from
60Co to 60Ni nuclei. Degree of
polarisation was measured
through the anisotropy of
gamma-rays.
• Beta particles from 60Co decay
were detected by a thin
anthracene crystal (scintillator)
placed above the 60Co source.
Scintillations were transmitted to
the photomultiplier tube (PMT)
on top of the cryostat.
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Wu’s results
• Graphs: top and middle - gamma
anisotropy (difference in counting
rate between two NaI crystals) control of polarisation; bottom - 
asymmetry - counting rate in the
anthracene crystal relative to the
rate without polarisation (after the
set up was warmed up) for two
orientations of magnetic field.
• Similar behaviour of gamma
anisotropy and beta asymmetry.
• Rate was different for the two
magnetic field orientations.
• Asymmetry disappeared when the
crystal was warmed up (the
magnetic field was still present):
connection of beta asymmetry with
spin orientation (not with magnetic
field).
• Beta asymmetry - Parity not
conserved
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Gamma Decays
• If something (beta/alpha decay or a reaction) places
a nucleus in an excited state, it drops to the lowest
energy through gamma emission
• excited states and decays similar to atoms
• conserve angular momentum and parity
• photon has spin =1 and parity = -1
• for orbital P= (-1)L
• first order is electric dipole moment (edm). Easier
to have higher order terms in nuclei than atoms
N  N 
*


L  0 , edm


L  1, e .quad .mom .
3  2 
2  0 
P final  P PN (  1)  (  1)(  1)(  1)  
L
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Gamma Decays
2
5

17
Cl
38
18
Ar
38
26%
E
MeV
gamma
11%
3
2
53%
gamma
0



0
N  N 
*



GT  L  1 ( P change )


 i  2 ; GT  L  1
2  2
2  0

L  0 ; edm


L  1; eqm
2  0 

2  3

3  2 
GT
L  1
P

conserve angular momentum and parity. lowest order is electric dipole
moment. then quadrapole and magnetic dipole
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Mossbauer Effect
• Gamma decays typically have lifetimes of around
10-10 sec (large range). Gives width:
  E 



10
10
 15
10
eVs
 10
5
eV
sec
• very precise
• if free nuclei decays, need to conserve momentum.
Shifts gamma energy to slightly lower value
A  A
*
p A  p  E  
M
2
A
*
M
2M
A
2
A
  M (1 
*
M
)
2M
• example. Very small shift but greater than natural
width
 M  . 13 MeV , M  191 * 931 . 5
 E   . 13 MeV  . 005 eV
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Mossbauer Effect II
• Energy shift means an emitted gamma won’t be
reabsorbed
A  A
E  . 13  . 000000005 MeV
A  A
E  . 13  . 000000005 MeV
*
*
• but if nucleus is in a crystal lattic, then entire lattice
recoils against photon. Mas(lattice)infinity and
Egamma=deltaM. Recoiless emission (or
Mossbauer)
• will have “wings” on photon energy due to lattice
vibrations
• Mossbauer effect can be used to study lattice enregies. Very
precise. Use as emitter or absorber. Vary energy by moving
source/target (Doppler shift) (use Iron. developed by R.
Preston, NIU)
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Nuclear Reactions, Fission and Fusion
•
•
•
•
2 Body reaction A+BC+D
elastic if C/D=A/B
inelastic if mass(C+D)>mass(A+B)
threshold energy for inelastic (B at rest)
M
2
 E tot  p tot  ( m C  m D )
2
2
 K th   Q
K th   Q
2
m A  m B  mC  m D
Q  M
2m B
mA  mB
( non  relativist ic )
mB
• for nuclei nonrelativistic usually OK
p H  H  H
3
2
2
Q  (1 . 007825  3 . 016049  2  2 . 014102 ) u   4 . 03 MeV
K th  4 (1  13 )  5 . 38 MeV ( non  rel )
K th  5 . 47 MeV ( rel )
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Nuclear Reactions (SKIP)
• A+BC+D
• measurement of kinematic quantities allows masses
of final states to be determined
• (p,E) initial A,B known
• 8 unknowns in final state (E,px,py,pz for C+D)
• but E,p conserved. 4 constraints4 unknowns
measure E,p (or mass) of D OR C gives rest
or measure pc and pd gives masses of both
• often easiest to look at angular distribution in C.M.
but can always convert
d
d
 CM
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Fission
• AB+C A heavy, B/C medium nuclei
• releases energy as binding energy/nucleon = 8.5
MeV for Fe and 7.3 MeV for Uranium
• spontaneous fission is like alpha decay but with
different mass, radii and Coulomb (Z/2)2 vs 2(Z-2).
Very low rate for U, higher for larger A
• induced fission n+AB+C. The neutron adds its
binding energy (~7 MeV) and can put nuclei in
excited state leading to fission
• even-even U(92,238). Adding n goes to even-odd
and less binding energy (about 1 MeV)
• even-odd U(92,235), U(92,233), Pu(94,239) adding
n goes to even-even and so more binding energy
(about 1 MeV)  2 MeV difference between U235
and U238
• fission in U235 can occur even if slow neutron
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Spontaneous Fission
•
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Induced Fission
•
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Neutron absorption
•
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Fusion
m ( H )  1 . 007825 u
1
m ( H )  2 . 014102 u
m ( He )  4  m ( H )
m ( He )  4 . 002603 u
m ( C )  3  m ( He )
2
4
4
12
1
4
m ( Be )  8 . 005305 u
8
m ( C )  12 . 00000 u
12
•
•
•
•
•
“nature” would like to convert lighter elements
into heavier. But:
no free neutrons
need to overcome electromagnetic repulsion 
high temperatures
mass Be > twice mass He. Suppresses fusion into
Carbon
Ideally use Deuterium and Tritium, =1 barn, but
little Tritium in Sun (ideal for fusion reactor)
2
H  H  He  n
3
4
Q  17 MeV
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Fusion in Sun
m ( H )  1 . 007825 u
1
m ( H )  2 . 014102 u
2
m ( He )  4 . 002603 u
4
m ( Be )  8 . 005305 u
8

p  p H  e  
2
p  H  He  
2
3
3
He  He  He  p  p
3
4
m ( C )  12 . 00000 u
12
•

•
•
•
rate limited by first reaction which has to convert a
p to a n and so is Weak
(pp) ~ 10-15 barn
partially determines lifetime of stars
can model interaction rate using tunneling – very
similar to Alpha decay (also done by Gamow)
tunneling probability increases with Energy
(Temperature) but particle probability decreases
with E (Boltzman). Have most probable (Gamow
Energy). About 15,000,000 K for Sun but Gamow
energy higher (50,000,000??)
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Fusion in Sun II
m ( H )  1 . 007825 u
4
He  He  Be
m ( H )  2 . 014102 u
8
Be  He  C  
1
2
m ( He )  4 . 002603 u
4
m ( Be )  8 . 005305 u
8
m ( C )  12 . 00000 u
12
4
8
4
12
m Be  2 m He  92 KeV
 Be  10
 12
sec
•
need He nuclei to have energy in order to make
Be. (there is a resonance in the  if have invariant
mass(He-He)=mass(Be))
• if the fusion window peak (the Gamow energy
weighted for different Z,mass) is near that
resonance that will enhance the Be production
• turns out they aren’t quite. But fusion to C start at
about T=100,000,000 K with <kT> about 10 KeV
each He. Gamow energy is higher then this.
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Fusion in Sun III
m ( H )  1 . 007825 u
4
He  He  Be
m ( H )  2 . 014102 u
8
Be  He  C  
1
2
m ( He )  4 . 002603 u
4
m ( Be )  8 . 005305 u
8
m ( C )  12 . 00000 u
12
•
4
8
4
12
m Be  2 m He  92 KeV
 Be  10
 12
sec
Be+HeC also enhanced if there is a resonance.
Turns out there is one at almost exactly the right
energy --- 7.65 MeV
 m  0 . 28 MeV
12
*
C 0

11 ,185 . 65 MeV
11 ,185 . 37  m Be  m He
7.65 MeV
11 ,185 . 27  3 m He
2

4.44 MeV
0

11 ,178 MeV
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