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.
Recalling the semi-emperical mass formula
M (Z , A)  Zm( H)  Nmn  B(Z , A) / c
1
2
where the binding energy was modeled by
2
(
A

2
Z
)
B  aV A  aS A2 / 3  aC Z ( Z  1) A1 / 3  asym

A
Note that for constant A this gives a parabola in M vs Z
which implies the mass takes on some minimum value!
Mass defect  (MeV)
A = 127
isobars
-80
-80
-84
-84
-86
-86
-88
-88
50
52
54
56
Atomic number, Z
For odd A the isobars
lie on a single parabola
as a function of Z.
In this case there is only
a single stable isobar
to which the other
members on the parabola
decay by electron or
positron emission
depending on whether
their Z value is lower
or higher than that
of the stable isobar.
For even A the pairing
term in the Mass formula
splits the parabola into
two, one for even Z and
one for odd Z.
The beta decay transition
switches from one
parabola to the other and
in this case there may be
two or even three stable
isobars.
Note also this implies
some even A nuclei
(eg 29Cu64) can decay
by either electron or
positron emission.
42Mo
A = 104 isobars

43Tc
103.912
?A

XN  Z+1
Y
+

?
? N-1
A
A
e-capture: ZX N + e  Z-1YN+1
A
-decay: Z
Mass, u
103.910
Odd Z
47Ag
103.908
45Rh
Even Z
103.906
48Cd
44Ru

46Pd
103.904
42
44
46
Atomic Number, Z
48
Electron emission
If Mc2(A,Z) is the rest mass energy of the parent atom
and Q is the energy released in the decay then
M(A,Z)c2 = M+(A,Z+1)c2 + mec2 + Q
where M+(A,Z+1)c2 is the rest mass energy of the positive ion produced.
Now since the ionization potential energy is relatively small we can write -
M+(A,Z+1)c2 + mec2 ~ M(A,Z+1)c2
Thus M(A,Z)c2 = M(A,Z+1)c2 + Q
energy will be released in the decay (ie Q > 0)
provided M(A,Z)c2 is greater than M(A,Z+1)c2.
What is the maximum energy of the electron emitted in the  decay of
3
1
The reaction is:
H?
3
3

1 H 2 He  e
Q  (M H  M He )c

2
 (3.016050u  3.016030u)(931.5MeV / u)
 0.0186MeV  KHe  Ke  K
Neglecting the kinetic energy of the nucleus, and mass of the neutrino
the Q is shared between e and . Ke at maximum when K0, so
max
Ke
 0.0186MeV
Positron emission
The charge on the nucleus is reduced
and the daughter atom has an excess electron.
Thus
M(A,Z)c2 = M(A,Z-1)c2 + 2mec2 + Q
The 2mec2 account for the emitted positron and
the excess atomic electron in the final state.
In this case
the decay process can only proceed (Q > 0) if
M(A,Z)c2 > M(A,Z-1)c2 + 2mec2.
i.e, the difference in rest mass energies must be at least 1 MeV.
Electron capture
Instead of p  n + e+ +  occuring inside the nucleus
it is possible for an atomic electron to be captured in the process
p + e  n + 
and the daughter atom produced has the correct compliment of electrons.
but in an excited state (with a vacancy in its electron configuration).
This process is often called K capture
because it is an inner shell atomic electron which is captured.
M(A,Z)c2 = M(A,Z-1)c2 + E* + Q
E* is the excitation energy of the daughter atom
(sufficiently small to be ignored).
Electron capture can proceed (Q > 0) if M(A,Z)c2 < M(A,Z-1)c2.
It strongly competes with positron decay but
is the only decay route available to nuclei with a proton excess
but a mass difference between the atoms of less than 1 MeV/c2.
-decay
-decay
Some Alpha Decay Energies and Half-lives
Isotope
232Th
238U
230Th
238Pu
230U
220Rn
222Ac
216Rn
212Po
216Rn
KE(MeV)
4.01
4.19
4.69
5.50
5.89
6.29
7.01
8.05
8.78
8.78
t1/2
1.41010 y
4.5109 y
8.0104 y
88 years
20.8 days
56 seconds
5 seconds
45.0 msec
0.30 msec
0.10 msec
l(sec-1)
1.61018
4.91018
2.81013
2.51010
3.9107
1.2102
0.14
1.5104
2.3106
6.9106
1930 Series of studies of nuclear beta decay, e.g.,
Potassium goes to calcium
Copper goes to zinc
Boron goes to carbon
Tritium goes to helium
 20Ca40
64
64
29Cu  30Zn
12  C12
B
5
6
3  He3
1H
2
19K
40
Potassium nucleus
Before decay:
After decay:
A
Which fragment has a greater
momentum?
energy?
B
A)
B)
C) both the same
1932 Once the neutron was discovered, included the more fundamental
np+e
For simple 2-body decay, conservation of energy and momentum
demand both the recoil of the nucleus and energy of the emitted
electron be fixed (by the energy released through the loss of mass)
to a single precise value.
Ee = (mA2 - mB2 + me2)c2/2mA
but this only seems to match
the maximum value
observed on a spectrum of beta
ray energies!
No. of counts per unit energy range
0
5
10
15
20
Electron kinetic energy in KeV
The beta decay spectrum of tritium ( H  He).
Source: G.M.Lewis, Neutrinos
(London: Wykeham, 1970), p.30)
1932
charge
mass
n  p + e + neutrino
0
+1
1
939.56563 938.27231 0.51099906
MeV
MeV
MeV
?
?
neutrino mass < 5.1 eV < me /100000
0
1936 Millikan’s group shows at earth’s surface
cosmic ray showers are dominated by
electrons, gammas, and
X-particles
capable of penetrating deep underground
(to lake bottom and deep tunnel experiments)
and yielding isolated single cloud chamber tracks
1937 Street and Stevenson
1938 Anderson and Neddermeyer
determine X-particles
•are charged
•have 206× the electron’s mass
•decay to electrons with
a mean lifetime of 2msec
0.000002 sec
1947 Lattes, Muirhead, Occhialini and Powell
observe pion decay

Cecil Powell (1947)
Bristol University
C.F.Powell, P.H. Fowler,
D.H.Perkins
Nature 159, 694 (1947)
Nature 163, 82 (1949)
Consistently
~600 microns
(0.6 mm)

Under the influence of a magnetic field
m

m+ energy
always
predictably fixed
by E
simple 2-body decay!
+ m+ + neutrino?
charge
+1
+1
?
n p + e + neutrino?
+ m+ + neutrino?
Then
p
m- e- + neutrino?
m
???
e
As in the case of decaying radioactive isotopes,
the electrons’s energy varied, with a maximum
cutoff (whose value was the 2-body prediction)
3 body decay!
m
e
2 neutrinos
1953, 1956, 1959
Savannah River (1000-MWatt)
Nuclear Reactor in South Carolina
looked for the inverse of the process
n  p + e- + neutrino
p + neutrino  n + e+
with estimate flux of
51013 neutrinos/cm2-sec
observed
2-3 p + neutrino events/hour
also looked for
Cowan & Reines
n + neutrino  p + e
but never observed!
Fermi Theory
starting from
Fermi’s Golden Rule
4
2 dN f
Wif 
M
h
dE
2
where M is the “matrix” element
M
*
f
H int i dV
for the transition and Hint is the interaction responsible for it
and
dN f
dE
as we’ve talked about before is the
“density of states”
Recall: The Golden Rule’s Wif is the rate of transition
from initial state i to final state f , i.e.
Wif = l
For a single particle we’re dealing with a 6-dimensional phase space –
three position coordinates and three momentum components.
The smallest element of volume a
state can occupy in this phase space
is h3.
(ignoring for now particle spin)
recall the uncertainty relation dxdpx ~ h
Thus (dx)3(dpx)3 ~ h3 and
the number of states in a the “volume” element d3x d3p is given by
3
3
d xd p
dN 
3
h
d3x  dx dy dz and d3p  dpx dpy dpz.
For a coordinate space volume V and
a momentum volume 4 p2dp
for both electron and neutrino sharing the same volume:
dNf = 162V2(pe)2dpe (p)2dp /(h6)
already integrating over d3xe and d3x
With three particles in the final state pe and p are independent.
Within the overall energy constraint the momenta is balanced
by that of the daughter nucleus. This means that any electron
momentum can be expressed in terms of the neutrino state.
Its the electron which can be directly detected, so often the
neutrino quantities are written in terms of
p = (Emax - Ee)/c = (Tmax-Te)/c
dp/dE = 1/c
Hence dNf/dE = 162V2(Tmax - Te)2(pe)2dpe/(h6c3)
To calculate the matrix element we replace Hint by a constant g
(coupling constant)
and assume e and  are ~constant inside the volume V
(and ~zero elsewhere).


i p r / 
3
 e ( r )  Ce
*
2
1


d
r

1

C

e
e

V



i q r / 
  ( r )  Ce
Actually:
e
e

i p r / 

i q r / 
 
ip  r
 1


 
iq  r
 1


For 1 MeV
electron
p/ħ=0.007 fm-1
and within
the nucleus
r<6 fm
42
2 dN f
Wif 
M
h
dE
dNf/dE = 162V2(Tmax - Te)2(pe)2dpe/(h6c3)
Intensity
Energy spectrum of beta
decay electrons from 210Bi
Kinetic energy, MeV
3
2
1


d
r

1

C

e
e

V
e

i p r / 
*
e

i q r / 
 
ip  r
 1


 
iq  r
 1


Then, under this approximation:
 i   A,Z
1 1
V V
 f   A,Z 1 e    A,Z 1
thus
g
g
*
M    A, Z 1 A, Z dV  M N
V
V
64
2
2 pe dpe
Wif  2 M N (Tmax  Te )
 P ( pe )dpe
7 3
g
hc
and
4
2
where P(pe) is the probability that
a beta particle with momentum pe
will be emitted in unit time.
The individual contributions, dl , per specific pe value:
64
2
2 pe
P ( pe )  2 M N (Tmax  Te ) 7 3
g
h c
2
4
64
2
2
2
 7 3 2 M N pe (Tmax  Te )
h c g
4
 C pe (Tmax  Te )
2
2
This quadratic equation has zeroes at pe
2
= 0 and Te = Tmax  Q
This phase space factor determines the decay electron momentum spectrum.
(shown below with the kinetic energy spectrum for the nuclide).
64
2
2 pe dpe
Wif  2 M N (Tmax  Te )

P
(
p
)
p
e
e
7 3
g
hc
2
4
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.
2
P ( pe )dpe  64 g M N F ( Z , pe )(Tmax
4
2
2
pe dpe
 Te )
7 3
hc
2
P (pe) is  the measured beta momentum spectrum
2
P ( pe )dpe  64 g M N F ( Z , pe )(Tmax
4
P ( pe )
2
pe F ( Z , pe )
2
2
pe dpe
 Te )
7 3
hc
 64 g M N (Tmax  Te )
P ( pe )
2
pe F ( Z , pe )
4 2

8 g
2
7 3
h c
2
2
1
2
7 3
h c
M N (Tmax  Te )
Including the measured probability Pmeas(pe) in
Pmeas ( pe )
2
p e F ( Z , pe )
which when plotted against Te yields a straight line
the Fermi-Kurie plot.
The overall decay probability l is obtained
by integrating P(pe)dpe from 0 to pmax :
l = {64 4g2 [MN]2/(h7c3)} ∫ F(Z,pe) (Tmax - Te)2(pe)2dpe
= {644g2m5c4 [MN]2/(h7)}f(Z,Tmax)
= G2 [MN]2 f(Z,Tmax)
where
f(Z,Tmax) =
∫ F(Z,pe) {(Tmax-Te)/(mc2)}2(pe/(mc))2dpe/(mc)
with energy and momenta expressed in units of mc2 and mc respectively.
The integral is often tabulated when it cannot be solved analytically.
l = G2 [MN]2 f(Z,Tmax)
Since the decay probability
l = ln(2)/t1/2
f(Z,Tmax) t1/2 = ln(2)/(G2 |MN|2)
This ‘ft' value provides
a measurement of
the nuclear matrix element MN
of the decay transition.
Note:
the nuclear matrix element depends
on how alike A,Z and A,Z±1 are.
the shortest half-lives (most common) -decays “super-allowed”
0+  0+
 10B*
14O  14N*
10C
1p1/2
1p3/2
1s1/2
The space parts of the initial and final wavefunctions are identical!
What differs?
The iso-spin space part (Chapter 11 and 18)
|MN|2 = 2
Table 9.2 ft values for “Superallowed” 0+0+ Decays
Decay
10C10B
14O14N
18Ne18F
22Mg22Na
26Al26Mg
26Si26Al
30S30P
34Cl34S
34Ar34Cl
38K38Ar
38Ca38K
42Sc42Ca
46V46Ti
50Mn50Cr
54Co54Fe
ft (seconds)
3100  31
3092  4
3084  76
3014  78
3081  4
3052  51
3120  82
3087  9
3101  20
3102  8
3145  138
3091  7
3082  13
3086  8
3091  5
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
J = 0 and
'the nuclear parity must not
change'.
Mirror Nuclei
two nuclei with the same closed shell core of nucleons but
one with a single odd proton outside this core,
the other a single odd neutron. 39
13
13
39
7
N6
6
C7
Ca19
20
K20
19
Such nuclear wavefunctions are identical except for small
coulomb effects.
This implies |MN|2
~1
thus any beta decay transition between the ground states of mirror nuclei
can be used to measure the coupling constant g.
It is found that log10(ft1/2) ~ 3.4
so g ~ 1.4 10-62 J m3
0.8810-4 MeV·fm3
11.13
width 0.26 MeV
3/2
9.7
3/2
7.48
5/2
6.56
width 1.0 MeV
10.79
width 0.29 MeV
3/2
width 1.2 MeV
5/2
9.2
7.19
5/2
6.51
4.63
7/2
4.55
7/2
0.478
0
1/2
3/2
0.431
0
1/2
3/2
Excited states of Li7
Excited states of Be7
The charge symmetry of nuclear forces is illustrated
by the existence of mirror nuclei like
Li7 and Be7
C14 and O14
Mirror Nuclei
two nuclei with the same closed shell core of nucleons but
one with a single odd proton outside this core,
the other a single odd neutron. 39
13
13
39
7
N6
6
C7
Ca19
20
K20
19
Such nuclear wavefunctions are identical except for small
coulomb effects.
This implies |MN|2
~1
thus any beta decay transition between the ground states of mirror nuclei
can be used to measure the coupling constant g.
It is found that log10(ft1/2) ~ 3.4
so g ~ 1.4 10-62 J m3
0.8810-4 MeV·fm3
j  total angular momentum of any single nucleon
I  total angular momentum of the entire nucleus
Often a single valence nucleon determines a nuclei’s properties
I=j
When necessary to consider 2 valence particles: I = j1 + j2
And when there’s an odd particle atop a filled core: I = jnucleon + jcore
All (hundreds) of known (stable and radio-active)
even-Z, even-N nuclei
have spin-0 ground states.
Powerful evidence for the nucleon-pairing we’ve described
as a fundamental part of nuclear structure.
Obviously the ground state of odd-A nuclei have
I = j of the odd proton or neutron.
Since individual p,n are fermions (with spin = 1/2 )
the individual j = s + ℓ
=½+ℓ
i.e. must be half-integral:
with z-components:
1 , 3 , 5 ,…
2 2 2
 1 ħ,  3 ħ,  5 ħ, …
2
2
2
Then for even-A nuclei: Iz = integral values
I is an integer!
odd-A nuclei: I = half-integer
even-A nuclei: I = integer
For nuclei, Parity is only in principal calculable,
In practice it is inferred
by studying the reactions nuclei participate in
or analyzing scattering distributions.

1 ip r / 
 e (r ) 
e

V

1 iqr / 
  (r ) 
e

V
1
V
1
V
 
 ip  r

  
1 



 
 iq  r

  
1 



1
V
1
V
An approximation based on the assumption that over the nuclear volume:
1/ 3
r  A r0  6 fm
 
p  r /   1
While fine for estimating energy differences,  spectra
over-simplifies picture w.r.t angular momentum considerations
Assumes, effectively, , e created at r=0 with ℓ=0 and j=s=1/2 for each.
1932
charge
mass
n  p + e + neutrino
0
+1
1
939.56563 938.27231 0.51099906
MeV
MeV
MeV
?0
?
neutrino mass < 5.1 eV < me /100000
0
but what else?
spin
½
½
½
?
½
together carry net momentum from the nucleus
Total S = 0 (anti-parallel spins)
Fermi Decays
Total S = 1 parallel spins)
Gamow-Teller Decays
Nuclear I = 0
Ii = If + 1
I = 0 or 1
If the electron and neutrino carry NO orbital (relative) angular momentum
With Pe, = (1)ℓ = +1
PA,Z = PA,Z1
I = 0,1 with no
P change
10C10B*
14O14N*
6He6Li
0+  0+
0+  1+
13B13C
3/2
 1/2
np
1/2+
 1/2
3H3He
1/2+
 1/2
13N13C
1/2
 1/2
Fermi Decays
Gamow-Teller Decays
e, pair account for I = 1 change
carried off by their parallel spins
Forbidden Decays
ℓ=1 “first forbidden”
With either Fermi decays s = 0
Gamow-Teller decays s = 1
I  0, 1, 2
with
17
Parity change!

N O
76

(1 / 2  5 / 2 )
17
Br Se




(1  0 )
76
Sb Sn (2  2 )
122
122
*
Forbidden Decays
even rarer!
ℓ=2 “second forbidden” With either Fermi decays s = 0
Gamow-Teller decays s = 1
I  2, 3
With no Parity change!
Fermi and Gamow-Teller already allow (account for)
I= 0, 1 with no parity change
and contributions from these deacys will dominate
22
Na  Ne
22
Cs  Ba
137
137


(3  0 )


(7 / 2  3 / 2 )