Phys. 649: Nuclear Instrumentation
Physics Department
Yarmouk University
Introduction
Energy Release in β Decay
Supplement 1:
Beta Decay
© Dr. Nidal M. Ershaidat
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Introduction
Electron emission, positron emission (1934, I. & F.
Joliot-Curie) and (orbital) electron capture (1938,
Alvarez) are all known as beta decay processes
−
n → p + e− + ν
→ + + νe
p → n + e+ + ν
p + e− → n + ν
β
β+
ε
The electron resulting from a β - decay process is
“created” thanks to the energy available.
(Electrons do not preexist inside a nucleus).
The processes involving protons occur only for
bound protons in nuclei (The presence of the
“nucleus field” is a sine qua non condition)
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
What really happens!
We know now that the
responsible for beta decay.
weak
interaction
is
n → p + e− + νe
Fig 1: β- decay
β -,
W-
In
The
mediates the interaction. One of the d
quarks of the neutron transform into a u quark.
The neutron, thus, becomes a proton. An electron
and its anti-neutrino are emitted in the process.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
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Typical Beta Decay Processes
Decay
Type
23Ne
→ 23Na + e-+
99Tc
→ 99Ru + e- +
νe
Q(MeV)
t1/2
β-
4.38
νe
β-
0.29
25Al
→ 25Mg + e+ + νe
β+
3.26
7.2 s
124I
→124Te + e+ + νe
β+
2.14
4.2 d
15O
e-
ε
2.75
1.22 s
ε
0.43
1.0×
×105 y
+
41Ca
→
15N
+ νe
+ e- → 41K + νe
38 s
2.1×
×105 y
Exercise: Check the Q-value for all these reactions.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
Energy Continuous Spectrum in β Decay
6
We expect to have mono-energetic electrons as we observe
the mono-energetic alpha’s in a decay. But instead we have
for the electrons resulting from a β decay a continuous
spectrum starting at 0 and ending at Emax (Endpoint energy)
which is the energy an electron should have in this decay!
Fig
1:
βdecay
spectrum, i.e. Energy
Distribution of the
electrons.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
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Energy Release in β Decay
Beta decay of 210Bi
210
210
83 Bi127 → 84 Po126
Energy Release in β Decay
+ e−
Q = ( 209.984095 - 209.982848) × 931.5002 - 0.511
= 0.650 MeV
Neglecting the recoil energy of the daughter,
the maximum energy an electron can have is :
Emax = 0.650 + 0.511 = 1.161 MeV
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
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The Neutrino
Experiments showed that the shape of the
spectrum of electrons emitted in β decay is
characteristic of the electron themselves.
In 1931 Pauli, suggested the presence of a
second particle emitted in the decay which can
carry a part of the available energy and linear
momentum. This particle should have a zero
mass, be neutral and interacts so weakly with
matter that detectors do not “see” it! Pauli
called it The ghost particle and Fermi gave it
the name neutrino (small neutron in Italian).
The neutrino was discovered in 1957!
Kinematics
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
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Kinematics
Consider the β - decay of a free neutron (t1/2 = 10 min)
−
−
n → p + e + νe
β
Q = (mn - mp - me - mν ) c2
In a frame attached to the decaying neutron, the
available energy will be shared by the 3 resulting
particles:
Q = T p + T − + Tν e
e
12
The β - Decay Electron is Relativistic
The energy carried by the electron in β- is of
the order of its rest mass energy Te/ me c2 > 0.1,
while the recoil energy is low and can be
taken non relativistically.
Neglecting the proton’s recoil energy Tp, which is
measured to be 0.3 keV, Q is essentially shared by
the electron and the antineutrino.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
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The electron-(anti)neutrino is massless
β- Decay Kinematics
A
Z XN
Q = mn c 2 − m p c 2 − m − c 2 − m νe c 2
e
= 939.573 − 938.280 − 0.511 − m νe c 2
= 0.782 MeV − m νe c
Q
β−
2
The measured value is Q = 0.782 ± 0.013 MeV. This
suggests that the mass of the antineutrino = 0
within the experimental error (13 keV). New
experiments give very much lower limits (few eV's)
Measurements of the linear momentum of the
electron and the proton indicate that a 3rd
particle should be present.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
→
A
−
Z + 1Y N − 1 + e + ν e −
= [m N
( X)−m (
A
Z
A
N Z + 1Y
)− m
e−
]c 2
where N indicates the nuclear mass and energy
and considering a massless antineutrino.
Neglecting the electrons binding energies we
have :
A
A
2
Q
= [m
Masses here
(tables)
are
β−
Q
( X )− m (
β−
Z
Z + 1Y
neutral
)]c
atomic
masses
= Te + Tνe
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
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β+ Decay Kinematics
ν −
Q is shared between e- and
e
This is the energy shared between the electron
and the antineutrino.We saw that in the case
of β- Decay of 210Bi,
Q = 1.161 MeV
is in good agreement with the measured value.
A
Z XN
Q
β+
=[m
→
A
Z − 1Y N − 1
( X ) − m(
A
Z
16
+ e+ + νe
A
Z −1Y
)− 2 m
e−
]c 2
This measurement is used to calculate the
mass of the 210Po isotope.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
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Electron Capture
A
Z XN
+ e− →
A
Z − 1Y N + 1 + ν e
The calculation of Q should take into account
the fact that the daughter nucleus is in an
excited state. The resulting X-ray (or X-rays)
should have the binding energy of the captured
electron
Qε = [m
( X ) − m(
A
Z
A
Z − 1Y
)]c
2
− Bn
Fermi Theory of β Decay
Where Bn is the binding energy of the captured
electron from shell n (K, L, M,…)
Note that here the neutrino is mono-energetic. And
if we neglect the recoil energy of AY , Eν = Q.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
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Fermi Theory of β Decay
The theory to explain β Decay should include the
following information :
1) The electron and the neutrino do not preexist in
the nucleus
2) The electron and the neutrino are relativistic.
3) The continuous distribution of electron energies
Enrico Fermi proposed, in 1934, a theory of β decay
based on Pauli’s neutrino hypothesis. The major
idea is that β Decay
is the result of a weak
interaction (compared to that responsible for the
quasi-stationary states). The characteristic times
in β decay is of the order of seconds or longer
where the nuclear characteristic time is of the
order of 10-20s.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
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Fermi Theory - Transition probability
The barrier potential used in alpha decay (See
Suppl2-Alpha decay) does not exist in the case of β
Decay. And even though if it exists, the
transmission probability is nearly 1.
Fermi’s idea was to consider β decay as a
perturbation forcing the quantum system (The
parent nucleus) in a transition.
The transition probability is given by :
λ=
2π
V fi
h
2
( )
ρ Ef
ρ(Ef) is the final density of final states = dn/dE
dn is the number of final sates per energy interval
dE and Vfi is the matrix element:
*
V fi = ∫ ψ f V ψ i dv
5
21
Fermi Theory - Perturbation Potential
Fermi had no idea about the weak interaction
potential so he tried all possible forms consistent
with special relativity, and showed that V can be
replaced by an operator OX, where X gives the form
of the operator O.
X = V (vector potential),
X = A (Axial vector)
X = S (Scalar)
X = Pseudoscalar
X = Tensor
Only experiment can help deciding which
transformation is the appropriate one. Now we
know that X is the so-called V-A (Vector-Axial
transformation).
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Supplement 1 Beta Decay
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Phys. 649: Nuclear Instrumentation
Physics Department
Yarmouk University
Basic α Decay Processes
(Introduction and Kinematics)
Supplement 2:
Alpha Decay
© Dr. Nidal M. Ershaidat
3
Introduction
Seven years after Becquerel’s discovery, Rutherford
(and Mme Curie) identified the naturally emitted α
particles as being less penetrating comparatively to
the other emitted ones (β
β & γ).
By simply using a deflecting magnetic field Marie Curie
demonstrated that a particles are doubly positively
charged.
γ
β
4
α Particles are Helium 4 Nuclei
Rutherford, using an evacuated
closed
chamber with a thin wall accumulated α
particles emitted by radium for several days,
proved by atomic spectroscopy that helium gas
is formed and thus that an α particle is in fact a
4He nucleus (or a doubly ionized helium atom).
This means that an α source ejects a cluster of
4 nucleons (2p & 2n).
α
Magnetic Field
Radioactive source
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Suppl. 2 Alpha Decay
Why?
One can imagine that it is easier and simpler
for an unstable nucleus to eject a single
nucleon or 2 nucleons.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Suppl. 2 Alpha Decay
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Q-Value
In a nuclear reaction
a+ A → B+b
The net energy release called the Q-value of the
reaction is given by:
(
)(
Q = ma c 2 + m A c 2 − mb c 2 + m B c 2
6
Q-Value - Generalization
)
The Q-value is simply the difference between
the available initial rest energy and the
resulting final rest energy.
More generally, in a nuclear reaction involving
s nuclei or nucleons and the result of which is
the creation of t nuclei or nucleons the
(general) definition of the Q-value is:
s
Q =
∑ mi c
t
2
−
i =1
∑mf
c2
f =1
The reaction is energetically possible if and
only if Q > 0.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Suppl. 2 Alpha Decay
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Suppl. 2 Alpha Decay
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α Decay Kinematics
In an α decay, an unstable nucleus X emits an α
(4He nucleus) and transforms to nucleus Y.
A
A− 4
4
Z X N → Z − 2Y N − 2 + 2 α 2
Example
226
88 Ra 136
→
222
86 Rn134
+α
E α = 4.8 MeV
t1/2(226Ra)=1600y
AX
If we consider that the nucleus
decays while at rest
(or equivalently we consider a referential attached to
this nucleus) then conservation of energy gives:
m X c 2 = mY c 2 + TY + m α c 2 + Tα
In these conditions the only available energy is the
rest mass energy of AX.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Suppl. 2 Alpha Decay
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Momentum and Kinetic Energy of α
Q = m X c 2 − m Y c 2 − m α c 2 = TY + T α
In the cm referential, X at rest (PX = 0), the
resulting nucleus Y and the α particle are
emitted in opposite directions and we have :
P 2 2 mα mY
T
PY = Pα ⇒ α = α2
=
>1
TY
P 2 mY mα
Y
The available energy Q is shared by Y and the α
particle inversely proportionally to their respective
masses, i.e. the α particle kinetic energy is much
bigger than the nucleus Y’s kinetic energy.
Q
4

Tα =
Q 1− 
1 + m α mY
A

A >> 4
→
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Suppl. 2 Alpha Decay
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Computing Q for “possible” decays of 232U
α Decay is energetically favored
Let’s look at the possible emission
decaying 232U nucleus (mass = 232.0366 u).
Emitted particle
(mass in MeV/c2)
Decay
232
231
1
92 U 140 → 92 U 139 + 0 n1
232
231
1
p (938.280)
92 U 140 → 91 Pa 140 + 1 H 0
230
2
2H (1875.628) 232
92 U 140 → 91 Pa 139 + 1 H 1
n (939.573)
4He
(3728.433)
232
92
U 140 →
228
Th138 + 24He 2
90
of
a
Decay
223
92 U 131
→
222
92 U 130
10
+ 01n1
Q = (232.0366 - 231.0357)×
×931.502 - 939.573= - 7.26 MeV
Q (MeV)
Decay
232
92 U 140
→
231
91 Pa 140
+ 11p0
- 7.26
Q = (232.0366 - 231.03558)×
×931.502 - 938.280= - 6.12 MeV
- 6.12
Decay
- 10.70
Q = (232.0366 - 230.03457)×
×931.502 - 1875.68 = - 10.70 MeV
Decay 232 U
→ 228Th + 4He
232
92 U 140
92
+ 5.41
→
140
230
91 Pa 139
90
138
+ 12H 1
2
2
Q = (232.0366 - 228.028715)×
×931.502 - 3727.409= + 5.41 MeV
Q =  m232 − m232 − m1  c2
92U139
0 n1 
 92U140
Q=(232.0366 - 231.0357) × 931.502 - 939.573 = - 7.26 MeV
Decay
232
92 U 140
→
226
89 Ac 137
+ 36Li 3
Q=(232.0366 - 226.02608- 6.0151)×
×931.502 = - 3.79 MeV
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Suppl. 2 Alpha Decay
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Suppl. 2 Alpha Decay
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Criteria for an α Decay to occur
Is Q > 0 the unique condition for an α Decay to occur?
12
The previous calculations show that only alpha
decay can occur spontaneously (Q > 0). But this is
not the only criterion. For example the following
decays are energetically possible(Q > 0)
1. Q-value should be > 0
232
92 U 140
232
92 U 140
technological limits to half-lives of the order of 1016
→
→
222
88 Pb136 +
220
86 Rn134 +
8
4 Be 4
12
6C6
Exercise : Check that!
However, nuclear spectroscopy shows that such
decays have vanishingly small partial decay
constants compared to a decay, and thus they are
not seen.
In some cases, beta decay is intense enough to
mask possible α decays.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Suppl. 2 Alpha Decay
2.The partial
enough This
decay constant should be large
corresponds, according to our
years.
3. α decay should not be masked by β decay.
Half of the unstable nuclei against alpha decay (A >
190 and many nuclei in the range 150 < A < 190)
verify this criterion.
© Dr. Nidal M. Ershaidat - Nuclear Instrumentation - Chapter 1: Radiation Sources - Suppl. 2 Alpha Decay
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Phys. 649: Nuclear Instrumentation
Physics Department
Yarmouk University
Systematics of α emission
Supplement 2:
Part 2- Systematics and
Theory of α emission
© Dr. Nidal M. Ershaidat
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Systematics of α Decay
1) Emitters with large disintegration energies are
short-lived and vice versa (Geiger & Nuttall, 1911)
Nuclide
232Th
1.4 x
218Th
Qα
Qα
(
(
218
232
t1/2
1.0 x
) = 2.41
Th)
Th
1010
10-7 s
Geiger & Nuttall
Fig. 6.1 : Qα vs. t1/2
Qα (MeV)
y
4.08
= 100 ns
9.80
t1 2
t1 2
(
(
218
232
) = 10
Th)
Th
− 24
!
For a factor 2 in energy, the half-lives are 10-24
times different !!
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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α Decay & Stability
2) Adding a neutron to an unstable nucleus
reduces the disintegration energy which
means, according to Geiger-Nuttall findings,
a longer lifetime or more stability.
α Decay & The Semiempirical Mass Formula
6
Qα = m(Z,A) c2 – m(Z-2, A-4) c2 - mα c2
Z
m(Z,A) c2 = (Z mpc2 + N mnc2– B(Z,A)) + Z me c2 +
∑ Bie
i =1
(See Fig for A > 212). There is a
distinguished
discontinuity
at
(N=126,
A=212). This is conform to the shell-model
and the existence of magic numbers.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
α Decay & The Semiempirical Mass Formula
B ( Z − 2, A − 4 ) = aV ( A − 4) − a S ( A − 4 )2 3
−1 3
− aC ( Z − 2 ) ( Z − 3 ) ( A − 4 )
(( A − 4) − ( Z − 2))2
( A − 4)
−3 4
+ a p ( A − 4)
− a Sym
Qα = B(4He) + B(Z-2, A-4) – B(Z,A)
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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B(Z-2,A-4) - B(Z,A)
- The volume term: aV (A - 4) - aV A = - 4 aV


- The Coulomb term :
B ( Z , A) = aV A − a S A 2 3 − aC Z ( Z − 1) A −1 3
− a Sym
( A − Z )2 + a
A
p
A− 3 4
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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4
 + aS A2 3
A
≅ aS A2 3 × 2 × 4 = + 8 aS A−1 3
A>>1
3 A
3
- The surface term: − aS A2 3  1 −
≅
A >> 1
Z >> 1

Z 

4 a C Z A − 1 3  1 −
3
A 

−1 3


2   4 
4 
 +( Z( Z
− 1−) 1A)−1 3
= − aC (ZZ −
A2−)1(3Z− 31)−A−1 3 (Z1 −− 3 )  1 ++ aC Z
Z   A 
3 A


© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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Qα = B (4He) + B(Z-2,A-4) - B(Z,A)
Qα = B(4He) + B(Z-2, A-4) – B(Z,A)

8
Z 

+ aS A−1 3 + 4 aC Z A−1 3  1 −
3
3
A 

2
 2Z 
−7 4
− 4 aSym  1−
 + 3 aP A
A


Qα = 28.3 − 4 aV
11
Explaining the Difference
The small difference (0.3 MeV) between the
2 values which comes from the fact that the
semiempirical mass formula’s parameters
are chosen so as to reproduce the maximum
number of nuclei masses, is not significant .
Qα = 28.3 - 62 + 7.35 + 36.90 - 3.81 - 0.0077 = 6.75 MeV
For 226Th this gives Q = 6.75 MeV, while the
measured value is 6.45 MeV.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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Theory of α Decay
Gamow and Condon & Gurney proposed
independently in 1928 a theory to explain
the alpha decay and its systematics.
Theory of α Emission
The theory is based on the idea that alpha
particles exist in the parent nucleus (or
considered to behave as if they do exist).
The interaction potential between an alpha
particle and the residual (daughter) nucleus
is represented by the following:
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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The Coulomb potential extends inward to a and then suffers
a cut off
V(r)
V (r ) ∝
15
1) A Spherical Potential Well
Nuclear Potential an α suffers
1
r
Three regions of interest appear :
Region I (r < a) : This is the spherical part
(potential well of depth V0) where the alpha
particle, with a kinetic energy E - V = Qα + V0 moves.
The radius a is the sum of the alpha particle and
the residual nucleus.
V(r)
Qα
V (r ) ∝
Qα
a
b
r
1
r
I
a
b
r
V0
V0
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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2) A Classical Accessible Region
2) A Barrier potential
Region II (a < r < b) : This is the annular shell
region which plays the role of a barrier potential
because the potential energy here is greater than
the particle’s total energy. Classically the particle
cannot enter this region (Kinetic energy = Q - V < 0).
V(r)
Region III (r > b) : This is a classically permitted
region. if the particle succeeds to escape! then it
can move freely in this 3rd region
V(r)
1
V (r ) ∝
r
Qα
I
V (r ) ∝
Qα
II
a
b
V0
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
r
I
1
r
II
a
b
III
r
V0
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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19
But alpha can escape!
3D Barrier Potential - Semi Classical Treatment
Classically the alpha particle rebounds each time it
hits the well at r = a. In quantum mechanics there
is a small probability that it “leaks” through the
wall. The wave function “digs” a tunnel and the
particle can escape the potential well !!
The escape (tunneling) probability which is related
to the decay constant depends on the penetrability
(P) of the barrier region.
The decay constant λ is given by: λ = f P
where f represents the frequency with which the
alpha particle presents itself at the barrier and P is
the penetrability, i.e. probability of transmission
from one side to another of the barrier.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
The transmission probability of a flux of incident
particles on a barrier potential (depth V0 and Length L)
can be easily calculated in quantum mechanics.
In a 3D problem with spherical symmetry (V depends
only on r) and zero angular momentum P the
probability to penetrate the complete barrier, also
called penetrability, is given by :
P ≈ e−γ
with
Here:
γ=
V (r ) =
12
2 r2
[2 m (V (r ) − E )] dr
∫
r
1
h
1 z Z ′ e 2 2.88 Z ′
( MeV )
=
4 π ε0
r
rin F
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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Height of the (Coulomb) Barrier Potential
The height of the barrier part is :
2.88 Z
B (r = a ) =
( MeV )
a in F
&
B(r = b) = 0
The height of the barrier varies from (B - Q)
above the particle’s energy at r = a to 0 at r = b.
We shall take as a representative height (the
energy difference (E - V0)) the average height :
1
(B − Q)
2
The average width which we shall take as L is :
1
(b − a )
2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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Parameter b
At a first approximation using the average
height and width the penetrability is :
P ≈ e − 2 k L = e − 2 k ((b − a ) 2 ) = e − k (b − a )
Typical values for a heavy parent (Z = 90) are :
Q = 6 MeV, a = 7.5 F
b can be calculated as being the radius at which
the alpha particle can leave the barrier. At this
point Q = V(b) or
1 z Z′ e2
2 .88 Z ′
(F)
=
Q =
4 π ε0
b
Q (MeV)
⇒ b
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
5
22
23
Penetrability
Which gives for Q = 6 MeV :
b=
Estimating f !
λ=f P
f is roughly of the order of v/a, where v is the
2.88 × 88
≅ 42 F
6
1

2 m c2  (B − Q)
2

k=
hc
2.88 × 88
= 33 .8 MeV
For a = 7.5 F we have : B = B (r = a ) =
7 .5
3727 .409 × (34 − 6 )
k =
= 1.64 F −1
197
relative velocity of the alpha particle.
for a = 7.5 F & Q = 6 MeV and V0 = (B) = 34 MeV:
f =
f =
Thus we have : P ≈ e − 2 k L = e − 2 k ((b − a ) 2 ) = e − k (b − a )
P ≈ e − 1.64 × (42 − 7.5 ) = 2.5 × 10 − 25
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
2
v 1 2 (Q + V0 ) / m c
c
=
a 2
a
2 × 40 / 3727.409
× c = 5.86 × 10 + 21 s −1
7.5
λ = 6 × 10+21 × 2.5 × 10-25 = 1.5 × 10-3 s-1 !
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
24
Gamow’s theory
t1 2
ln 2
=
≈ 700 s
λ
If we Change Q from 6 MeV to 5 MeV then P
becomes 1 × 10-30 and t1/2 = 108 s
The theory, although using a semi-classical
treatment explain remarkably the major
observation by Geiger and Nuttall.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
25
Quantum Theory of α Decay
Refined calculation is achieved using quantum
mechanics principles. First we consider the system
(Daughter, alpha) in the a cm (center of mass)
where the problem reduces to considering a
particle with the reduced mass :
1
1
1
mα MY
→ µ =
=
+
MY mα
mα + MY
µ
Then one divides the Coulomb barrier potential
into spherical shells of radii (r,r+dr). The
transmission probability between r and r + dr is

 
2µ  1 z Z e 2
dP = exp  − 2
− Q α  dr 
2

 

h  4 π ε0 r
 

© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
6
26
27
The Gamow factor G
P in Quantum Mechanics
The penetrability is: P = e -2G
For x << 1 ( Q << B or a << b), which is the case
in the case for most decays of interest, G
becomes:
2
Where G (the Gamow factor) is:
1
 2
1 b   1 z Z ′ e2
G = ∫a ≡ R 2 µ 
− Qα  dr


h
  4 π ε0 r

G=
Thus we have :

1 z Z′ e2
P = exp − 2
 4 π ε 0 h c
Taking x = a/b = Q/B, the integral gives :
G =
1 z Z′e2
cos −1
4 π ε0 h v
[(
)
x − x (1 − x )
1 z Z′e π

−2 x
4 π ε 0 h v  2

]

14.4 × Z ′
P = exp − 4 ×
197

Exercise : Check that!
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
2µ c2
Q
π
Q  
 −2

B 
2

2 × 3727.409  π
Q  
 −2

Q
B  
2
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
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29
Half-life in Quantum Mechanics
t1 2 =
For the even-even isotopes of Th (Z =90), the previous
calculations give the following table :
ln 2 ln 2
=
λ
fP
t1/2 (s)
2
v 1 2 (Q + V0 ) /m c
f = =
c
a 2
a
t1 2

µ c2
exp  + 25 .25 × Z ′
2 (V0 + Q )

Qα(MeV)
Measured
Calculated
220
8.95
10-5
3.3 x 10-7
222
8.13
2.8 x 10-3
6.3 x 10-5
224
7.31
1.04
3.3 x 10-2
226
6.45
1854
6.0 x 101
A
with f given by:
a
= 0 .693
c
Half-life in Quantum Mechanics
π

2
1
1  
−2

Q
B  
107
2.4 x 106
228
5.52
6.0 x
230
4.77
2.5 x 1012
1.0 x 1011
232
4.08
4.4 x 1017
2.6 x 1016
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
7
32
The Real Situation
The rough and approximate calculations give good
results (discrepancies are within 1 to 2 orders of
magnitude over a range of more than 20 orders of
magnitude).
To improve the calculations we should have taken into
consideration the following arguments :
• Fermi’s Golden Rule : The initial and final wave
functions of the transition,
• The fact that an alpha particle carries an angular
momentum,
• The non spherical shape of the majority of nuclei.
• For the highly deformed nuclei (A > 230) , the
differences become very significant. We use this fact in
the reverse order! Life-times are used to have an
approximation of the nuclei radii.
33
Application of Gamow’s Theory
• Measurements of
deformed nuclei.
the
radii
of
highly
• Calculation and Prediction of heavier nuclei
(12C ) emission
• Calculation and Prediction of single-proton
decay processes.
See Introductory Nuclear Physics
(Kenneth. S. Krane) pages 254-257.
© Dr. Nidal M. Ershaidat - Nuclear Techniques - Chapter 1: Radiation Sources - Supplement 2 Alpha Decay
8
Phys. 649: Nuclear Instrumentation
Physics Department
Yarmouk University
Introduction
Supplement 3:
Gamma Decay
© Dr. Nidal M. Ershaidat
3
Introduction
The gamma decay is the emission of an energetic
photon when an excited state of a nucleus decays
to a lower energy state. In general a series of γ
decays is necessary to reach the stable ground
state.
These decays are encountered each time a nucleus
is excited. This excitation could be the result of a
or b decays or a nuclear reaction.
The resulting photons are just like the atomic Xrays in nature, i.e. they are electromagnetic
radiations, but are more energetic (roughly 0.1 to
10 MeV, wavelengths are between 104 F and 100 F*)
4
Importance of γ spectroscopy
The Study of excited states is very rich in
information about the nuclei properties.
γ photons are relatively easy to detect which
makes them a very popular tool for
spectroscopists.
The study of the competitive process to
gamma
decay
namely,
the
internal
conversion, is an excellent tool to obtain the
spins and parities of nuclear states!
* λ(F) = hc/E = 1240/E(MeV)
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
1
5
6
Energetics of γ Decay
Eγ ≅ ∆E
Consider a nucleus, at rest, in an excited state
Ei decaying to a lower energy state Ef , a γ
photon (Eγ= pγ c) is emitted: A* → A + γ
Conservation of energy:
Ei = Ef + Eγ + TR
(1)
The symbol R stands for "recoil" of the parent.
Eγ
r
r
⇒ pR = − p γ ⇒ pR = p γ =
c
Expanding the square root we have:
2


∆E
1  ∆ E   

(5)
+
Eγ = M c 2  − 1 +  1 +


M c 2 2  M c 2   



From 3 (or 5), if we neglect the term ∆E/Mc2, we
have Eγ = ∆E.
(2)
Equation 1 can be written as:
∆E = E i − E f = E γ −
E γ2
2 M c2
1 2


∆ E  
(4)
E γ = M c 2  − 1 ±  1 + 2

M c 2  



∆E is typically of the order of 1 MeV. The rest
energy Mc2 is of the order of A×
×103 MeV, i.e. ∆E <<
Mc2
Conservation of momenta: in the cm we have:
r
r
0 = pR + p γ
Eγ is the solution of the quadratic equation 3, i.e.
(3)
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
7
Eγ ≅ ∆E
The correction to Eγ due to the recoil energy
[(∆
∆E)2/2Mc2] is negligible (10-5) and except a
special case in nuclear spectroscopy*, Eγ is
simply taken as equal to ∆E
E γ = ∆E −
(∆ E ) 2
2 M c2
(6)
Lifetimes for γ Emission,
Selection Rules
* Mössbauer Spectroscopy which is dedicated to the use of this correction
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
2
9
10
Example: 72Se 3rd excited state
Weisskopf Estimates vs. Experiment
We’ll have a closer look on the 3rd excited state of
1) Evaluation of the partial decay rate for γ emission
Example: Fig. 1 shows the energy levels of the
(even-even) 72Se (Z=34) isotope
Fig. 1
72Se.
Spin-parity = 2+
Energy = 1317 keV
t1/2 = 8.7x10-12 s
Transition 3rd-2nd = 380 keV
Transition 3rd-1st = 455 keV
The measured relative
intensities (or branching
ratios) are
λγ,1317: λγ,445:λ
λγ,380 = 51:39:10
Energies and γ transition
energies are given in keV
The decay constant λt = ln2/t 1/2 = 8.0×
×1010 s-1
All details are shown in an energy level scheme: spinparity, energies and/or γ transition energies and halflives of excited states.
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
Neglecting the (internal) conversion factors, λt is simply the
sum of the decay rates of the three transition that depopulate
this excited state!, i.e. λt = λg,1317 + λg,445 + λg,380
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
11
Comparison with Weisskopf Estimates
The partial decay
(1317,455,380) are:
rates
of
the
3
12
Weisskopf Estimates for E = 1317 keV
transitions
λγ,1317 = 0.51 * λt = 4.1×
×1010 s-1
L
λγ,1317 = 4.1×
×1010 s-1
λγ,455 = 0.39 * λt = 3.1×
×1010 s-1
λγ,380 = 0.10 * λt = 0.8×
×1010 s-1
The following tables give the calculations of λ(EL)
and λ(ML) for the energies involved in this example
(A=72)
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
λ(EL) (s-1)
λ(ML) (s-1)
1
λ(E1) = 3.95×
×1015
λ(M1) = 1.28×
×1014
2
λ(E2) = 8.70×
×1010
λ(M2) = 2.40×
×109
3
λ(E3) = 1.20×
×106
λ(M3) = 3.30×
×104
Weisskopf Estimates for E = 455 keV
L
λγ,455 = 3.1×
×1010 s-1
λ(EL) (s-1)
1.63×
×1014
λ(ML) (s-1)
1
λ(E1) =
2
λ(E2) = 4.26×
×108
λ(M2) = 1.18×
×107
3
λ(E3) = 7.11×
×102
λ(M3) = 1.94×
×101
λ(M1) = 5.28×
×1012
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
3
13
Weisskopf Estimates for E = 380 keV
L
λγ,380 = 0.8×
×1010 s-1
λ(EL) (s-1)
λ(ML) (s-1)
1
λ(E1) = 9.50×
×1013
λ(M1) = 3.07×
×1012
2
λ(E2) = 1.73×
×108
λ(M2) = 4.80×
×106
3
λ(E3) = 2.02×
×102
λ(M3) = 5.48
The previous calculations indicate that the favored
transitions are the E2 ones.
But they also show that the measured values are one
order of magnitude greater than Weisskopf Estimate.
There is a strong evidence for the collective structure
of the nucleus!, since Weisskopf used the shell (a
single individual particle) model to make his
estimations.
14
The M4 Transitions (Systematics) Case
Fig. 2 represents the experimental data for different
nuclei.
The
straight
line
represents
Weisskopf
estimate
τ(M4) = 1.54×
×105 A-2 E-9
Fig. 2 : log(ττ A2) vs. E(in keV)
This figure shows, in particular, the good agreement
with the expected E-9 dependence.
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
16
Multipoles and Angular Momenta
An em field produced by oscillations of charges
and currents produces also angular momentum. In
QM this angular momentum carried by the quanta
of energy (photons of energy E = h ν) is quantized.
3 : Selection Rules
The rate at which this angular momentum is radiated,
is proportional to the rate at which energy is radiated.
The proportionality is preserved if each emitted
photon carries a definite angular momentum.
A multipole operator of order L includes a spherical
harmonic Ylm(θ
θ,φ
φ), which is associated with an angular
momentum L.
Conclusion: a multipole of order L transfers an
angular momentum of L per photon.
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
4
17
Angular Momentum and Parity Selection Rules
18
Parity and EM Transitions
Consider a γ transition from an initial excited state of angular
momentum Ii and parity pi to a final state (If, πf). Assume Ii ≠ If.
πf
π
(Spin-parity for these states are I i i and I f respectively)
The following table resumes what we know about the
parity associated to electric and magnetic transition.
Parity
Conservation of angular momentum is expressed by:
r
r
r
Ii = I f + L
7
Thanks to the rules of addition of angular momenta we know that
L could only have restricted values.
8
| Ii – If | ≤ L ≤ Ii + If
Example: For Ii = 3/2 and If = 5/2, the possible values for L are:
1, 2, 3 and 4 and the radiated field would be a mixture of dipole,
quadrupole, octupole and hexadecapole radiation!
The relative parity of the initial and final levels determine the type
of the emitted radiation (electric or magnetic). The following table
resumes the parities related to em radiations
L
Electric Transition
Even
Odd
+
-
Magnetic Transition
+
9
The following notations are used when studying parity changes –
see Table 10
Ii
If
+
+
-
+
+
L
∆π
∆π = no
∆π = yes
+
+
-
Even
10
Odd
The two tables are used to determine the type of the emitted
radiation (electric or magnetic).
19
20
Example
Selection Rules
Let’s take again the previous example Ii = 3/2 and If = 5/2.
For L = 1 and ∆π = no, the transition cannot be electric
because the associated parity is (-)1 negative and this would
give a final parity different from the initial one. In this case
the transition is the magnetic dipole M1 transition.
In the case ∆π = yes, the transition cannot be magnetic; the
associated parity is (-)1+1 positive. In this case the transition
is the electric dipole E1 transition. See Table 18 for the other
∆π = yes
∆π = no
cases.
L
Transition
L
Transition
1
2
3
4
M1
E2
M3
E4
1
2
3
4
E1
M2
E3
M4
The precedent example suggests restrictions on
the possible transitions. These restrictions are
called selection rules
| Ii – If | ≤ L ≤ Ii + If ; Ii ≠ If.
∆π = no
even electric, odd magnetic
∆π = yes
odd electric, even magnetic
12
11
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
5
21
22
Pure Multipole Transition
The Case Ii = If.
This exception to the previous selection rules
occurs because it would mean that L = 0 is a
possible value, and there are no monopole
transitions in which a single photon is
emitted!*
The lowest multipole order allowed in the case
where Ii ≠ If. is L = 1
Another interesting case is when (Ii≠0, If.=0) or
(Ii=0, If ≠ 0)
L is equal to Ii for the first type of transition (or If
for the second one).
For an even Z – even N nucleus (like 72Se) the first
excited state 2+ decays to the ground state 0+
through the emission of a pure E2 (quadrupole)
transition.
* The magnetic monopole does not exist. For an electric
monopole, a spherical distribution of charge (like a single point
charge) the Coulomb field is not affected by radial oscillations
and thus no corresponding radiation is produced.
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
23
Exercise: Pure Multipole Transition
24
The Ii = If = 0 Case – Internal Conversion
Here the only possible value for L is zero. This case is
not permitted for radiative transitions.
Find the type of transitions for
the decay from the fourth
state to the ground state in
the case of 72Se
The transition between 2 (excited) states with spin
= 0* occurs through the competitive process we
mentioned before, namely the internal conversion.
Internal Conversion
In this process, the available energy is transmitted
to an orbital electron which in turn is ejected.
Orbital electrons with wave functions penetrating
the nucleus field are concerned.
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
* A few even Z – even N nuclei have 0+ first excited states.
Those are forbidden to decay to the 0+ ground state by γ
emission.
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
6
25
26
Selection Rules and Weisskopf Estimates
Selection Rules and Weisskopf Estimates
In table 18 we showed which transitions are
possible between the 2 states Ii = 3/2 and If =
5/2. Several multipoles are permitted. For
example in the ∆π = no case, M1, E2, M3 and E4
are allowed.
To decide which are the ones we observe, we
shall make a simple
calculation using
Weisskopf estimates.
Let’s assume a medium-weight nucleus
(A=125)* and E = 1 MeV.
According
to
Weisskopf,
the
transition
probabilities are as follows:
σL)/λ
λ(M1)
λ(σ
σL, A=125, E=1 MeV) λ(σ
λ(M1) = 5.6×
×1013 s-1
λ(E2) = 1.83×
×109 s-1
λ(M3) =
1.00×
×104 s-1
λ(E4) = 1.72 s-1
1
1.4×
×10-3
≅2.1×
×10-10
1.3×
×10-13
∆π = no
L Transition
1
M1
2
E2
3
M3
4
E4
* A2/3 = 25, A4/3 = 625
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
27
Selection Rules and Weisskopf Estimates
28
Expectations based on the single-particle Estimates
The previous results show that the lower orders
are dominant. They also indicate that this
transition could be composed of M1 radiation with
possibly a small mixture of E2.
1) The lowest permitted multipoles usually dominate
For the ∆π = yes case, the calculations show that the E1 is
dominant and the other modes are most likely not to occur!
3) Emission of multipole L+1 is less probable than
emission of multipole L by a factor of the order of 10-5
4) Points 2 and 3 combined give the following relations
(L’ = L +1)
σL)/λ
λ(E1)
λ(σ
σL, A=125,E=1 MeV) λ(σ
16
-1
1
λ(E1) = 1.25×
×10 s
2.3×
×10-7
λ(M2) = 4.375×
×109 s-1
λ(E3) = 5.31×
×104 s-1
λ(M4) = 0.70
s-1
≅
2.1×
×10-10
2.1×
×10-17
∆π = yes
Transition
1
E1
L
2
3
4
M2
E3
M4
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
2) Electric multipole emission is 2 orders of magnitude
more probable than the same magnetic multipole
emission.
λ ( EL ') λ ( EL ') λ ( EL )
=
×
= 10 − 5 × 10 2 = 10 − 3
λ ( ML ) λ ( EL ) λ ( ML )
λ ( ML ') λ ( ML ') λ ( ML )
=
×
= 10 − 5 × 10 − 2 = 10 − 7
λ ( EL )
λ ( ML ) λ ( EL )
13
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
7
29
Important Remark
The precedent calculations are based on a
single
particle
model
with
simple
approximations. We observe in the lab.
transitions in which λ(E1) > λ(M1) especially in
transitions between vibrational and rotational
collective states.
© Dr. Nidal M. Ershaidat - Nuclear Instrumnentation - Chapter 1: Radiation Sources - Suppl. 3 Gamma Decay
8