Nuclear and Particle Physics
Radioactive decay
Change The Narrative with Dhruvita
A Focus for Future | Email: edhruvita@gmail.com
Alpha Decay
1
1. The Atomic Nucleus -by R. D. Evans
2. Modern Physics by -Arthur Beiser
3. Introduction to Nuclear Physics – by Keneth S Krane
4. Concepts of Nuclear Physics – by B L Cohen
5. Nuclear Physics – by R R Roy & B P Nigam
6 Nuclear Physics ; Principles & Applications – by Lilley
7. Basic Ideas & Concepts in Nuclear Physics – by K Heyde
8. Introductory Nuclear Physics – by Samuel S M Wong
References:………………………..To enjoy the nucleus
2
- emitter
−
+
→ e
p
n
+ emitters/EC
+
+
→ e
n
p
3
N=Z
N>Z
N/Z Curve
https://www.nndc.bnl.gov/
4
Radioactive decay ……………….what is ?
Radioactive decay is a process in which an unstable nucleus
spontaneously loses its energy by emitting ionizing particles and/or
radiation. This decay, or loss of energy, results in a nucleus of one
type, called the parent nucleus, transforming to a nucleus of a
different type, named the daughter nucleus .
Modes of decay
1- Alpha
2- Beta(EC, - and B+ )
3- Gamma
Modes of decay?
Over N/Z plot
Basics of Alpha –decay processes
Electron
Proton
Neutron
2
2
2
2
N c
c
.
)c
X
( c
Zm
Nm
m
Z
m e
n
p
A
Z +
+
=
2
2
2
2
N c
c
.
)c
X
( c
Zm
Nm
m
Z
m e
n
p
A
Z +
+

2
.c
m

Mass defect
B. E.
Expected
Found
)
X
( N
A
Z
m
Energy
5
E=mc2
Facts about Alpha particle
Mass (m)=4.00300 u 0r 3.7 GeV/c2
Charge =+2e
Spin and parity=1+
Binding Energy(E. B. )=28.7 MeV
B. E.
2
N
2
2
2
)c
X
(
)
c
c
.
( A
Z
e
n
p m
c
Zm
Nm
m
Z −
+
+
B. E.=
2
2
N
A
Z
atomic
2
n
2
1
1
atomic )c
X
(
m
c
Nm
H)c
(
m
Z
(B.E.) −
+

=
Alpha particle : is a nucleus consists two
protons and two neutrons
Alpha particle
6
Decay of Alpha Particle:-
)
X
(
Nucleus
Parent
N
A
Z
)
Y
(
Nucleus
Daughter
2
-
N
4
A
2
Z
−
−
He)
(4
2
Energy(Q)
Can be written as ;
Nucleus
Parent
)
X
( N
A
Z
Nucleus
Daughter
Q(energy)
)
Y
( 2
-
N
4
A
2
Z +
+
−
− Alpha
Alpha decay is Radioactive decay
process in which an unstable
nucleus spontaneously loses its
energy by emitting an alpha
particle..
7
Decay of Alpha Particle:………………….Observations from N/Z and B.E. Curve
▪ A large of nuclides(A>60) occurring in
the nature are unstable and tend to
lose its excessive energies and charges
by emitting α- particles.
▪ Another observation is the mean life
time of -emitting nuclides
▪ Eg
Why so…...
Why are not decaying by another
structure like Li, Be, C etc
To understand the decay of alpha
particle from the natural occurring
nuclides . We need valid reasons. Hence,
1. We will examine the Energetics (Q-
values and B.E) and
2. Applications of quantum mechanics
for the theory

+
→

=
 Ra
Th
for
years
10
2.03 228
88
232
90
10
α
Pb
Po
onds for
. +
→

=
 − 208
82
212
84
7
sec
10
3
4
24 orders of magnitude
Dramatic variation in mean life time
Classically -decay impossible
Note: Classical mechanics can
not explain the alpha decay
8
❑ An α- particle being kicked out from the unstable parent nucleus is due to
the effect of Coulomb repulsion.
❑ A heavy nucleus with too many protons can reduce some Coulomb
repulsion energy by emitting an α- particle.
❑ An α- particle is less massive and grater binding energy. (EB=28.3 MeV) than
parent nucleus.
❑ Q-Value is the energy available for the reaction i.e. given by the following
Alpha decay ………………….Examining Q-value
2
2
]c
Particle
Alpha
of
Mass
Nucleus
dauchter
of
Mass
Nucleus
Parent
of
[Mass
)]
4
,
2
(
)
4
,
2
(
)
,
(
[
−
−
=
−
−
−
−
=


Q
or
c
M
A
Z
M
A
Z
M
Q
The available energy Qα goes into the kinetic energies of the α-
particle and of the recoil of the daughter nucleus.
If Qα > 0, α-decay is energetically possible
Alpha decay ………………….Examining B.E
9
)
,
(
)
4
,
2
(
)
4
,
2
( A
Z
B
B
A
Z
B
Q −
+
−
−
=

He
)
4
,
2
(
)
,
( 4
2
+
−
−
→ A
Z
A
Z
Thus α-decay is energetically allowed if (Q>0)






+
=

−
−
−

A
B
A
A
B
A
A
B
A
Z
B
A
Z
B
B
d
)
/
(
d
4
d
d
4
)
4
,
2
(
)
,
(
)
4
,
2
(
A
B
A
B
A
A
A
A
A
B
A
B
A
A
A
B
A
A
A
B
+






=






+






=






=
d
d
d
d
d
d
d
d
d
d
Alpha decay can also express as;
A=4 , Z=2 for Alpha
(2)
(1)
(3)
A = 151
Average
binding
Energy
per
nucleon
B
ave
=BE/A
(MeV/A)
B. E. curve







−
= −
A
A
B 3
10
7
.
7
4
3
.
28 which is A
A
B 3
10
7
.
7
075
.
7 −

+
=
Form the B. E Curve slop for
A ≈ 120, d(B/A)/dA is about
−7.7×10−3 MeV.
From mass energy difference
B(2,4), is 28.3 MeV
so the for particular A must
satisfy the following relation:






+
=

−
−
−

A
B
A
A
B
A
A
B
A
Z
B
A
Z
B
B
d
)
/
(
d
4
d
d
4
)
4
,
2
(
)
,
(
)
4
,
2
(
(5)
Above this A the inequality of equation (3) is satisfied by most nuclei and α-
decay becomes, in principle, energetically possible. In fact from A = 144 to A
= 206, 7 α-emitters are known amongst the naturally occurring nuclides.
Alpha decay ………………….Examining B.E
(4)
10
11
From A = 144 to A =206, there are 7 α-emitters of naturally occurring
nuclides. When α-emitters are found in this range of A, the energies of the
emitted α-particle are normally less than 3 MeV. It is known that the lower
the energy release the greater is the lifetime. Their existence implies
mean lifetimes comparable to or greater than the age of the earth (about 4
× 109 years). Most nuclei in this range on the line stability may be
energetically able to decay by α-emission. They do not do so at a
detectable level because the transition rate is too small.
From A = 144 to A =206
Most of the heavy nuclei to be found on earth were
probably produced in one or more supernova
explosions of early massive stars. Such explosions
can produce very heavy nuclei including trans-
uranic elements (Z > 92) and their subsequent
decay by α-emission will take them down the
periodic table in steps of ΔA = −4. Each α-decay
increases the ration N/Z until a β- decay intervenes
to restore the nucleus closer to the line of stability.
Above Z = 82 many naturally occurring α-emitters are found, many with
short lives.
Why are they to be found when their lifetime is so short?
Above Z = 82 (A > 206)
Alpha decay ……………….Observations
12
α- emitter Natural Abundance Mean life τ
144Nd84 23.8% 1.04×1016 years
147Sm125 15.1% 2.74×1011 years
190Pt112 0.0127% 8.51×1011 years
192Pt114 0.78% ≈ 1015 years
209Bi126 100% 3×1017 years
232Th142 100% 2×1010 years
238U146 99.2739% 6.3×109 years
7 α-emitters of naturally occurring nuclides.
The age of the earth is ~ 4×109 years.
Alpha decay ……………….Observations
Book Used : Krane and Besier
Alpha decay ………………….Examining B.E (clear view)
(4)







−
= −
A
A
B 3
10
7
.
7
4
3
.
28
which is
A
A
B 3
10
7
.
7
075
.
7 −

+
=
Form the B. E Curve slop for
A ≈ 150, d(B/A)/dA is about −7.7×10−3
MeV.
From mass energy difference
B(2,4), is `~ 28 MeV
so the for particular A must satisfy
the following relation:






+
=

−
−
−

A
B
A
A
B
A
A
B
A
Z
B
A
Z
B
B
d
)
/
(
d
4
d
d
4
)
4
,
2
(
)
,
(
)
4
,
2
(
(5)
Above this A the inequality of equation (3) is satisfied by most
nuclei and α-decay becomes, in principle, energetically possible.
In fact from A = 144 to A = 206, 7 α-emitters are known amongst
the naturally occurring nuclides.
13
Alpha decay ………………….Examining B.E (clear view)
(4)







−
= −
A
A
B 3
10
7
.
7
4
3
.
28 which is A
A
B 3
10
7
.
7
075
.
7 −

+
=
Form the B. E Curve slop
A ≈ 150, d(B/A)/dA is abo
−7.7×10−3 MeV.
From mass energy differ
B(2,4), is `~ 28 MeV
so the for particular A
satisfy the following rela






+
=

−
−
−

A
B
A
A
B
A
A
B
A
Z
B
A
Z
B
B
d
)
/
(
d
4
d
d
4
)
4
,
2
(
)
,
(
)
4
,
2
(
(5)
Above this A the inequality of equation (3) is satisfied by most nuclei and α-
decay becomes, in principle, energetically possible.
In fact from A = 144 to A = 206, 7 α-emitters are known amongst the naturally
occurring nuclides.
(4)
14

alpha decay-1-upload.pdf

  • 1.
    Nuclear and ParticlePhysics Radioactive decay Change The Narrative with Dhruvita A Focus for Future | Email: edhruvita@gmail.com Alpha Decay 1
  • 2.
    1. The AtomicNucleus -by R. D. Evans 2. Modern Physics by -Arthur Beiser 3. Introduction to Nuclear Physics – by Keneth S Krane 4. Concepts of Nuclear Physics – by B L Cohen 5. Nuclear Physics – by R R Roy & B P Nigam 6 Nuclear Physics ; Principles & Applications – by Lilley 7. Basic Ideas & Concepts in Nuclear Physics – by K Heyde 8. Introductory Nuclear Physics – by Samuel S M Wong References:………………………..To enjoy the nucleus 2
  • 3.
    - emitter − + → e p n +emitters/EC + + → e n p 3 N=Z N>Z N/Z Curve https://www.nndc.bnl.gov/
  • 4.
    4 Radioactive decay ……………….whatis ? Radioactive decay is a process in which an unstable nucleus spontaneously loses its energy by emitting ionizing particles and/or radiation. This decay, or loss of energy, results in a nucleus of one type, called the parent nucleus, transforming to a nucleus of a different type, named the daughter nucleus . Modes of decay 1- Alpha 2- Beta(EC, - and B+ ) 3- Gamma Modes of decay? Over N/Z plot
  • 5.
    Basics of Alpha–decay processes Electron Proton Neutron 2 2 2 2 N c c . )c X ( c Zm Nm m Z m e n p A Z + + = 2 2 2 2 N c c . )c X ( c Zm Nm m Z m e n p A Z + +  2 .c m  Mass defect B. E. Expected Found ) X ( N A Z m Energy 5 E=mc2 Facts about Alpha particle Mass (m)=4.00300 u 0r 3.7 GeV/c2 Charge =+2e Spin and parity=1+ Binding Energy(E. B. )=28.7 MeV B. E. 2 N 2 2 2 )c X ( ) c c . ( A Z e n p m c Zm Nm m Z − + + B. E.= 2 2 N A Z atomic 2 n 2 1 1 atomic )c X ( m c Nm H)c ( m Z (B.E.) − +  = Alpha particle : is a nucleus consists two protons and two neutrons Alpha particle
  • 6.
    6 Decay of AlphaParticle:- ) X ( Nucleus Parent N A Z ) Y ( Nucleus Daughter 2 - N 4 A 2 Z − − He) (4 2 Energy(Q) Can be written as ; Nucleus Parent ) X ( N A Z Nucleus Daughter Q(energy) ) Y ( 2 - N 4 A 2 Z + + − − Alpha Alpha decay is Radioactive decay process in which an unstable nucleus spontaneously loses its energy by emitting an alpha particle..
  • 7.
    7 Decay of AlphaParticle:………………….Observations from N/Z and B.E. Curve ▪ A large of nuclides(A>60) occurring in the nature are unstable and tend to lose its excessive energies and charges by emitting α- particles. ▪ Another observation is the mean life time of -emitting nuclides ▪ Eg Why so…... Why are not decaying by another structure like Li, Be, C etc To understand the decay of alpha particle from the natural occurring nuclides . We need valid reasons. Hence, 1. We will examine the Energetics (Q- values and B.E) and 2. Applications of quantum mechanics for the theory  + →  =  Ra Th for years 10 2.03 228 88 232 90 10 α Pb Po onds for . + →  =  − 208 82 212 84 7 sec 10 3 4 24 orders of magnitude Dramatic variation in mean life time Classically -decay impossible Note: Classical mechanics can not explain the alpha decay
  • 8.
    8 ❑ An α-particle being kicked out from the unstable parent nucleus is due to the effect of Coulomb repulsion. ❑ A heavy nucleus with too many protons can reduce some Coulomb repulsion energy by emitting an α- particle. ❑ An α- particle is less massive and grater binding energy. (EB=28.3 MeV) than parent nucleus. ❑ Q-Value is the energy available for the reaction i.e. given by the following Alpha decay ………………….Examining Q-value 2 2 ]c Particle Alpha of Mass Nucleus dauchter of Mass Nucleus Parent of [Mass )] 4 , 2 ( ) 4 , 2 ( ) , ( [ − − = − − − − =   Q or c M A Z M A Z M Q The available energy Qα goes into the kinetic energies of the α- particle and of the recoil of the daughter nucleus. If Qα > 0, α-decay is energetically possible
  • 9.
    Alpha decay ………………….ExaminingB.E 9 ) , ( ) 4 , 2 ( ) 4 , 2 ( A Z B B A Z B Q − + − − =  He ) 4 , 2 ( ) , ( 4 2 + − − → A Z A Z Thus α-decay is energetically allowed if (Q>0)       + =  − − −  A B A A B A A B A Z B A Z B B d ) / ( d 4 d d 4 ) 4 , 2 ( ) , ( ) 4 , 2 ( A B A B A A A A A B A B A A A B A A A B +       =       +       =       = d d d d d d d d d d Alpha decay can also express as; A=4 , Z=2 for Alpha (2) (1) (3)
  • 10.
    A = 151 Average binding Energy per nucleon B ave =BE/A (MeV/A) B.E. curve        − = − A A B 3 10 7 . 7 4 3 . 28 which is A A B 3 10 7 . 7 075 . 7 −  + = Form the B. E Curve slop for A ≈ 120, d(B/A)/dA is about −7.7×10−3 MeV. From mass energy difference B(2,4), is 28.3 MeV so the for particular A must satisfy the following relation:       + =  − − −  A B A A B A A B A Z B A Z B B d ) / ( d 4 d d 4 ) 4 , 2 ( ) , ( ) 4 , 2 ( (5) Above this A the inequality of equation (3) is satisfied by most nuclei and α- decay becomes, in principle, energetically possible. In fact from A = 144 to A = 206, 7 α-emitters are known amongst the naturally occurring nuclides. Alpha decay ………………….Examining B.E (4) 10
  • 11.
    11 From A =144 to A =206, there are 7 α-emitters of naturally occurring nuclides. When α-emitters are found in this range of A, the energies of the emitted α-particle are normally less than 3 MeV. It is known that the lower the energy release the greater is the lifetime. Their existence implies mean lifetimes comparable to or greater than the age of the earth (about 4 × 109 years). Most nuclei in this range on the line stability may be energetically able to decay by α-emission. They do not do so at a detectable level because the transition rate is too small. From A = 144 to A =206 Most of the heavy nuclei to be found on earth were probably produced in one or more supernova explosions of early massive stars. Such explosions can produce very heavy nuclei including trans- uranic elements (Z > 92) and their subsequent decay by α-emission will take them down the periodic table in steps of ΔA = −4. Each α-decay increases the ration N/Z until a β- decay intervenes to restore the nucleus closer to the line of stability. Above Z = 82 many naturally occurring α-emitters are found, many with short lives. Why are they to be found when their lifetime is so short? Above Z = 82 (A > 206) Alpha decay ……………….Observations
  • 12.
    12 α- emitter NaturalAbundance Mean life τ 144Nd84 23.8% 1.04×1016 years 147Sm125 15.1% 2.74×1011 years 190Pt112 0.0127% 8.51×1011 years 192Pt114 0.78% ≈ 1015 years 209Bi126 100% 3×1017 years 232Th142 100% 2×1010 years 238U146 99.2739% 6.3×109 years 7 α-emitters of naturally occurring nuclides. The age of the earth is ~ 4×109 years. Alpha decay ……………….Observations Book Used : Krane and Besier
  • 13.
    Alpha decay ………………….ExaminingB.E (clear view) (4)        − = − A A B 3 10 7 . 7 4 3 . 28 which is A A B 3 10 7 . 7 075 . 7 −  + = Form the B. E Curve slop for A ≈ 150, d(B/A)/dA is about −7.7×10−3 MeV. From mass energy difference B(2,4), is `~ 28 MeV so the for particular A must satisfy the following relation:       + =  − − −  A B A A B A A B A Z B A Z B B d ) / ( d 4 d d 4 ) 4 , 2 ( ) , ( ) 4 , 2 ( (5) Above this A the inequality of equation (3) is satisfied by most nuclei and α-decay becomes, in principle, energetically possible. In fact from A = 144 to A = 206, 7 α-emitters are known amongst the naturally occurring nuclides. 13
  • 14.
    Alpha decay ………………….ExaminingB.E (clear view) (4)        − = − A A B 3 10 7 . 7 4 3 . 28 which is A A B 3 10 7 . 7 075 . 7 −  + = Form the B. E Curve slop A ≈ 150, d(B/A)/dA is abo −7.7×10−3 MeV. From mass energy differ B(2,4), is `~ 28 MeV so the for particular A satisfy the following rela       + =  − − −  A B A A B A A B A Z B A Z B B d ) / ( d 4 d d 4 ) 4 , 2 ( ) , ( ) 4 , 2 ( (5) Above this A the inequality of equation (3) is satisfied by most nuclei and α- decay becomes, in principle, energetically possible. In fact from A = 144 to A = 206, 7 α-emitters are known amongst the naturally occurring nuclides. (4) 14