alpha decay-1-upload.pdf

Nuclear Physics
Alpha decay

1. Nuclear and Particle Physics
Radioactive decay
Change The Narrative withDhruvita
A Focus for Future | Email: [email protected]

Alpha Decay
1

2. 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

3. 
-
emitter−
+→ epn

+
emitters/EC+
+→ enp
3
N=Z
N>Z
N/Z Curve
https://www.nndc.bnl.gov/

4. 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

5. Basics of Alpha –decay processes
Electron
Proton
Neutron2222
N
cc.)cX( cZmNmmZm
enp
A
Z
++= 2222
N
cc.)cX( cZmNmmZm
enp
A
Z
++ 2
.cm
Mass defect
B. E.
Expected
Found)X(
N
A
Z
m
Energy
5
E=mc
2
Facts about Alpha particle
Mass (m
)=4.00300 u 0r 3.7 GeV/c
2
Charge =+2e
Spin and parity=1
+
Binding Energy(E. B. )=28.7 MeV
B. E. 2
N
222
)cX()cc.(
A
Zenp
mcZmNmmZ −++
B. E.=2
2
N
A
Zatomic
2
n
21
1atomic
)cX(mcNmH)c(mZ(B.E.) −+=
Alpha particle : is a nucleus consists two
protons and two neutrons
Alpha particle

6. 6
Decay of Alpha Particle:-)X(
NucleusParent
N
A
Z )Y(
NucleusDaughter
2-N
4A
2Z
−
− He)(
4
2 Energy(Q)
Can be written as ;NucleusParent
)X(
N
A
Z NucleusDaughter
Q(energy) )Y(
2-N
4A
2Z
++
−
− 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 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 +→=  RaThfor years10 2.03
228
88
232
90
10 αPbPo onds for . +→= 
− 208
82
212
84
7
sec1034
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. (E
B=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-value2
2
]cParticle Alpha of MassNucleusdauchter of MassNucleusParent of [Mass
)]4,2()4,2(),([
−−=
−−−−=


Q
or
cMAZMAZMQ
The available energy Q
α goes into the kinetic energies of the α-
particle and of the recoil of the daughter nucleus.
If Q
α > 0, α-decay is energetically possible

9. Alpha decay ………………….Examining B.E
9),()4,2()4,2( AZBBAZBQ −+−−=
 He)4,2(),(
4
2
+−−→ AZAZ
Thus α-decay is energetically allowed if (Q
>0)





+=
−−−
A
B
A
AB
A
A
B
AZBAZBB
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
AA
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 = 151Average binding Energy per nucleon
B
ave
=BE/A (MeV/A)
B. E. curve 





−=
−
A
A
B
3
107.743.28
which is A
A
B
3
107.7075.7
−
+=
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
AB
A
A
B
AZBAZBB
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
× 10
9
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 Natural Abundance Mean life τ
144
Nd
84 23.8% 1.04×10
16
years
147
Sm
125 15.1% 2.74×10
11
years
190
Pt
112 0.0127% 8.51×10
11
years
192
Pt
114 0.78% ≈ 10
15
years
209
Bi
126 100% 3×10
17
years
232
Th
142 100% 2×10
10
years
238
U
146 99.2739% 6.3×10
9
years
7 α-emitters of naturally occurring nuclides.
The age of the earth is ~ 4×10
9
years.
Alpha decay ……………….Observations
Book Used : Krane and Besier

13. Alpha decay ………………….Examining B.E (clear view)
(4)





−=
−
A
A
B
3
107.743.28
which is A
A
B
3
107.7075.7
−
+=
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
AB
A
A
B
AZBAZBB
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 ………………….Examining B.E (clear view)
(4)





−=
−
A
A
B
3
107.743.28
which is A
A
B
3
107.7075.7
−
+=
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
AB
A
A
B
AZBAZBB
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