Transcript alpha decay I - Department of Physics, HKU

Alpha Decay
basics
[Sec. 7.1/7.2/8.2/8.3 Dunlap]
Alpha decay Example
Parent nucleus Cm-244.
The daughter isotope is Pu-240
244
Cm
96
240
94Pu
Why alpha particle instead of other light nuclei
Energy Q associated with the emission of various particles from a 235U nucleus.
There are always two questions that
can be asked about any decay in
atomic, nuclear or particle physics: (i)
How much kinetic energy was
released? and (ii) How quickly did it
happen? (i.e. Energy? and Time?).
Lets look at both of these questions for
 decay.
Energy Released Q Experiments
The above diagram (right) shows the experimental
energy of release. The above diagram (left) shows the
abundance of alpha emitters. Both diagrams are as a
function of A. Can you see the relationship?
The Energy of the α-particle, Tα
A
Z
Mass of X
X
Q
Mass of Y
+  particle
A
Z
A 4
Z 2
X  Y  He  Q
4
2
A 4
Z 2
Y
And the energy released in the
decay is simply given by energy
    Y  M  Hec
Q  M X  M
A
Z
A 4
Z 2
4
2
2
The Energy of the α-particle, Tα
Conserving energy and momentum one finds:
2
2
p2  4
A 4


Q 


 1  T 


8M  A 
2 AM 8M
 A 
p
p
BEFORE
-p,
P2/2AM
+p,
p2/8M
AFTER
 A 
T  Q 

 A4
Q
1
m

m
D
Energy Released Q.
This can be estimated from the SEMF by realizing that the B(Z,A) curve is
rather smooth at large Z, and A and differential calculus can be used to
calculate the B due to a change of 2 in Z and a change of 4 in A. Starting from
(8.2) we also have:
 


Q  B 4 He  B( NAX N )  B( ZA42YN 2 )


Q  B 24 He  2
B  aV A  a S A
2/3
B
B
B
B
4
 28.3MeV  2
4
Z
A
Z
A
 aC
Z2
A1 / 3
( A  2Z ) 2
Z2
Z2
2/3
 aA
 aV A  a S A  a C 1 / 3  a A A  4a A Z  4a A
A
A
A
B
Z
Z
 2a C 1 / 3  4a A  8a A
Z
A
A
B
2
1
 aV  a S A1 / 3  a C
A
3
3
8
Q  28.3MeV  4aV  a S
3
Z2
4/3
 a a  4a A
Z2
A
A2
2
1
Z 
Z
Z



4
a
1


4
a
1

2
C
A
1/ 3
1/ 3 

3
A
A 
A
A 


There can be multiple alpha energies
This diagram shows the
alpha decay to the 240Pu
daughter nucleus – and this
nucleus is PROLATE and
able to ROTATE collectively.
Alpha decay can occur to
any one of the excited
states although not with the
same probability.
For each decay:
Q  Q 0  E
where E is the excited
state energy
Total angular momentum and parity need be conserved
Total angular momentum and parity need be conserved
5/2-
243Am
239Np
9/2-
0.172 MeV
7/2-
0.118 MeV
5/2-
0.075 MeV
7/2+
0.031 MeV
5/2+
0 MeV
How fast did it happen?
The mean life (often called just “the lifetime”) is defined simply as 1/ λ. That is the
time required to decay to 1/e of the original population. We get:
The first Decay Rate Experiments The Geiger –Nuttal Law
The first Decay Rate Experiments The Geiger –Nuttal Law
As early as 1907,
Rutherford and coworkers
had discovered that the particles emitted from shortlived isotopes were more
penetrating (i.e. had more
energy). By 1912 his
coworkers Geiger and
Nuttal had established the
connection between particle
range R and emitter halflife . It was of the form:
The first Decay Rate Experiments The Geiger –Nuttal Law
The one-body model of α-decay assumes that the
α-particle is preformed in the nucleus, and confined
to the nuclear interior by the Coulomb potential
barrier. In the classical picture, if the kinetic energy
of the -particle is less than the potential energy
represented by the barrier height, the α-particle
cannot leave the nucleus.
vα
Qα
r=R
 
  
 2R


 PT


r=b
In the quantum-mechanical picture, however, there is
a finite probability that the -particle will tunnel
through the barrier and leave the nucleus.
The α-decay constant is then a product of the
frequency of collisions with the barrier, or ``knocking
frequency'‘ (vα/2R), and the barrier penetration
probability PT.
How high and wide the barrier?
2Ze 2
1
V (r ) 
 2Z .c.
(4 0 )r
r
The height of the barrier is:
Emax
2 Z . .c

R
The width of the barrier is
30MeV
w
Lets calculate these for
Emax
w  bR 
235
92
U
taking R0=1.2F, we have
2 x 92 x197 MeV .F

 36 MeV
137 x 7.4 F
2Z . . c
R
Q
R  1.2x(235)1/ 3  7.4F
2x92 x197 MeV .F
w
 7.4 F  49 F
137 x 4.68 MeV