Alpha Decay basics
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Transcript of Alpha Decay basics
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Alpha Decaybasics
[Sec. 7.1/7.2/8.2/8.3 Dunlap]
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Alpha decay Example
Parent nucleus Cm-244. The daughter isotope is Pu-240
96Cm244
94Pu240
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Why alpha particle instead of other light nuclei
Energy Q associated with the emission of various particles from a 235U nucleus.
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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.
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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?
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The Energy of the α-particle, Tα
Mass of X
Mass of Y+ particle YA
Z42
XAZ
Q
QHeYX AZ
AZ
42
42
And the energy released in the decay is simply given by energy
242
42 cHeMYMXMQ A
ZAZ
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The Energy of the α-particle, Tα
Conserving energy and momentum one finds:
A
AT
AMp
M
p
AM
pQ
41
48
2
8
2
2
2
Dm
mQ
A
AQT
14
BEFOREAFTER
-p, P2/2AM
+p, p2/8M
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Energy Released Q.
)()( 242
4
N
AZN
AN YBXBHeBQ
AB
ZBMeV
AB
ZBHeBQ
423.284242
AZaZaAa
AZaAaAa
AZAa
AZaAaAaB AAACSVACSV
2
3/1
23/2
2
3/1
23/2 44)2(
AZaa
AZa
ZB
AAC 842 3/1
2
2
3/4
23/1 4
31
32
AZaa
AZaAaa
AB
AaCSV
2
3/13/1 2143
1413843.28
AZa
AZ
AZa
AaaMeVQ ACSV
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:
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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:
EQQ 0
where E is the excited state energy
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Total angular momentum and parity need be conserved
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Total angular momentum and parity need be conserved
243Am5/2-
9/2-
7/2-
5/2-
7/2+
5/2+
0.172 MeV
0.118 MeV0.075 MeV
0.031 MeV
0 MeV239Np
1.1%%10.6%
%
88%%
0.12%%
0.16%%
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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:
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The first Decay Rate Experiments - The Geiger –Nuttal Law
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The first Decay Rate Experiments - The Geiger –Nuttal Law
As early as 1907, Rutherford and coworkers had discovered that the -particles emitted from short-lived 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 half-life . It was of the form:
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The first Decay Rate Experiments - The Geiger –Nuttal Law
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TPR
2
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.
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.
vα
r=br=R
Qα
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How high and wide the barrier?
rcZ
rZerV 1..2
)4(2)(
0
2
The height of the barrier is:
RcZE ..2
max
The width of the barrier is
2 . .w b Z cR RQ
w
Lets calculate these for taking R0=1.2F, we have U23592 FR 4.7)235(x2.1 3/1
MeVF
FMeVE 364.7x137
.197x92x2max FF
MeVFMeV 494.7
68.4x137.197x92x2w
30MeV