Alpha Decaybasics

[Sec. 7.1/7.2/8.2/8.3 Dunlap]

Alpha decay Example

Parent nucleus Cm-244. The daughter isotope is Pu-240

96Cm244

94Pu240

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

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

The Energy of the α-particle, Tα

Conserving energy and momentum one finds:

A

AT

AM

p

M

p

AM

pQ

41

4

8

2

8

2

2

2

Dm

m

Q

A

AQT

14

BEFORE

AFTER-p, P2/2AM

+p, p2/8M

Energy Released Q.

)()( 242

4

N

AZN

AN YBXBHeBQ

A

B

Z

BMeV

A

B

Z

BHeBQ

423.284242

A

ZaZaAa

A

ZaAaAa

A

ZAa

A

ZaAaAaB AAACSVACSV

2

3/1

23/2

2

3/1

23/2 44

)2(

A

Zaa

A

Za

Z

BAAC 842

3/1

2

2

3/4

23/1 4

3

1

3

2

A

Zaa

A

ZaAaa

A

BAaCSV

2

3/13/1214

314

1

3

843.28

A

Za

A

Z

A

Za

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:

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

Total angular momentum and parity need be conserved

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 MeV

0.075 MeV

0.031 MeV

0 MeV239Np

1.1%%10.6%

%

88%%

0.12%%

0.16%%

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 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:

The first Decay Rate Experiments - The Geiger –Nuttal Law

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

How high and wide the barrier?

rcZ

r

ZerV

1..2

)4(

2)(

0

2

The height of the barrier is:

R

cZE

..2max

The width of the barrier is

2 . .w b

Z cR R

Q

w

Lets calculate these for taking R0=1.2F, we have U23592 FR 4.7)235(x2.1 3/1

MeVF

FMeVE 36

4.7x137

.197x92x2max FF

MeV

FMeV494.7

68.4x137

.197x92x2w

30MeV