WELL COME

•Guide By•GYANRAO DHOTE

•PRESENTED BY•PINKI YADAV

ALPHA PARTICAL EMISSION

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.

Alpha decay ()• Decay by the emission of doubly

charged helium nuclei 4He2+.• 238U234Th + 4He• Z = -2, N=-2, A=-4• All nuclei with Z≥83 decay by -

decay as do some rare earth nuclei.Alpha decay is also known in the 100Sn region.

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.

Important Features of Alpha Decay

• Generally energy of decay increases with increasing Z, but in any case the energy of the emitted -particle is less than the Coulomb barrier for the -nucleus interaction.

• For e-e nuclei, decay leads to gs of daughter. For odd A nuclei, decay is not to the gs but a low-lying excited state.

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

AMp

M

p

AM

pQ

41

48

2

8

2

2

2

Dm

mQ

A

AQT

14

BEFOREAFTER

-p, P2/2AM

+p, p2/8M

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:

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

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

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

Thank You