Alpha decay basics
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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. 96 Cm 244. 94 Pu 240. Why alpha particle instead of other light nuclei. Energy Q associated with the emission of various particles from a 235 U nucleus.

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Alpha Decay basics

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Alpha decay basics

Alpha Decaybasics

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


Alpha decay basics

Alpha decay Example

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

96Cm244

94Pu240


Alpha decay basics

Why alpha particle instead of other light nuclei

Energy Q associated with the emission of various particles from a 235U nucleus.


Alpha decay basics

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

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 particle t

The Energy of the α-particle, Tα

Mass of X

Q

Mass of Y

+  particle

And the energy released in the decay is simply given by energy


The energy of the particle t1

The Energy of the α-particle, Tα

Conserving energy and momentum one finds:

BEFORE

AFTER

+p, p2/8M

-p, P2/2AM


Energy released q

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:


There can be multiple alpha energies

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:

where E is the excited state energy


Total angular momentum and parity need be conserved

Total angular momentum and parity need be conserved


Total angular momentum and parity need be conserved1

243Am

5/2-

Total angular momentum and parity need be conserved

1.1%%

10.6%%

88%%

9/2-

0.172 MeV

0.12%%

0.16%%

7/2-

0.118 MeV

5/2-

0.075 MeV

7/2+

0.031 MeV

5/2+

0 MeV

239Np


How fast did it happen

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


The first decay rate experiments the geiger nuttal law1

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 law2

The first Decay Rate Experiments - The Geiger –Nuttal Law


Alpha decay basics

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.

r=R

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.


Alpha decay basics

How high and wide the barrier?

The height of the barrier is:

The width of the barrier is

w

30MeV

Lets calculate these for taking R0=1.2F, we have


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