Nov 2006, Lectures 9 Nuclear Physics Lectures, Dr. Armin Reichold 1 Lectures 8 9 decay theory.

Nov 2006, Lectures &9 Nuclear Physics Lectures, Dr. Armin Reichold 1

Lectures 8 & 9 decay theory

Nov 2006, Lectures &9 Nuclear Physics Lectures, Dr. Armin Reichold

2

8.0 Overview 8.1 QM tunnelling and

decays 8.2 Fermi theory of decay

and electron capture 8.3 The Cowan and Reines

Experiment 8.4 The Wu experiment 8.5 decays (very brief)


3

Decay Theory Previously looked at kinematics and energetics

now study the dynamics i.e. the interesting bit.

Will need this to calculate life times Will get to understand variations in lifetimes


4

8.1 Decay Theory Consider 232Th, Z=90, with radius of R=7.6 fm It alpha decays with Ea=4.08 MeV at r= But at R=7.6 fm the potential energy of the alpha

would be E,pot=34 MeV if we believe:

Question: How does the escape from the Th nucleus?

Answer: by QM tunnelling

2

1 204

e cE Z Zc R

which we really should!


5

r

nucleus inside barrier (negative KE)

small flux of real α

8.1 Decay Theory

I II III

potential energy of

total energy of

Exponential decay of

radial wave function in alpha decay in 3 regions oscillatory oscillatory

r=t

r=R •see also Williams, p.85 to 89


6

8.1 QM Tunnelling through a square well (the easy bit)

Boundary condition for and d/dx at r=0 and r=t give 4 equations for times such that Kt>>1 and approximating k≈K we get

transmission probability: T=|D|2~exp(-2Kt) [Williams, p.85]

exp( ) exp( )

exp( ) exp( )

exp( )

I

II

III

ikr A ikr

B Kr C Kr

D ikr

0

2

2 ( )kin

kin

k mE

K m V E

in regions I and III

in region II

unit incoming oscillatory wavereflected wave of amplitude A

two exponential decaying waves of amplitude B and C

transmitted oscillatory wave of amplitude D

Wave vector Ansatz:

Stationary Wavefunction Ansatz:

Etot

Potential :V

r=0 r=t

V=V0

I II IIIrV=0

4 unknowns !


7

8.1 -decay

Neutrons

Protons

Alphas

Ebind(42)=28.3 MeV > 4*6MeV

Esep≈6MeV per nucleon for heavy nuclei


8

8.1 Tunnelling in -decay Assume there is no recoil in the remnant nucleus Assume we can approximate the Coulomb potential by sequence of

many square wells of thickness r with variable height Vi

Transmission probability is then product of many T factors where the K inside T is a function of the potential:

The region between R and Rexit is defined via: V(r)>Ekin

Inserting K into the above gives:

We call G the Gamov factor

22 ( )2

00

lim

Rexit

iR

k r drN K r

trans ri

P T e e

2 2exp 2 ( ( ) ) exp( )exitR

kinR

T m V r E dr G


9

2 21 2 1 2

0 0

( )4 4kin

exit

Z Z e Z Z eV r Er R

1/ 21/ 22

2

0

2

0

2 1/ 1/2

cos 2 cos sin0

exitR

exitR

exit exit

exit

mZ eG r R dr

r R dr R dr R r R

substitutingand and

0 0

1/ 221/ 2 2 22

0 0 00 0 0

4 sin sin (1/ 2) sin cos2 exit

mZ eG R d d

where

8.1 Tunnelling in -decay Use the Coulomb potential for an a particle of charge Z1 and a nucleus of

charge Z2 for V(r)the latter defines the relation between the exit radius and the alpha particles kinetic energy

2exp( ) exp 2 ( ( ) )exitR

kinR

G m V r E dr

inserted into: and Z1=2 gives


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8.1 Tunnelling in -decay How can we simplify this ?

for nuclei that actually do a-decay we know typical decay energies and sizes

Rtyp≈10 fm, Etyp ≈ 5 MeV, Ztyp ≈ 80 Rexit,typ ≈ 60 fm >>Rtyp

since

Inserting all this into G gives:

And further expressing Rexit via Ekin gives:

1/ 22

04exitmZe R

G

0 0 0cos / cos 0 / 2exitR R

1/ 22 2 2

0 0

24 4 2exit

kin

Ze e mZR GE E


11

8.1 -decay Rates How can we turn the tunnelling probability into a decay rate? We need to estimate the “number of hits” that an makes onto

the inside surface of a nucleus. Assume:

the a already exists in the nucleus it has a velocity v0=(2Ekin/m)1/2

it will cross the nucleus in t=2R/v0

it will hit the surface with a rate of 0=v0/2R Decay rate is then “rate of hits” x tunnelling probability

Note: 0 is a very rough plausibility estimate! Williams tells you how to do it better but he can’t do it either!

1/ 22 2

0

(2 / ) (2 / )exp( ) exp

2 2 4 2kin kin

kin

E m E m e mZGR R E


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8.1 -decay experimental tests

Predict exponential decay rate proportional to (Ekin)1/2

Agrees approximately with data for even-even nuclei. But angular momentum effects complicate the picture:

Additional angular momentum barrier (as in atomic physics)

El is small compared to ECoulomb E.g. l=1, R=15 fm El~0.05 MeV compared to Z=90 Ecoulomb~17 MeV. but still generates noticeable extra exponential suppression.

Spin (J) and parity (P) change from parent to daughterJ=L P=(-1)L

2

2 2

( 1)( )2l

l l cEmc r


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8.1 -decay experimental

tests We expect:

ln(d

ecay

rate

)

.,,

,/ /red daughter

daughter daughter kintot kin

mZ Z EE m

1/ 22 2

0

(2 / )exp

2 4 2kin

kin

E m e mZR E


14

8.2 Fermi Decay Theory Consider simplest case:

of -decay, i.e. n decay At quark level: du+W

followed by decay of virtual W to electron + anti-neutrino

this section is close to Cottingham & Greenwood p.166 - ff

but also check that you understand Williams p. 292 - ff

(782 )e

e

n pe Q keV

d ue

or at quark level

W-

e-

( ) e

d

u u d

u d

np


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8.2 Fermi Theory 4 point interaction

Energy of virtual W << mW life time is negligible assume interaction is described by only a single number we call this number the Fermi constant of beta decay G also assume that p is heavy and does not recoil (it is often

bound into an even heavier nucleus for other -decays) We ignore parity non-conservation

* * * 3( ) ( ) ( ) ( )fi e p nH r r r G r d r 1/ 2 1/ 2( ) exp( . ) ; ( ) exp( . ) ;e e er V ik r r V ik r q k k

* 3

~ 1 / ~ 5 ~ 1/ 40 exp( . ) 1

0

( ) ( )

n p

fi p n f

q MeV c R fm q riq r

L R L L L

H G r r d r G H

From nuclear observations we know :

which is only applicable for as otherwise can be larger

which is just a numbe

n pr since and are at rest

e- ( ) e

d u u d

u d

np

Nov 2006, Lectures &9 16

8.2 Fermi Theory

2 3 2 34 / 4 /e e e edn p dp h dn p dp h n n n ; ;

2 2 3 2 34 / 4 /e ed n p dp h p dp h

0

/ ( ) / ; / 1/

f

f e

f e f

EE

E E E

p E c E E c p E c

= total energy released in the decay == total energy of the final state

= mass deficit= total kinetic energy + rest masses of the final state

as we neglect nuclear recoil energy electron energy distribution is determined by density of states

but pe and p or Ee and E are correlated to conserve energy we can not leave them both variable

2 22 2

6 3

16 ( )e f ef e

d n p E EdE dp h c


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8.2 Fermi Theory Kurie Plot

2 2

22

( ) ( )

( ) ( )

e e f ee

ef e

e

dR I p Ap E Edp

I p A E Ep

FGR to get a decay rate and insert previous results:2

22

22 2 26 3

42 2

7 3

2

2

2 16 ( )

64 ( )

fif

fie f e

fi e f e

f e f e

dnR HdE

dR d nHdP dE dP

H p E Eh c

G H p E Eh c

* 3( ) ( )fi p n fH G r r d r G H

2 22 2

6 3

16 ( )e f ef e

d n p E EdE dp h c

A

let’s plot that from real data


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8.2 Electron Spectrum Observe electron kinetic energy

spectrum in tritium decay Implant tritium directly into a biased

silicon detector Observe internal ionisation (electron

hole pairs) generated from the emerging electron as current pulse in the detector

number of pairs proportional to electron energy

Observe continuous spectrum neutrino has to carrie the rest of the energy

End point of this spectrum is function of neutrino mass

But this form of spectrum is bad for determining the endpoint accurately

Ekin,e (keV)

Rel

ativ

e In

tens

ity

Simple Spectrum


19

8.2 Kurie Plot A plot of: should be linear

…but it does not! Why? …because that’s off syllabus! But if you really must know … Electron notices Coulomb field of nucleus e gets enhanced near to proton (nucleus) The lower Ee the bigger this effect We compensate with a “Fudge Factor”

scientifically aka “Fermi Function” K(Z,pe) Can be calculated but we don’t have means

to do so We can’t integrate I(pe) to give a total rate

2

( )ee

e

I p Ep

vs.

(I(p

)/p2 K

(Z,p

))1/

2

Ekin,e (keV)

Kurie-Plot


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8.2 Selection Rules Fermi Transitions:

ecouple to give spin Se=0 “Allowed transitions” Le=0 Jnp=0.

Gamow-Teller transitions: ecouple to give spin Se=1 “Allowed transitions” Le=0 Jnp=0 or ±1

“Forbidden” transitions

See arguments on slide 15 Higher order terms correspond to non-zero L.

Therefore suppressed depending on (q.r)2L

Usual QM rules give: Jnp=Le+Se

...).().(1).exp( 2 rqOrqirqi


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8.2 Electron Capture

capture atomic electron Can compete with + decay. Use FGR again and first look at matrix element

* * 3

0

( ) ( ) ( ) ( )R

fi e p nH G r r r r d r

* 3

0

(0) (0) ( ) ( )R

fi e p nH G r r d r

3/ 221/ 2

20

exp( . ) 1(0) ; ( ) (0)4

ee

Zm e ik rrV V

32 22 2

204e

fi f

G Zm eH HV

* 3

0

( ) ( )R

F n pH r r d r

ee p n

For “allowed” transitions we consider e and const.

Only le=0 has non vanishing e(r=0) and for ne=1 this is largest.


22

8.2 Electron Capture Density of states easier now

only a 2-body final state (n,) n is assumed approximately stationary only matters final state energy = E

2

3

2

3

4 ; ;

4

dN dN dN dqq V E q cdq h dE dq dE

dN q VdE h c

2

32 2 22 2

4 3 20

2

164

ffi

f

eF

dNH

dE

E Zm eG Mh c

apply Fermi’s Golden Rule AGAIN:


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8.3 Anti-neutrino Discovery Inverse Beta Decay

Assume again no recoil on n But have to treat positron fully relativistic Same matrix elements as -decay

because all wave functions assume to be plane waves

Fermi’s Golden Rule (only positron moves in final state!)

2 222 2e efi F

e e

dN dNH G H

dE dE

;e en pe p ne - decay : inverse - decay :

2 22 2fi FH G H V


24

8.3 Anti-neutrino Discovery Phase space factor:

Neglect neutron recoil:

Combine with FGR22

3 2

42 e eF

p EG Hh c V

2

3

4e e

e e

dN dpp VdE h dE

2 2 2 2 4 2; /ee e e e

e

dpE p c m c E pc

dE

/ ;F c V R F

322

4 3

16 e eF

p EG H

h c


25

8.3 The Cowan & Reines Experiment

for inverse -decay @ E ~ 1MeV ~10-47 cm2

Pauli’s prediction verified by Cowan and Reines.

1 GW Nuclear Reactor

PMT

H20+CdCl2

Liquid Scint.

2e p ne

e en Cd several

(on protons from the water)

(prompt : shortly after inverse beta decay) (9MeV,delayed coincidence)

Shielding

original proposalwanted to use abomb instead!

Liquid Scint. PMT

-beam

all this well under groundto reduce cosmic rays!

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8.4 Parity Definitions

Parity transforms from a left to a right handed co-ordinate system and vice versa

Eigenvalues of parity are +/- 1. If parity is conserved: [H,P]=0 eigenstates of H are

eigenstates of parity all observables have a defined parity If Parity is conserved all result of an experiment should be

unchanged by parity operation If parity is violated we can measure observables with mixed

parity, i.e. not eigenstates of parity best read Bowler, Nuclear Physics, chapter 2.3 on parity!

2

1 2

1

; [ ( )] ( )

[ ( )] ( ) 1

( ) ;

( )

. ( )

.

i

r r P r r

P r r P Eignevalues

v P v v

L r x p P L L

s v v P s s

O v L

let be a true vector: let be an axial vector: let be a true scalar: let be a pseudo scal

( )P O Oar:


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Parity Conservation If parity is conserved for reaction a + b c + d.

Absolute parity of states that can be singly produced from vacuum (e.g. photons = -1) can be defined wrt. vacuum

For other particles we can define relative parity. e.g. arbitrarily define p=+1, n=+1 then we can determine parity of other nuclei wrt. this definition

parity of anti-particle is opposite particle’s parity Parity is a hermitian operator as it has real eigenvalues! If parity is conserved <pseudo-scalar>=0 (see next

transparency). Nuclei are Eigenstates of parity

( 1) ( 1) finalinitial LLa b c d

x x

where are intrinsic parities of particle


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* 3 * 2 3p p pO O d r P O d r

* 3p pO PO P d r

2 * 3( )p p pO O d r * 3

p pO O d r

Parity Conservation Let Op be an observable pseudo scalar operator, i.e. [H, Op]=0 Let parity be conserved [H, P]=0 [P, Op]=0 Let be Eigenfunctions of P and H with intrinsic parity p

<Op> = - <Op> = 0 QED it is often useful to think of parity violation as a non

vanishing expectation value of a pseudo scalar operator

insert Unity

as POp=-OpP since [P, Op]=0

use E.V. of under parity


29

Q: Is Parity Conserved In Nature?

A1: Yes for all electromagnetic and strong interactions.

Feynman lost his 100$ bet that parity was conserved everywhere. In 1956 that was a lot of money!

A2: Big surprise was that parity is violated in weak interactions.

How was this found out? can’t find this by just looking at nuclei. They are parity

eigenstates (defined via their nuclear and EM interactions) must look at properties of leptons in beta decay which are

born in the weak interaction see Bowler, Nuclear Physics, chapter 3.13


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Mme. Wu’s “Cool” Experiment

Adiabatic demagnetisation to get T ~ 10 mK Align spins of 60Co with magnetic field. Measure angular distribution of electrons and

photons relative to B field. Clear forward-backward asymmetry of the electron

direction (forward=direction of B) Parity violation. Note:

Spin S= axial vector Magnetic field B = axial vector Momentum p = real vector Parity will only flip p not B and S

60 60 * 60 * 60( 5) ( 4) ;eCo J Ni J e Ni Ni

5+

0+

4+

2+- allowedGamov Teller decay J=1

2.51 MeV1.33 MeV

0 MeV

ExcitationEnergy

60Ni60Co

~100%

Nov 2006, Lectures &9 31

The Wu Experiment

’s from late cascadedecays of Ni* measure degree of polarisationof Ni* and thus of Cogamma det. signals

summed over bothB orientations!

scintillator signal

electron signal showsasymmetry of theelectron distribution

see also Burcham & Jobes, P.370

sample warms up asymmetry disappears


32

Interpreting the Wu Experiment

Let’s make an observable pseudo scalar Op: Op=JCo * pe = Polarisation (axial vector dot real

vector) If parity were conserved this would have a

vanishing expectation value But we see that pe prefers to be anti-

parallel to B and thus to JCo Thus: parity is violated


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Improved Wu-Experiment Polar diagram of angular

dependence of electron intensity

is angle of electron momentum wrt spin of 60Co or B

using many detectors at many angles

points indicate measurements

if P conserved this would have been a circle centred on the origin


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8.5 decays When do they occur?

Nuclei have excited states similar to atoms. Don’t worry about details E,JP (need a proper shell model to understand).

EM interaction less strong then the strong (nuclear) interaction Low energy excited states E<6 MeV above ground state can’t

usually decay by nuclear interaction -decays -decays important in cascade decays following and

decays. Practical consequences

Fission. Significant energy released in decays (see later lectures)

Radiotherapy: from Co60 decays Medical imaging eg Tc (see next slide)


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Energy Levels for Mo and Tc Make Mo-99 in an accelerator attach it to a bio-compatible molecule inject that into a patient and observe where the patient emits -rays don’t need to “eat” the detector as ’s penetrate the body call this substance a tracer

both decay leaves Tc in excited state.

MeV

MeV

interesting metastable state

Nov 2006, Lectures 9 Nuclear Physics Lectures, Dr. Armin Reichold 1 Lectures 8 9 decay theory.

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Transcript of Nov 2006, Lectures 9 Nuclear Physics Lectures, Dr. Armin Reichold 1 Lectures 8 9 decay theory.