1

1

Nuclear Physics

(7th lecture) Content

• Gamma decay (contd.)

• Sum rules

• Measuring methods of the gamma decay constant

• Gamma-gamma angular correlation (multipolarity meas.)

• Nuclear models #1: liquid drop model

• Nuclear models #2: The Fermi-gas model

2

Gamma-decay

1) Multipole expansion of electromagnetic waves

,,

)(

,

m

llml

E

ml Yfb LkrrB )(

,

)(

,

E

ml

E

mlk

iBrE and „electric”

,,

)(

,

m

llml

M

ml Yfa LkrrE )(

,

)(

,

M

ml

M

mlk

iErB and „magnetic”

These waves have good angular momentum and parity!

Electric transitions parity change: l1

Magnetic transitions parity change: 11

l

2) Selection rules Angular momentum: fifi IIlII

Parity: 21

Summary of previous lecture:

Note: since the „intrinsic” angular momentum of a photon is 1,

0 → 0 transitions are strictly forbidden. 1l

3

The gamma-decay (contd.)

Decay constant (transition probability)

Complicated quantum-electrodynamic calculation.

Approximation: only one unit charge changes its state.

Result:

2

,

12

2!!12

18, i

E

mlf

l

E Mcll

lml

for electric transitions

Similar for magnetic transitions

Further approximations:

- The and wave functions contain spherical

harmonics, these can be integrated with the operator

→ S(Ii, If, l) „statistical” factor

- The m quantum numbers averaged (no direct observation)

- The radial part of the wave functions = constant (!)

,m

lY

E final and initial wave functions

interaction operator

f iE

mlM ,

4

The gamma-decay (contd.) Finally we get the Weisskopf units:

lIISRl

E

ll

ll fi

l

l

E ,,3

3

197!!12

1104,4 2

2

2

21

lIISRl

E

ll

ll fi

l

l

M ,,3

3

197!!12

1109,1 22

12

2

21

Since this is dependent on units, R should be in fm, E in MeV.

The results are in 1/s.

How do the transition probability change with multipolarity?

ll

ll RRhcRE

22

22

Take A = 125 nucleus, and E = 0,5 MeV

fm 104R

fm 4004

2

1025,2

R

In atomic physics and fm 108 8 10

2

10

R

fm 61252,1 3 R

Only E1 transitions occur for the atoms! Role of collisions!

2

5

The gamma-decay (contd.)

The trend of the

Weisskopf units

6

4) More complicated transitions

2

,,

,

1

21

2112

1,

mmm p

E

ml pMI

IIElB

Reduced

transition

probability

Initial and

final state

„directions”

Single particle

matrix elements

The B(El) values are usually given in Weisskopf units (W.u.).

If B(EL) ~ 1 W.u. → single particle excitations

If B(El) >> 1 W.u. → collective excitation (involving many particles)

Gamma-decay (contd.)

7

Sum rules

Excitation of the nucleus

0

The quadrupole operator: ΩQ 2

2 ˆˆˆ Yr

The possible

quadrupole excitations: 0Q̂q ?

q

Usually | q > is not an eigenfunction of the

Hamiltonian, and even not normalized!

However, it can be expanded!

If | f > is a complete normalized set of eigenfunctions then

f

ifif

f

fifi cccq ,fff

Multiply this by < fi | f

ffcq f

| q > can be normalized if Sqci

i

i

i

22f

8

Finally we have: Si

i 2

0Q̂f

Sum rules (contd.)

This is the Thomas-Kuhn sum rule.

Here can represent any multipole operator (not only quadrupole)

What does the sum rule mean physically?

22

0Q̂ii q ff ~ how strongly the < fi | nuclear state

can be excited from the ground state

with the operator

Q̂

Q̂

It can be calculated theoretically using simple assumptions!

For example for the dipole operator:

602

022

2

A

NZ

A

NZ

mc

edEED D

f

f

f [MeV∙mb]

3

9

There can be many excitations

with the same multipolarity!

10ˆ

2

S

i Qf describes the contribution of one particular

excitation to the sum rule (e.g. in %)

The sum rule describes the

total possible strength of

excitation (with any energy)

for a multipole operator

Sum rules (contd.)

10

If only one < fi | contributed to S, then this particular state would

„exhaust” 100% of the sum rule.

excitation strength

Few Weisskopf units

B(El) values

Sum rule

Single particle transitions

Giant resonances

collective transitions

Giant dipole resonance

https://inspirehep.net/record/1242152/files/DCS_TST.png

Giant multipole vibrations

Sum rules (contd.)

„breathing”

mode

11

Measuring the gamma-decay constant

The exponential decay law t

t eAeAtA

00

1) Direct measurement of life time

The method of delayed coincidences

TAC = time to amplitude converter

(also time to digital converter TDC could be used)

s 10 10

12

Measuring the gamma-decay constant

2) Using the Doppler shift s 1010 912

stopper foil

target

projectile

beam

d

Flight time: v

dt

During the flight: Doppler shift

After the stop: No Doppler shift

4

13

Measuring the gamma-decay constant

3) Doppler shift in the target s 10510 1213 target

projectile

beam Detector 1

Dete

cto

r 2

Doppler shift &broadening

Doppler

broadening

only

The mechanism of the stopping in the

target must be well known

(usually simulation)

4) From line-width E (if E > detector resolution)

14

Gamma-gamma angular correlation A method for determining

the multipolarity

The idea:

The angular distribution of different

multipolarities are different (e.g. dipole antenna)

The problem:

The nuclei in a sample are not aligned

→ observation of single gamma-rays is isotropic

The solution:

Detect two consecutive gamma-rays! The detection of the first

gamma-ray fixes a direction → observation of the second

gamma-ray will not be isotropic respective to the direction of

the first one!

The angle between the two -rays depends on their multipolarity!

15

det

ecto

r #

1

&

Coincidence experiment

Example: angular correlation result

https://wiki.umn.edu/pub/MXP/AngularCorrelationofGammaRaysfromCobalt60/graph.jpg

i

i iaW 2cos

The number and the values of the

ai coefficients are characteristic

for the multipolarities involved

Gamma-gamma angular correlation (contd.)

movable

detector

source

16

The problem:

Unlike the atom, the nucleus cannot be described „exactly”,

because…

a) … the nucleus is held together by the two-body interactions

between its constituents; there is no central source for the

nuclear potential;

b) …we don't know exactly the nucleon-nucleon interaction.

Nuclear models

The solution:

We use simplified models. The pay-off is that these models

describe the behaviour only partly.

5

17

1) The liquid drop model

Assumptions:

- Nuclear forces are attractive and short-ranged

→ the density is constant (like in a liquid)

- The nucleus is sphere-shaped (because of the surface tension)

→ the surface energy is proportional to the surface

- The nucleus is homogeneously charged

→ Coulomb term

- The constituents are fermions (Pauli principle)

→ asymmetry energy term

- Pairing energy can be empirically taken into account

2

12

31

2

32

AbA

ZNb

A

ZbAbAbB PACFV

We have treated it already. That led to the Weizsäcker-formula.

It describes well the binding energy. But only that!

18

2) The Fermi-gas model

Since the nucleons are fermions, we can try to set up a model of

fermion „gas”.

Assumption: the fermions are „closed” in a spherical potential

well, but they move „freely” inside .

What can be expected from this model?

- Describe only the kinetic energy contribution of the nucleons

- Since quantum mechanics is used, the „asymmetry” term

might be described

- Some surface effects can also be expected

The ground state:

- Since they are fermions, only one particle

in a state (Pauli exclusion principle)

- Since ground state → the lowest energy

states are filled

- p =ħk

f

f

pp

pppn

if ,0

if ,1

p pf

1

19

The number of states in the phase-space:

p

pp

d

2

dd2

2

dd2d

3

23

3

33

r

prN

Because of the spin

3

4

22d4

22dd2d

3

3

0

2

3

33 f

ppV

ppV

pnpn

f

prNN

For the number of protons and of neutrons we have

3

4

22

3

3

fppVZ

3

4

22

3

3

fnpVN

Using and we get : ArRV

3

0

3

3

4

3

4 30 ArR

A

Z

rr

A

Zp fp

3

0

2

3

0

3

3 3

3

42

2

, and

A

N

rp fn

3

0

23 3

20

To get some numerical estimation we approximate:

0

31

2

2

3

rpp fnfp

The highest kinetic energy at the Fermi-level: MeV 332

2

M

pE

f

f

Since the average binding

energy of a nucleon is ~ 7 MeV,

we get the following picture for

the nuclear potential:

Ef

-40

33

7

R E

2

1

A

N

A

Z

we get Contains only known constants!

fm 2,10 r