Semi-Empirical Mass Formula

Applications - I

[Sec. 4.2 Dunlap]

Pairing Energy

Deuteron Triton - particle

The saturated sub-unit in the nucleus consists of 2 protons and 2 neutrons. This suggests, in conjunction with the Pauli Exclusion Principle (PEP) and in analogy with electronic shells in an atom, that the basic quantum state – an S-(J=0) state – consists of 2 protons and 2 neutrons with antiparallel spins as shown.

5He does not exist as a bound state (this state breaks up ~10-21s. The PEP allows us to put at most 2p and 2n in a relative S-state. Any additional nucleon must go into a higher spatial quantum state.

Because the -particle and not the deuteron (2H) is the saturated sub-unit, this shows that the force between nucleons is attractive in both the singlet () and triplet () states.

Pairing Energy

FUSIONFISSION

(Scale not linear)

Note that extra strong binding occurs for:

B/A

(MeV)

Mg ,Ne ,O ,N ,CBe, He, 2412

2010

168

147

126

84

42

Pairing Energy

),,( r

Why does the pairing energy “drop off” as A-3/4. This is something to do with the fact that nucleons do move on trajectories around the nuclear volume and do interact with other nucleons. The larger the nucleus the less the effect of the nucleon-nucleon interaction within the alpha sub-unit. A deeper understanding of the pairing energy will come when we study the SHELL MODEL.

Mass Parabolas

4/3

2

3/1

23/2 1)2(

),(A

aA

ZAa

A

ZaAaAaZAB PACSV

Let’s remind ourselves on the full form of the SEMF. The mass M(A,Z) of the nucleus is given by:

ZABc

mmZmZAZAX epnAZ ,

1)()(),(MM

2

XAZ

Cf. Eq. 4.12

or as one equation:

24/32

2

23/1

2

2

3/2

2

)2(

)()(),(M

cA

a

Ac

ZAa

cA

Za

c

Aa

c

AammZmZAZA

PA

CSVepn

Collecting together powers of Z, it is seen that this expression is quadratic in Z

Mass Parabolas

2

2

23/1

2

2

3/2

2

)2()(),(M

Ac

ZAa

cA

Za

c

Aa

c

AaZmmZAZA ACSV

Hn

Consider the case of odd A when aP=0

collecting terms

223/12

2

23/122

4

4

.),(M

ZcA

a

Ac

a

Zmmc

a

AcA

a

c

a

c

amZA

CA

HnA

SAVn

. )(M),(M 2A ZZAZZA

4

4-

23/12

2

23/122

cA

a

Ac

a

mmc

acA

a

c

a

c

am

CA

HnA

SAVn

with

we have a mass parabola !

M(A,Z)

Z

Mass Parabolas

Neutron number

Proton number Z=N

Z increasing

ISOBARS A=Z+N=const.

Mass Parabolas – Odd AFig 4.3 Mass parabola for A=135 showing One Stable Nuclide with Z=56

We can find from the SEMF mass parabola an equation for the minimum of the MA(Z) curve

Physically one is always having tight binding on either the neutron side or the proton side of the nucleus.

Neutron rich Proton rich

+, EC

Mass Parabolas - Odd A

... )(M),(M 2A ZZAZZA

To minimize

Set the derivative wrt. Z = 0

MA(Z)

Z

2

02

0

Z

ZZ

M A

0ZZ

23/12

2

23/122

4

4-

cA

a

Ac

a

mmc

acA

a

c

a

c

am

CA

HnA

SAVn

Remembering that:

3/1

2

0 42

)(4

A

a

Aa

cmmaZ

CA

HnA

Note that Dunlap 4.13 has a mistake:

Mass Parabolas - Odd A

3/1

2

0 42

)(4

A

a

Aa

cmmaZ

CA

HnA

Lets calculate for A=135:

MeVa

MeVa

MeVcmm

A

C

Hn

2.23

72.0

78.0)( 2

5.56]14.0687.0[2

78.02.93

)135(72.0

1352.23x4

2

78.02.23x4

3/1

0

MeV

MeVZ

We may have hoped for slightly better agreement – experimentally the value of Z0 =55.7. But this shows that the global parameters for the SEMF have only limited accuracy. However remember that we are some way from Z=A/2=67.5

Mass Parabolas - Odd A

3/1

2

0 42

)(4

A

a

Aa

cmmaZ

CA

HnA

Note that to a good approximation we can neglect compared to 4aA

So that to a good approximation we get:

2)( cmm Hn

3/20 1

1

2 A

AZ

where: 310x76.7

2.23x4

72.0

4

MeV

MeV

a

a

A

C

Applying this to the case of A=135 we get:

05.56)135(x10x76.71

15.67

66666.030

Z

which is quite good because using the full expression gave Z0=56.5

Mass Parabola – Odd A

Note that the energy released in either - (neutron rich) or EC decay (proton rich) can be expressed in terms of the parabolicity and Z0

You can work these expressions out yourself – it is easy.

(looking down the valley of stability, i.e. decreasing A)

Mass Parabolas - even ANote that for even A there exist two mass parabolas – the top one for low pairing energy binding (ODD-ODD) and the bottom one for high binding energy (EVEN-EVEN)

odd-odd even-even

A=140

Note that some decay such as

140Nd140Pr

have quite low Q energy, while other such as

140Pr140Ce

have large Q energy

(looking down the valley of stability – A decreasing)

Mass Parabolas - even A

A=128Sometimes the positioning of the isobars is such that one can get TWO STABLE ISOBARS

Eg. 128Te and 128Xe

and the strange phenomenon that a nuclide such as 128 I can both + and - decay!

The displacement of the parabolas is of course

4/3

22

A

aP

(looking down the valley of stability – A decreasing)

Mass Parabolas – Even AThe decay energies are given by the same expressions as for Odd A – except now one either subtracts or adds a 2

Beta minus decay – Q value

Q-

Ze-Ze-

β+ Decay

eeeZYZeX

eZeYZeX

NA

ZNAZ

NA

ZNAZ

)1(11

11

ELECTRON CAPTURE

DECAY

A beta emitter– and beta + emitter

1 12.7h

Zn6430

5782.0QMeVcme 022.12 2

EC

653.0022.16749.1

6749.1

Q

QEC

Cu6429

Ni6428

These are what will be quoted

00

Half - life

Spin and parity of nucleus

Beta plus and Beta minus spectra

The momentum spectra for beta plus (right) and beta minus (left) are shown for 64 Cu. The end-point energies for these decays are approximately the same (0.654 MeV) – Beta Plus and (0.578 MeV) –Beta minus.

Note though (i) the spectrum are continuous because of the sharing of energy between three particles, and (ii) that the Beta plus spectrum is skewed to higher momentum (the beta minus to lower momenta).

Neutron separation energy

Energy

Sn

+

21

1

)()( cXMmYMS

nYX

NAZnN

AZn

NAZN

AZ