Semi-Empirical Mass Formula

part IClassical Terms

[Sec. 4.2 Dunlap]

THE FAMOUS B/A (binding energy per nucleon) CURVE

The Bethe – Weizsacher Mass Formula

The Semi-Empirical Mass Formula is sometimes referred to as:

Hans Bethe (1906 -2005) Carl F. von Weisächer (1912 -2007)

Both interested in production of energy inside stars – both involved in A-bomb

Or just “The Mass Formula”

The SEMFLets take a look at it:

24/32

2

23/1

2

2

3/2

2AZ

)2()()(M

cA

a

Ac

ZAa

cA

Za

c

Aa

c

AammZmZAX pACSVepn

[Eq. 4.12]

We see an expected general form of:

XBc

mmZmZAZAX AZepn

AZ 2

1)()(),(MM

where is the binding energy of the nucleus – given by: XB AZ

Where are constant/parameters found empiricallyPACSV aaaaa , , , ,

=Mass constituents – [Binding Energy/c2]

4/3

2

3/1

23/2 )2(

),(A

a

A

ZAa

A

ZaAaAaZABXB pAC

SVAZ

The SEMF gives the form for B/A

4/72

2

3/4

2

3/1

4/3

2

3/1

23/2

)2(

)2(),(

A

a

A

ZAa

A

Za

A

aa

A

B

A

a

A

ZAa

A

ZaAaAaZABXB

PACSV

pACSV

AZ

= Volume E – Surface E – Coulomb E – Asymmetry E – Pairing E

In terms of different components.

Volume Surface Coulomb Asymmetry Pairing

A

The Volume Term . . . ),( AaZAB V

To the first approximation the nucleus can be considered as a LIQUID made up of nucleons (neutrons and protons). In a molecular classical liquid one has to put in LATENT HEAT (L) per kg of liquid evaporated. Why? Because each molecule has to break the same number of molecular bonds on leaving the liquid. It needs energy q (eV) – depending only on nearest neighbor bonds

The energy for removing A molecules is:

ALmq

ALmA ).(

The Volume Term

What is the latent heat for a nucleon?

Like a molecule in a liquid – the nucleon is only bound by nearest neighbors because the STRONG FORCE is a SHORT RANGE interaction.

An approximate treatment takes there to be 12 nearest neighbors. If each bond has U (MeV) of B.E. then the total amount of B.E. is 6U (MeV) – Not 12 because we must not double count.

MeVaq V 5.15

U=2.6MeV per bond

The Surface Term

q

If we say that the total B.E. of the nucleus is aVA then we make an ERROR

(i) The bonding of nucleons on surface is ~50% less than those in the bulk

(ii) The density of nucleons in the “skin thickness” is ~50% less [remember electron scattering findings]R

Number of nucleons in R =0

2

2

1..4 RR where 3

033/1

03

34 4

3

)(4

3

RAR

A

R

A

3/2

030

2

.2

3

2

3A

R

R

R

RRN surf

Let the number of bonds for a surface nucleon be only 6 (not 12) – B.E = 3U

3/23/2

0

..2

9.3 AaA

R

RUNUB Ssurfsurf

Taking U=2.6MeV, R0=1.2F, R=2.4F, aS=9U=23MeV. . …EXPERIMENTAL VALUE = 18MeV

Adding the Surface TermSo far with volume and surface term we have:

3/2),( AaAaZAB SV NOTE: This same expression would apply to a molecular liquid drop

The way to maximize the binding of a liquid drop is to minimize the surface area - that is why liquid drops tend to be SPHERICAL

NOTE: So far the binding energy depends only on the number of nucleons and not on the charge Z.

The Coulomb TermThis term gives the contribution to the energy of the nucleus due to the potential energy of PROTONS in the nucleus

This in ELECTROSTATIC energy – originating form the EM FORCE - It is a NEGATIVE B.E. because its effect is to give out energy.

Lets assume that the mean distance between two nucleons in the nucleus is 5 F, then how much electrostatic energy is involved.

MeVF

FMeVFV

r

c

rc

cerV

3.05

.197.)5(

.

.)4(

.)(

1371

0

2

But some protons are much closer ~ 2F

V(2F) = 0.7 MeV

The Coulomb TermSo how can we estimate the Coulomb energy for a nucleus

We can assume that in the first approximation the nucleus has a UNIFORM density of PROTONS out to radius R.

The we perform an electrostatics thought experiment where we bring up small charge dq from infinity to fill up the shell between r and r+dr

R

Infinity

334

2

where

.4

R

Ze

drrdq

Q

The amount of charge we are “pushing against” is

3

3

R

rZeQ

The Coulomb Term

R

Infinity

334

2

where

.4

R

Ze

drrdq

Q

3

3

R

rZeQ

Small work done =

drrR

e

R

Zedrr

R

Zer

drrr

QdqrVdW

46

0

22

32

30

2

2

0

.)4(

3Z

4

3.4.

)4(

.4.)4(

).(

The Coulomb Term

R

Infinity

drrR

e

R

Zedrr

R

Zer

drrr

QdqrVdW

46

0

22

32

30

2

2

0

.)4(

3Z

4

3.4.

)4(

.4.)4(

).(

Now we are ready to build the whole nucleus from r=0 to r=R

3/1

2

3/1

2

0

3/10

2

0

22

0

46

0

22

.

5

3

.

5

3

)4(5

3

)4(

3

A

Za

A

Z

R

c

AR

cZ

R

eZdrr

R

eZW

S

R

Inclusion of the Coulomb Term

3/1

23/2),(

A

ZaAaAaZAB CSV

NOTE: Left like this the nucleus would tend to become totally neutrons – NO PROTONS.

MeVFx

FMeVxx

R

caC 72.0

2.15

.197137

13.

5

3

0

Which is very close to the experimental value