06 mass formula theory i

8/11/2019 06 Mass Formula Theory I

1/14

Semi-Empirical Mass Formula

part I

Classical Terms

[Sec. 4.2 Dunlap]


2/14

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


3/14

The BetheWeizsacher Mass Formula

The Semi-Empirical Mass Formula

is sometimes referred to as:

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

Both interested in production of energy inside starsboth involved in A-bomb

Or just The Mass Formula


4/14

The SEMFLets take a look at it:

24/32

2

23/1

2

2

3/2

2

A

Z

)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 AZepnA

Z 2

1)()(),(MM

where is the binding energy of the nucleusgiven by:XB A

Z

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

pACSV

A

Z


5/14

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

Za

AaAaZABXB

PACSV

pAC

SV

A

Z

= Volume ESurface ECoulomb EAsymmetry EPairing E

In terms of different components.

Volume Surface Coulomb Asymmetry Pairing

A


6/14

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).(


7/14

The Volume Term

What is the latent heat for a nucleon?

Like a molecule in a liquidthe nucleon is only bound by nearest neighborsbecause 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


8/14

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 [rememberelectron scattering findings]R

Number of nucleons in R =0

2

2

1..4 RR where 3

0

33/1

0

3

34 4

3

)(4

3

RAR

A

R

A

3/2

030

2

.2

3

2

3

AR

R

R

RR

Nsurf

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


9/14

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.


10/14

The Coulomb Term

This term gives the contribution to the energy of the nucleus due to the

potential energy of PROTONS in the nucleus

This in ELECTROSTATIC energyoriginating 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.

MeF

FMeVFV

r

c

rc

cerV

3.05

.197.)5(

.

.)4(

.)(

137

1

0

2

But some protons are much closer ~ 2F

V(2F) = 0.7 MeV


11/14

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

We can assume that in the first approximation the nucleus has aUNIFORM 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

3

3

4

2

where

.4

R

Ze

drrdq

Q

The amount of charge we are pushing against is

3

3

R

rZeQ


12/14

The Coulomb Term

R

Infinity

3

3

4

2

where

.4

R

Ze

drrdq

Q

3

3

R

rZeQ

Small work done =

drrR

e

R

Zedrr

R

Zer

drr

r

QdqrVdW

4

60

22

3

2

3

0

2

2

0

.)4(

3Z

4

3.4.

)4(

.4.

)4(

).(


13/14

The Coulomb Term

R

Infinity

drrRe

R

Zedrr

R

Zer

drrr

QdqrVdW

4

6

0

22

3

2

3

0

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

3/1

0

2

0

22

0

4

6

0

22

.

5

3

.

5

3

)4(5

3

)4(

3

A

Z

aA

Z

R

c

AR

cZR

eZdrr

R

eZW

S

R


14/14

Inclusion of the Coulomb Term

3/1

2

3/2),(AZaAaAaZAB CSV

NOTE: Left like this the nucleus would tend to become

totally neutronsNO PROTONS.

MeVFx

FMeVxx

RcaC 72.0

2.15

.197137

13

.53

0

Which is very close to the experimental value

06 mass formula theory i

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