Semi-Empirical Mass Formula
part I
Classical Terms
[Sec. 4.2 Dunlap]
THE FAMOUS B/A (binding energy per nucleon) CURVE
The Semi-Empirical Mass Formula
is sometimes referred to as:
The Bethe – Weizsacher Mass Formula
Or just “The Mass Formula”
Hans Bethe (1906 -2005)
Carl F. von Weisächer (1912 -2007)
Both interested in production of energy inside stars – both involved in A-bomb
The SEMF
Lets take a look at it:
2/3
2
2
ap
a
A
a
A
a
Z
a
(
A

2
Z
)
A
V
S
C
A
M Z X  ( A  Z )mn  Z (m p  me )  2 
 1/ 3 2 
 3/ 4 2
2
2
c
c
A c
Ac
A c
 
aV , aS , aC , aA , aP are constant/parameters found empirically [Eq. 4.12]
Where
We see an expected general form of:
M
 X   M( A, Z )  ( A  Z )m
n  Z ( m p  me ) 
A
Z
=Mass constituents – [Binding Energy/c2]
where
B

A
Z
1
B
2
c
 X
A
Z
 
B ZA X is the binding energy of the nucleus – given by:

X  B( A, Z )  aV A  aS A
2/3
ap
aC Z 2 a A ( A  2Z ) 2
 1/ 3 
 3/ 4
A
A
A
The SEMF gives the form for B/A
A
In terms of different components.
Volume
Surface
Coulomb
Asymmetry
Pairing
2
2
ap
a
Z
a
(
A

2
Z
)
A
2/3
C
A
B Z X  B( A, Z )  aV A  aS A  1/ 3 
 3/ 4
A
A
A
2
2
a
a
Z
B
a A ( A  2Z )
aP
S
C
 aV  1/ 3  4 / 3 
 7/4
2
A
A
A
A
A
 
= Volume E – Surface E – Coulomb E – Asymmetry E – Pairing E
The Volume Term
B( A, Z )  aV A . . .
q  LmA
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:
( LmA ).A
The Volume Term
q  aV  15.5MeV
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.
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
R
Number of nucleons in R =

N surf
(ii) The density of nucleons in the “skin
thickness” is ~50% less [remember
electron scattering findings]
4R 2 .R.
A
3A
3
1

 0 where   4 3 
4 ( R0 A1/ 3 )3 4R03
2
3 R
3 R 2 R
3 R 2 / 3


.A
2 R03
2 R0
Let the number of bonds for a surface nucleon be only 6 (not 12) – B.E = 3U
9 R 2 / 3
 Bsurf  3U .N surf  U .
. A  a S A2 / 3
2
R0
Taking U=2.6MeV, R0=1.2F, R=2.4F,
aS=9U=23MeV.
. …EXPERIMENTAL VALUE = 18MeV
Adding the Surface Term
So far with volume and surface term we have:
B( A, Z )  aV A  aS A
2/ 3
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 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 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.
e 2 .c
 .c
V (r ) 

(4 0 )c.r
r
V (5 F ) 
1
137
.197MeV .F
 0.3MeV
5F
But some protons are much closer ~ 2F
V(2F) = 0.7 MeV
The Coulomb Term
So 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
Q
Infinity
d q  4r 2 d r.
wh ere
Ze
  4
3

R
3
The amount of charge we are “pushing against” is
r3
Q  Ze 3
R
The Coulomb Term
R
Q
d q  4r 2 d r.
wh ere
Ze
  4
3

R
3
r3
Q  Ze 3
R
Small work done =
dW  V (r ).dq 
Q
(4 0 )r
.4r 2 dr.
Zer 2
3Ze
2

.4r dr.
3
(4 0 ) R
4R 3
3Z2 e 2
4

.
r
dr
6
(4 0 ) R
Infinity
The Coulomb Term
dW  V (r ).dq 
R
Q
(4 0 )r
.4r 2 dr.
Zer 2
3Ze
2

.4r dr.
3
(4 0 ) R
4R 3
3Z2 e 2
4

.
r
dr
6
(4 0 ) R
Now we are ready to build the whole nucleus from r=0 to r=R
R
2 2
3Z 2 e 2
3
Z
e
4
W
r dr 
6 
(4 0 ) R 0
5 (4 0 ) R
3 Z 2 .c

5 R0 A1/ 3
3  .c Z 2
Z2

 aS 1/ 3
1/ 3
5 R0 A
A
Infinity
Inclusion of the Coulomb Term
2
Z
2/3
B( A, Z )  aV A  aS A  aC 1/ 3
A
1
3x
x197MeV .F
3  .c
aC 
 137
 0.72MeV
5 R0
5 x1.2 F
Which is very close to the experimental value
NOTE: Left like this the nucleus would tend to become
totally neutrons – NO PROTONS.