Transcript Mass Formula theory I - Department of Physics, HKU
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 Weis ächer (1912 -2007) Both interested in production of energy inside stars – both involved in A-bomb
The SEMF
Lets take a look at it: M
X
(
A
Z
)
m n
Z
(
m p
m e
)
a V A c
2
a S A
2 / 3
c
2
a C Z
2
A
1 / 3
c
2
a A
(
A
2
Z
) 2
Ac
2
a p A
3 / 4
c
2
a
,
a
,
a
,
a
,
a
[Eq. 4.12] We see an expected general form of: M
Z
M (
A
,
Z
) (
A
Z
)
m n
Z
(
m p
m e
) 1 2
B
Z c
=Mass constituents – [Binding Energy/c 2 ]
B
– given by:
B
Z
B
(
A
,
Z
)
a V A
a S A
2 / 3
a C Z
2
A
1 / 3
a A
(
A
2
Z
) 2
A
a p A
3 / 4
The SEMF gives the form for B/A
A In terms of different components.
Volume Surface Coulomb Asymmetry Pairing
B
Z
B A
a V
B
(
A
,
Z
)
a A
1
S
/ 3
a C
Z A
4 /
a V
3 2
A
a S A
2 / 3
A
2
a C Z a A
(
A
2
Z
) 2
A
1 / 3 2
a
P A
7 / 4
a A
(
A
2
Z
) 2
A
a p A
3 / 4 = Volume E – Surface E – Coulomb E – Asymmetry E – Pairing E
The Volume Term
B
(
A
,
Z
)
a V A
.
.
.
q
Lm A
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 : (
Lm A
).
A
The Volume Term
q
a V
15 .
5
MeV
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 R If we say that the total B.E. of the nucleus is
a V A
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] Number of nucleons in R =
N surf
4 3 2
R R
2
R
2 .
R
.
R
3 0 1 2 3 2 0
R
0 where
R
.
A
2 / 3 4 3
A R
3 4 ( 3
R
0
A A
1 / Let the number of bonds for a surface nucleon be only 6 (not 12) – B.E = 3U 3 ) 3 3 4
R
0 3
B surf
3
U
.
N surf
9 2
U
.
R
.
A
2 / 3
R
0
a S A
2 / 3 Taking U=2.6MeV, R 0 =1.2F, R=2.4F, a S =9U=23MeV. . … EXPERIMENTAL VALUE = 18MeV
Adding the Surface Term
So far with volume and surface term we have:
B
(
A
,
Z
)
a V A
a S 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.
V
(
r
)
e
2 .
c
( 4 0 )
c
.
r
.
c r V
( 5
F
) 1 137 .
197
MeV
.
F
5
F
0 .
But some protons are much closer ~ 2F 3
MeV V
(2
F
) = 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 to fill up the shell between r and r+dr Q R
dq
4
r
2
dr
.
where 4 3
Ze
R
3 The amount of charge we are “pushing against” is
r
3
Q
Ze R
3
The Coulomb Term
Q R
dq
4
r
2
dr
.
where 4 3
Ze
R
3 Small work done =
Q
Ze r
3
R
3
dW
V
(
r
).
dq
Q
( 4 0 )
r
.
4
r
2
dr
.
Zer
2 ( 4 0 )
R
3 .
4
r
2
dr
.
3
Ze
4
R
3 ( 3Z 2 4 0
e
2 )
R
6 .
r
4
dr
Infinity
The Coulomb Term
R
dW
V
(
r
).
dq
Q
( 4 0 )
r
.
4
r
2
dr
.
Zer
2 ( 4 0 )
R
3 .
4
r
2
dr
.
3
Ze
4
R
3 ( 3Z 2 4 0
e
2 )
R
6 .
r
4
dr
Now we are ready to build the whole nucleus from r=0 to r=R 2 2
R
2 2
W
( 4 3
Z
0
e
)
R
6 0
r
4
dr
3 5 ( 4
Z
e
0 )
R
3 5
Z
2 .
c R
0
A
1 / 3 3 5 .
c R
0
Z
2
A
1 / 3
a S Z
2
A
1 / 3 Infinity
Inclusion of the Coulomb Term
B
(
A
,
Z
)
a V A
a S A
2 / 3
a C Z
2
A
1 / 3
a C
3 5 .
c R
0 1 3
x
137
x
197
MeV
.
F
5
x
1 .
2
F
0 .
72
MeV
Which is very close to the experimental value NOTE: Left like this the nucleus would tend to become totally neutrons – NO PROTONS.