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06 Mass Formula Theory I
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Transcript of 06 Mass Formula Theory I
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Semi-Empirical Mass Formula
part I
Classical Terms
[Sec. 4.2 Dunlap]
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THE FAMOUS B/A (binding energy per nucleon) CURVE
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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
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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
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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
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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).(
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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
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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
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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.
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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
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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
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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(
).(
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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
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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