SEMF application I - Department of Physics, HKU

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Semi-Empirical Mass Formula
Applications - I
[Sec. 4.2 Dunlap]
Pairing Energy
Deuteron
Triton
 - particle
The saturated sub-unit in the nucleus consists of 2 protons and 2 neutrons. This
suggests, in conjunction with the Pauli Exclusion Principle (PEP) and in analogy with
electronic shells in an atom, that the basic quantum state – an S-(J=0) state – consists
of 2 protons and 2 neutrons with antiparallel spins as shown.
5He
does not exist as a bound state (this state breaks up ~10-21s. The PEP allows us to
put at most 2p and 2n in a relative S-state. Any additional nucleon must go into a higher
spatial quantum state.
Because the -particle and not the deuteron (2H) is the saturated sub-unit, this shows
that the force between nucleons is attractive in both the singlet () and triplet ()
states.
Pairing Energy
B/A
(MeV)
FUSION
FISSION
Note that extra strong binding
occurs for:
4
2
He,
8
4
Be,
12
6
C,
14
7
N,
16
8
O,
20
10
Ne ,
(Scale not linear)
24
12
Mg
Pairing Energy
Why does the pairing energy “drop off” as A-3/4. This is something to do
with the fact that nucleons do move on trajectories around the nuclear
volume and do interact with other nucleons. The larger the nucleus the
less the effect of the nucleon-nucleon interaction within the alpha sub ( r , ,  )
unit. A deeper understanding of the pairing energy will come when we
study the SHELL MODEL.
Mass Parabolas
Let’s remind ourselves on the full form of the SEMF. The
mass M(A,Z) of the nucleus ZA X is given by:
1
A
M  Z X   M ( A, Z )  ( A  Z )m n  Z (m p  m e )  2 B  A, Z 
c
B ( A , Z )  aV A  a S A
2/3
 aC
Z
A
2
 aA
1/ 3
( A  2Z )
A
2
 aP
1
A
3/4
or as one equation:
M ( A, Z )  ( A  Z )m n  Z (m p  m e ) 

aA ( A  2Z )
Ac
2
2
aP

A
3/4
c
2
aV A
c
2

aS A
c
2
2/3

aC Z
A
1/ 3
2
c
2
Cf. Eq. 4.12
Collecting together powers of Z, it is seen that this expression is quadratic in Z
Mass Parabolas
Consider the case of odd A when aP=0
aV A
M ( A , Z )  ( A  Z ) m n  Zm H  2 
c
collecting terms
a
a
a


M ( A , Z )   m n  V2  2A  1 / 3S 2  . A
c
c
A c 

aS A
c
2
2/3

aC Z
A
1/ 3
2
c
2

aA ( A  2Z )
Ac
2
 4a

  2 A   m n  m H  Z
 c


aC  2
 4a A

 1/ 3 2  Z
2
A c 
 Ac
2
M ( A, Z )  M A ( Z )   . A   Z  Z
we have a mass parabola !
with
  mn 
aV
c
2

aA
c
2
 4a
aS

A
1/ 3
c
2
M(A,Z)

  -  2 A   m n  m H 
 c

 
4a A
Ac
2
aC

A
1/ 3
c
2
Z
2
Mass Parabolas
Proton number
Z=N
Z increasing
ISOBARS A=Z+N=const.
Neutron number
Mass Parabolas – Odd A
Neutron rich
Proton rich
Fig 4.3 Mass parabola for
A=135 showing One Stable
Nuclide with Z=56
+, EC
We can find from the
SEMF mass parabola an
equation for the minimum
of the MA(Z) curve
Physically one is always
having tight binding on
either the neutron side
or the proton side of the
nucleus.
Mass Parabolas - Odd A
To minimize

M ( A , Z )  M A ( Z )   . A   .Z   .Z
MA(Z)
2
Set the derivative wrt. Z = 0
M
A
Z

   2Z  0
Z0  
Z

Z  Z0
2
Remembering that:
  mn 
aV
c
2

aA
c
2
 4a
aS

A
1/ 3

  -  2 A   m n  m H 
 c

 
4a A
Ac
2
aC

A
1/ 3
c
2
c
2

Z0 
4 a A  (m n  m H )c
2
aC 
 4a A
2

1/ 3 
A
A


Note that Dunlap 4.13 has a mistake:
Mass Parabolas - Odd A
Z0 
4 a A  (m n  m H )c
2
aC 
 4a A
2
 1/ 3 
A 
 A
Lets calculate for A=135:
( m n  m H ) c  0 . 78 MeV
2
a C  0 . 72 MeV
a A  23 . 2 MeV
Z0 
4 x 23 . 2  0 . 78 MeV
 4 x 23 . 2
0 . 72 
2

MeV
1/ 3 
135
(
135
)



93 . 2  0 . 78
2[ 0 . 687  0 . 14 ]
 56 . 5
We may have hoped for slightly better agreement – experimentally the value of
Z0 =55.7. But this shows that the global parameters for the SEMF have only
limited accuracy. However remember that we are some way from Z=A/2=67.5
Mass Parabolas - Odd A
Z0 
4 a A  (m n  m H )c
2
aC 
 4a A
2
 1/ 3 
A 
 A
Note that to a good approximation we can neglect ( m n  m H ) c compared to 4aA
2
So that to a good approximation we get:
Z0 
where:  
A

1
2 1  A
aC

4a A
2/3

0 . 72 MeV
 7 . 76 x 10
3
4 x 23 . 2 MeV
Applying this to the case of A=135 we get:
Z 0  67 . 5
1
1  7 . 76 x 10
3
x (135 )
0 . 66666
 56 . 05
which is quite good because using the full expression gave Z0=56.5
Mass Parabola – Odd A
(looking down the valley of stability, i.e. decreasing A)
Note that the energy
released in either (neutron rich) or EC
decay (proton rich)
can be expressed in
terms of the
parabolicity  and Z0
You can work
these expressions
out yourself – it is
easy.
Mass Parabolas - even A
(looking down the valley of stability – A decreasing)
Note that for even A there exist
two mass parabolas – the top
one for low pairing energy
A=140
binding (ODD-ODD) and the
bottom one for high binding
energy (EVEN-EVEN)
odd-odd
even-even
Note that some decay such as
140Nd140Pr
have quite low Q energy, while
other such as
140Pr140Ce
have large Q energy
Mass Parabolas - even A
(looking down the valley of stability – A decreasing)
A=128
Sometimes the positioning
of the isobars is such that
one can get TWO STABLE
ISOBARS
Eg.
128Te
and 128Xe
and the strange
phenomenon that a nuclide
such as 128 I can both +
and - decay!
The displacement of the
parabolas is of course
2 
2aP
A
3/4
Mass Parabolas – Even A
The decay energies
are given by the same
expressions as for
Odd A – except now
one either subtracts or
adds a 2
Beta minus decay – Q value
Ze-
Ze-
Q-
β+ Decay

A

A


A
Z
X N  Ze  Z 1Y N  1  Ze  e  
A
Z
X N  Ze  Z 1Y N  1  ( Z  1) e  e  e  



ELECTRON CAPTURE
DECAY
A beta emitter– and beta + emitter
1

0


Half - life
12.7h
64
29

Cu

0
Spin and parity
of nucleus

64
30
EC
2 m e c  1 . 022 MeV

Zn
2
64
28
Q    0 . 5782
Ni
Q EC  1 . 6749
Q    1 . 6749  1 . 022  0 . 653
These are what will be
quoted
Beta plus and Beta minus spectra
The momentum spectra for beta plus (right) and beta minus (left) are shown
for 64 Cu. The end-point energies for these decays are approximately the
same (0.654 MeV) – Beta Plus and (0.578 MeV) –Beta minus.
Note though (i) the spectrum are continuous because of the sharing of energy
between three particles, and (ii) that the Beta plus spectrum is skewed to
higher momentum (the beta minus to lower momenta).
Neutron separation energy
Energy
+
Sn
A
Z
X

 Z Y N 1  n
A
N

S n  M ( Y N 1 )  m n  M ( X
A
Z
A
Z
N

)c
2
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