08 Mass Formula Applications Idfgd
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Transcript of 08 Mass Formula Applications Idfgd
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
FUSIONFISSION
(Scale not linear)
Note that extra strong binding occurs for:
B/A
(MeV)
Mg ,Ne ,O ,N ,CBe, He, 2412
2010
168
147
126
84
42
Pairing Energy
),,( r
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-unit. A deeper understanding of the pairing energy will come when we study the SHELL MODEL.
Mass Parabolas
4/3
2
3/1
23/2 1)2(),(
Aa
AZAa
AZaAaAaZAB PACSV
Let’s remind ourselves on the full form of the SEMF. The mass M(A,Z) of the nucleus is given by:
ZABc
mmZmZAZAX epnAZ ,1)()(),(MM 2
XAZ
Cf. Eq. 4.12
or as one equation:
24/32
2
23/1
2
2
3/2
2
)2(
)()(),(M
cAa
AcZAa
cAZa
cAa
cAammZmZAZA
PA
CSVepn
Collecting together powers of Z, it is seen that this expression is quadratic in Z
Mass Parabolas
2
2
23/1
2
2
3/2
2
)2()(),(MAc
ZAacAZa
cAa
cAaZmmZAZA ACSV
Hn
Consider the case of odd A when aP=0
collecting terms
223/12
2
23/122
4
4
.),(M
ZcA
aAca
Zmmca
AcA
aca
camZA
CA
HnA
SAVn
. )(M),(M 2A ZZAZZA
4
4-
23/12
2
23/122
cAa
Aca
mmca
cAa
ca
cam
CA
HnA
SAVn
with
we have a mass parabola !
M(A,Z)
Z
Mass Parabolas
Neutron number
Proton number Z=N
Z increasing
ISOBARS A=Z+N=const.
Mass Parabolas – Odd AFig 4.3 Mass parabola for A=135 showing One Stable Nuclide with Z=56
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.
Neutron rich Proton rich
+, EC
Mass Parabolas - Odd A
... )(M),(M 2A ZZAZZA
To minimize
Set the derivative wrt. Z = 0
MA(Z)
Z
2
02
0
Z
ZZM A
0ZZ
23/12
2
23/122
4
4-
cAa
Aca
mmca
cAa
ca
cam
CA
HnA
SAVn
Remembering that:
3/1
2
0 42
)(4
Aa
Aa
cmmaZCA
HnA
Note that Dunlap 4.13 has a mistake:
Mass Parabolas - Odd A
3/1
2
0 42
)(4
Aa
Aa
cmmaZCA
HnA
Lets calculate for A=135:
MeVaMeVa
MeVcmm
A
C
Hn
2.2372.0
78.0)( 2
5.56]14.0687.0[2
78.02.93
)135(72.0
1352.23x42
78.02.23x4
3/1
0
MeV
MeVZ
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
3/1
2
0 42
)(4
Aa
Aa
cmmaZCA
HnA
Note that to a good approximation we can neglect compared to 4aA
So that to a good approximation we get:
2)( cmm Hn
3/20 11
2 AAZ
where: 310x76.7
2.23x472.0
4
MeVMeV
aa
A
C
Applying this to the case of A=135 we get:
05.56)135(x10x76.71
15.67 66666.030
Z
which is quite good because using the full expression gave Z0=56.5
Mass Parabola – Odd ANote 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.
(looking down the valley of stability, i.e. decreasing A)
Mass Parabolas - even ANote that for even A there exist two mass parabolas – the top one for low pairing energy binding (ODD-ODD) and the bottom one for high binding energy (EVEN-EVEN)
odd-odd even-even
A=140
Note that some decay such as 140Nd140Pr
have quite low Q energy, while other such as140Pr140Ce
have large Q energy
(looking down the valley of stability – A decreasing)
Mass Parabolas - even A
A=128Sometimes 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
4/3
22AaP
(looking down the valley of stability – A decreasing)
Mass Parabolas – Even AThe 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
Q-
Ze- Ze-
β+ Decay
eeeZYZeX
eZeYZeX
NA
ZNAZ
NA
ZNAZ
)1(11
11
ELECTRON CAPTURE
DECAY
A beta emitter– and beta + emitter1 12.7h
Zn6430
5782.0Q
MeVcme 022.12 2 EC
653.0022.16749.16749.1
QQEC
Cu6429
Ni6428
These are what will be quoted
00
Half - life
Spin and parity of nucleus
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
+
21
1
)()( cXMmYMS
nYX
NAZnN
AZn
NAZN
AZ