Transcript nuclei properties

Properties of Nuclei
• Z protons and N neutrons held together with
a short-ranged force  gives binding energy
mp  938.3 MeV / c 2
M nucleus  Zmp
with
mn  939.6 MeV / c 2
 Nmn  Ebind  AmN
A Z N
mN  u  9315
. MeV / c 2
• P and n made from quarks. Most of the mass
due to the strong interactions binding them
together. Recent JLAB results show masses
inside nucleus might be slightly smaller than
free particles
• P and n are about 1 Fermi in size and the
strong force doesn’t compress. Size ~ range
of strong force  all nuclei have the same
density and higher A nuclei are bigger
(unlike atoms)
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Protons vs Neutron
• neutron slightly heavier than proton and so it
decays. No reason “why” just observation
proton
m ass
neutron
938.27 MeV / c 2
lifetim e  1035 yrs(all modes)
 103133 yrs( specific)
939.56 MeV / c 2
886s
n  p  e 
p  e   0 ,     0 
• quark content: n = udd and p = uud (plus g,
qqbar) Mass up and down quarks 5-10 Mev
• three generations of quarks. Only top quark
ever observed as “bare” quark. Somehow up
quark seems to be slightly lighter than down
quark
 23   u   c   t 
 1        
  3   d   s  b
mt  mb
mc  ms
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md  mu ????
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Nuclei Force
• Strong force binds together nucleons
• Strong force nominally carried by gluons. But
internucleons carried by pions (quark-antiquark
bound states) as effective range too large for
gluons
n
p

p
n
p   n n   p
p 0  p n 0  n
• Each p/n surrounded by virtual pions. Strong
force identical p-p, p-n, n-n (except for
symmetry/Pauli exclusion effects)
• Range of 1 F due to pion mass
 E t     t 
x 

 10  25 s
m
c 197 MeVF

 1F
m
135 MeV
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Nuclear Sizes and Densities
• Use e + A  e + A scattering completely EM
• pe = 1000 MeV/c  wavelength = 1.2 F now
JLAB, in 60s/70s SLAC up to 20 GeV( mapped
out quarks)
• Measurement of angular dependence of cross
section gives charge distribution (Fourier
transform)
• Can also scatter neutral particles (n, KL) in
strong interactions to give n,p distributions
• Find density ~same for all but the lowest A
nucleii
(0)
18
3
(r ) 

10
kg
/
m
( r  a )/ b
1 e
a  107
. A1/ 3 F
b  0.55F
 skin depth
P461 - Nuclei I
volume  A
surface area  A 2 / 3
radius  A1/ 3
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Nuclear Densities
• can write density as an energy density
R  1.2 FA1 / 3  m p  1.7  1027 kg  938MeV / c 2
 E / vol  0.14GeV / F 3
• Note Quark-Gluon Plasma occurs if
E / vol  1GeV / F
3
1017 kg
1028 kg
14
3
 10 g / cm 
3
m
F3
1 F  1015 m
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Nuclear Densities
 ( K L  A  anything)
A  C, Al, Cu, Sn, Pb
PRL 42,9(1979)
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Nuclear Densities
 A
.84
 ( K L  A  anything)
A  C, Al, Cu, Sn, Pb
PRL 42,9(1979)
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P461 Model of Nuclei
• “billiard ball” or “liquid drop”
• Adjacent nucleons have force between them but not
“permanent” (like a liquid). Gives total attractive
energy proportional to A (the volume) – a surface
term (liquid drop)
• Repulsive electromagnetic force between protons
grows as Z2
• Gives semi-empirical mass formula whose terms
can be found by fitting observed masses
• Pauli exclusion as spin ½  two (interacting)
Fermi gases which can be used to model energy
and momentum density of states
• Potential well is mostly spherically symmetric so
quantum states with J/L/S have good quantum
numbers. The radial part is different than H but
partially solvable  shell model of valence states
and nuclear spins
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Semiempirical Mass Formula
•
•
•
•
•
•
•
•
M(Z,A)=f0 + f1 +f2 + f3 + f4 + f5
f0 = mp Z + mn (A-Z)
mass of constituents
f1 = -a1A A ~ volume  binding energy/nucleon
f2 = +a2A2/3 surface area. If on surface, fewer
neighbors and less binding energy
f3 = +a3Z2/A1/3 Coulomb repulsion ~ 1/r
f4 = +a4(Z-A/2)2/2 ad hoc term. Fermi gas gives
equal filling of n, p levels
f5 = -f(A) Z, N both even
=
0
Z even, N odd or Z odd, N even
= +f(A) Z., N both odd f(A) = a5A-.5
want to pair terms (up+down) so nuclear spin = 0
Binding energy from term f1-f5. Find the constants
(ai’s) by fitting the measured nuclei masses
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Semiempirical Mass Formula
volume
surface
Coulomb
Eb=
E/A
Total
N/Z asymmetry
A
• the larger the binding energy Eb, the greater the
stability. Iron is the most stable
• can fit for terms
• good for making quick calculations; understanding
a small region of the nuclides.
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www.metasynthesis.com/webbook/33_segre/segre.html
most stable (valley)
number of protons
Number of neutrons
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Semiempirical Mass
• the “f5” term is a paring term. For nuclei near U
there is about a 0.7 MeV difference between
having both n and p paired up (even A), odd A
(and so one unpaired), and another 0.7 MeV for
neither n or p being paired spin (even A)
U 235  n  U 236
Eb (n)  M 92, 235  M 0,1  M 92,236  6.6 MeV
f5 
11
 0.7 MeV
236
• so ~5.9 MeV from binding of extra n plus 0.7
MeV from magnetic coupling
U 238  n  U 239
Eb (n)  6.6 MeV  2  .7  5.2 MeV
• easier for neutron capture to cause a fission in
U235. U236 likelier to be in an excited state.
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Fermi Gas Model
• p,n spin ½ form two Fermi gases of
indistinguishable particles  p n through
beta decays (like neutron stars) and p/n ratio due
to matching Fermi energy
• In finite 3D well with radius of nucleus.
Familiar:
8V
N ( p)  p  D( E )  3 (2m3 ) 1/ 2 E 1/ 2
h
2
• Fermi energy from density and N/A=0.6
h2  3 
EF 
 
8m   
2/3
 E F  43 MeV
N
 4
a  11
.F
3
3 a A
pF 
2mE F  270 MeV
• Slightly lower proton density but shifted due to
electromagnetic repulsion
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Fermi Gas Model II
• V = depth of well = F(A) ~ 50 MeV
• Fermi energy same for all nuclei as density =
constant
• Binding energy B = energy to remove p/n from
top of well ~ 7-10 MeV V = EF + B
• Start filling up states in Fermi sea (separate for
p/n)
• Scattering inhibited 1 + 2  1’ + 2’ as states 1’
and 2’ must be in unfilled states  nucleons are
quasifree
n  p  e  
B
vs
V
(ignore Coulomb)
n p
P461 - Nuclei I
n p
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Nuclei
• If ignore Coulomb repulsion, as n<->p through beta
decay, lowest energy will have N=Z (gives (N-Z)
term in mass formula)
• proton shifted higher due to Coulomb repulsion.
Both p,n fill to top with p<->n coupled by Weak
interactions so both at ~same level (Fermi energy
for p impacted by n)
n  p  e  
p  n  e  
p  e  n 
U 238  92 p,146n
n
p
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Nuclei: Fermi motion
• if p,n were motionless, then the energy thresholds
for some neutrino interactions are:
e  p  n  e 
E  1.8 MeV
   n     p E  120 MeV
• but Fermi momentum allows reactions to occur at
lower neutrino energy.
E F  40 MeV
dN/dp
pF  2mE F  280 MeV
p
  p  n  e
Ethrsh  1.8MeV
mp
Ep  pp
P461 - Nuclei I
 1.4 MeV
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Nuclei:Fermi motion
•
electron energy loss
PRL 26,445(1971)
e pe p
free p
eC e p X
( Ni, Pb)
solid lines are
modified Fermi gas
calculation (tails
due to interactions)
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•
  n    p
n in C nucleus
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Nuclei:Pauli Suppression
• But also have filled energy levels and need to give
enough energy to p/n so that there is an unfilled
state available. Simplest to say “above” Fermi
Energy
• similar effect in solids. Superconductivity mostly
involves electrons at the “top” of the Fermi well
e  p  n  e 
  n     p
  e   e
•
at low energy transfers (<40 MeV) only some p/n
will be able to change states. Those at “top” of
well.
• Gives different cross section off free protons than
off of bound protons. Suppression at low energy
transfers if target is Carbon, Oxygen, Iron...
• In SN1987, most observed events were from
antineutrinos (or off electrons) even though (I
think) 1000 times more
neutrinos. Detectors were
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water…..
Physics Reports
1972 C.H.
Llewellen-Smith
C
1-Suppression
factor
Fermi gas
“shell” model
includes spin
effects
Fe
energy transfer
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Nuclei: Fermi Suppression and
Pauli Exclusion
• important for neutrino energies less than 1 GeV. prevents
accurate measurement of nuetrino energy in detector
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