Transcript Nuclear Binding Energy

2 convenient units
Unified mass unit, u
1
 12 the mass of a neutral C12 atom
by definition (of a mole) 1 mole, C12 = 12 grams
1 gram
-24gram = 1.6604310-27 kg
=
1.6604310
1u =
6.022521023
the electron volt, eV
Energy acquired by a particle with 1e charge when
accelerated through a potential difference of 1 volt.
1eV = 1.60210-19 Coul-volt = 1.60210-19 joule
1eV = 1.60210-19 Coul-volt = 1.60210-19 joule
1u
c2
=
1.6604310-27
c2
kg
1.60210-19 joule/eV
8 m/sec) 2
(2.99810
= 1.6604310-27 kg
1.60210-19 joule/eV
1u c 2 = 9.315825108 eV = 931.58 MeV
1u
= 931.58 MeV/c 2
Rest mass energy can be measure in terms of eV or MeV.
Sometime mass is measure in the convenient unit of MeV/c2.
1 keV = 103 eV
1 MeV = 106 eV
1 GeV = 109 eV
1 TeV = 1012 eV
• 1012
• 109
• 106
• 103
tera
giga
mega
kilo
(trillion)
(billion, “80 Gigabyte hard drive”)
(million, “128 Megabytes of RAM”)
(thousand, 1 kilogram = 2.2 pounds)
• 10– 2
• 10– 3
• 10– 6
• 10– 9
• 10– 12
• 10– 15
centi
milli
micro
nano
pico
femto
(hundredth, 1 in. = 2.54 cm)
(thousandth, “milliVolt”)
(millionth, micron=micrometer=10-6m)
(billionth, “nanosecond”)
(trillionth)
With the nuclear volume  the number of nucleons A
the nuclear force is not sufficient to contract
the nucleus to higher densities at its center.
the nuclear force saturates
the density of nucleons in the nucleus ~ constant
Liquid Drop Model
1935-36 Bohr & von Weizsäcker
Cross Section measurements help fix
the effective range of the nuclear force
Precision mass measurements reveal binding energies that
tell us even more about the strength of the nuclear force
Particle
symbol
rest energy in MeV
or mass in MeV/c2
electron
e
muon

105.658357
neutral pion
0
134.9766
charged pion

139.57018
proton
p
938.27200
neutron
n
939.56533
deuteron
H2
1875.580
triton
H3
2808.873
alpha
 (He4)
3727.315
0.5109989
Know from relativity E and m intimately related
Rest mass energy: E
=moc2
For a free particle, total energy: E
with p2
E2
or
–
= T + moc2
= moc2 = Eo
=  2m2v2 =  2 Eo2v2/c4 then
p 2c 2
=
2
2E 2
o
E2 = p2c2 + Eo2
v
( 1 2 )
c
2 

v
 2  1 / 1  2 
 c 
and so:
T  KE = moc2 (  - 1) = c2 Dm
Work done by an unbalanced force increases an object’s energy from
a rest mass of
moc2  moc2
or increases it’s mass from
mo   mo
If balanced or resisted by a restoring force: increases an object’s
POTENTIAL ENERGY
Proton billiards: consider the head-on collision between 2 protons:
m0
m0
Einitial  m0c  m0c
2
' ( 2m0 )c
2
E final  m0c  m0c
2
' ( 2m0 )c  (1   )m0c
2
2
2
2
'  (1   ) / 2
revealing the intermediate state of “bound” protons to have a mass of
' ( 2m0 )  (1   )m0
> 2m0 !!!
or consider the same collision in the center of momentum frame:
E=m0c2
E=m0c2
Einitial  m0c  m0c
2
2
so here can’t have E=2m0c2
Einitial  m0c  m0c
2
2
E  2m0c  2m0c
2
2
The intermediate system of protons has more mass than two protons!
For a repulsive potential
combined rest mass > sum of the individual rest masses
For an attractive potential
combined rest mass < sum of the individual rest masses
Rest energies
of a proton + neutron = 938.27200+939.56533 = 1877.83733 MeV
while of a deuteron H2 = 1875.61269 MeV
2.22464 MeV
A deuteron cannot spontaneously decay into a proton and a neutron.
To break up a deuteron 2.22464 MeV
must be added by bombarding deuterons
with an energetic beam of particles or
electromagnetic radiation.
Neutrons from a reactor striking protons are sometimes “captured”
through the reaction
n+pd+
For the alpha particle Dm= 0.0304 u which gives
28.3 MeV binding energy!
For “bound states” the binding energy, B>0
representing the energy required to disintegrate the
nucleus (into individual neutrons and protons)
B

Zm

(
A

Z
)
m

M
(
A
,
Z
)
p
n
nucleus
2
c
B

Z
(
m

m
)

(
A

Z
)
m

M
(
A
,
Z
)
p
e
n
neutral
atom
2
c
B

Zm

(
A

Z
)
m

M
(
A
,
Z
)
H
n
neutral
atom
2
c
Working in atomic mass units, u
the difference between atomic rest mass energy M(A,Z)
and atomic mass number A  u
is called the MASS DEFECT (or “mass excess”)
D  M ( A, Z )  Au
Notice with Au  Zmp + (A-Z)mn
this is essentially the same as the Binding Energy
D follows the conventions of mass spectroscopy measurements
Peaks at ~8.795 MeV near A=60
Binding energy per nuclear
particle (nucleon) in MeV
for A>50
~constant
8-9 MeV
Mass Number, A
This confirms the short range ascribed to the nuclear force…
it must involve only nearest neighbor nuclei
If the binding involved contributions from all nucleons
 the total number of pair-wise bonds
A( A  1)  A2
then
B
A
A
(not constant!)
Nuclear Binding Energy Curve
The binding energy curve is obtained by dividing the total nuclear binding energy by the
number of nucleons. The fact peak in the binding energy curve near iron means that
either the breakup of heavier nuclei (fission) or the combining of lighter nuclei (fusion)
will yield nuclei which are more tightly bound (less mass per nucleon).
The binding energies of nucleons are in the range of millions of electron volts compared
to tens of eV for atomic electrons. Whereas an atomic transition might emit a photon in
the range of a few electron volts, perhaps in the visible light region, nuclear transitions
can emit gamma-rays with quantum energies in the MeV range.
The iron limit:
The buildup of heavy elements by the nuclear fusion processes in stars is limited to
elements below iron, since the fusion of iron would subtract energy rather than provide
it. Iron-56 is abundant in stellar processes, and with a binding energy per nucleon of 8.8
MeV, it is the third most tightly bound of the nuclides. Its average binding energy per
nucleon is exceeded only by 58Fe and 62Ni, the nickel isotope being the most tightly
bound of the nuclides.
The Most Tightly Bound Nuclei
62Ni
M. P. Fewell, American Journal of Physics.
(56Fe actually comes in a close third)
B/A (keV/A)
62Ni
8794.60 +/- 0.03
58Fe
8792.23 +/- 0.03
56Fe
8790.36 +/- 0.03
60Ni
8780.79 +/- 0.03
The most tightly bound nuclides are all even-even nuclei.
The high binding energy of the “iron group” of around A=60 is significant in
the understanding of the synthesis of heavy elements in the stars.
The Semi-emperical Mass Formula
is an approximate fit to all this data
To 0th order:
M ( A, Z )  Zm p  ( A  Z )mn
The 1st correction to this (the “mass defect” or binding energy)
~proportional to volume
B/A (MeV)
EV   av A
true enough for
very small A !
A
But the nuclei at the surface have fewer nearest-neighbor bonds!
So we’ve obviously over-estimated!
by something  the nucleus’ surface area
Es   a s A
2/3
recall
r  A1 / 3
With these effects alone,
all isobars might be expected to be stable
though clearly we’ve seen
a narrow band of stability in the N vs A plane
for light nuclides: N=Z (A = 2Z)
for heavy nuclides: A>2Z
There’s a repulsive Coulomb energy building up with all those protons!
Coulomb Energy
The work done in collecting and compacting
total charge Q into a sphere of volume 4 R 3.
3
potential at the surface of a spherical concentration of charge
kq
R
with charge density:  
Q
( 4 / 3)r 3
so at r:
3
2


k ( 4r 3 / 3)
Q
4

r
/
3
r


  kQ
V
 k

r
R3
 4R 3 / 3  r 
Bringing an additional dq=4r2dr to the surface at r requires:
kQ 2
kQ 4 5
2
Vdq  3 r  4r dr 
r dr
3
R
R
Coulomb Energy
kQ4
Q 1 5
kQ 4 R 5
R
Integrating:
0 r dr 
3
3
3
R 4R / 3 5
R

 1  3Q 2 1

 
 4 0  5 R
Motivating us to add a correction term for this Coulomb energy
e2
Z(Z-1)e2  Q2
 Z ( Z  1)
R
number of pairings
Ec  ac Z ( Z  1) A1 / 3
RA1/3
B  av A  as A  ac Z ( Z  1) A
1 / 3
2/3
volume term
grows with
addition of
each interacting
nuclei
surface term
corrects for
nuclei at the
surface (not
completely
surrounded by
nearest neighbors)
Coulomb term
accounts for
(repulsive) energy
built up in the
accumulation
of protons
Mass defect D (MeV)
A = 127
isobars
-80
-80
-84
-84
-86
-86
-88
-88
50
52
54
56
Atomic number, Z
Think of
the stability band
as a contour map
of a valley:
A consequence of the exclusion principal: less economical (more
expensive in energy) to have more of one type of nucleon than the other:
(A/ 2  Z)


a
symmetry
sym
A
2
Like the “closed shell” electronic configurations of the noble gases
we see “magic numbers” recurring…marking tightly bound,
extraordinarily stable nuclear configurations.
=4He (with Z=N=2) and 16O (Z=N=8) are doubly magic!
including to a lesser extent, marked stability for nuclei with
paired protons and/or paired neutrons
(like the closed energy levels of atomic orbitals
where electrons with opposite spins cancel)
Since nuclei with paired protons or paired neutrons
tend to be more tightly bound, we introduce the
pairing energy term:

ap
A3 / 4
 ( A)  0

ap
A3 / 4
even Z, N
Z=N=0
odd Z, N
volume term
grows as
nuclei added
surface term
corrects for
surface nuclei
B  av A  a s A
2/3
 ac Z ( Z  1) A
(A/ 2  Z)
 asym
A
symmetry term
drives number of
protons ≈ neutrons
Coulomb term
repulsion due
to protons
1 / 3
2
  ( A)
pairing energy term
increased stability
for paired nuclei
M ( A, Z )  Zm p  ( A  Z )mn  B( A, Z )c 2
B( A, Z )  av A  as A
2/3
1 / 3
 ac Z ( Z  1) A
 asym ( A / 2  Z ) / A   ( A, Z )
2
av=15.5 MeV
ac=0.72 MeV
ap=34.0 MeV
as=16.8 MeV
asym=23.0 MeV
Atomic shell model
Nuclear shell model
each nucleon, separately occupies its own energy level,
with each nucleon-type (p,n)
filling energy levels separately
Each energy level, n exists with an angular quantum number:
s p d f g …
ℓ = 0, 1, 2, 3, 4, …
each with a different
angular wave function
Yℓ,m(,)
Because of spin (2 possible orientations of each nucleon)
any energy level n,ℓ accommodates
2ℓ+1 protons
2ℓ+1 neutrons
s: ℓ = 0
p: ℓ = 1
d: ℓ = 2
f: ℓ = 3
These are different particles –
They don’t
mutually exclude each other!
coupled with spin:
s1/2
p1/2
p3/2
d3/2
d5/2
f 5/2
f 7/2
J=0+½
The total (magnitude)
of J is limited to
-|ℓ-s|…|ℓ+s|
in unit steps!
There is a 1f5/2 energy level that can hold
5
2( ) ????
1  6 protons with different mj values
2
and another that can hold 6 neutrons.
Consider:
27
13
Al
(ground state)
2
4
2
Protons: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
2
4
2
5
Neutrons: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
6
1 un-paired proton with j=5/2
so aluminum: I=5/2
Consider:
9
Be
(ground state)
2
2
2
3
p: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
n: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
1 un-paired neutron with j=3/2 so: I=3/2
Consider:
9
B
(ground state)
2
3
2
2
p: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
n: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
1 un-paired proton with j=3/2
so: I=3/2
Magnetic Dipole Moment
B
classically
energy Emag = -B
=
1
c
current  area
For a point particle q, velocity v, circular orbital radius r
v
I q
2r
frequency!
1
 = c current  area
 
q
1 qvr
 v 
2
( mvr )
=
q
  r = c
2mc
2
 2r 
L
q 


L but maybe too classical!
2mc
 
We still expect   J quantum mechanically
and write
e  g measures the

g
J
(expected)
2mc
deviation from
1
= c
quantum mechanically
J is quantized!
J  j
e
 g
j
2mc

 g 0 j
0 = “magneton”
For atomic physics (involving electrons)
“Bohr magneton”
e
14
B 
 0.5788 10 MeV / G
2me c
5
 5.7885 10 eV / T
For nuclear physics
“nuclear magneton”
e
N 
 3.1525 10 18 MeV / G
2m p c
 3.15255 10 8 eV / T
For the neutron: g= 3.826083 0.0000018  0 !