Post on 04-Oct-2020
transcript
By: Sushil Kumar
HMV Jalandhar
2 convenient units
Atomic mass unit, u thmass of a neutral C12 atom 1
12
by definition (of a mole) 1 mole, C12 = 12 grams
1u = 1 gram
6.022521023 = 1.6604310-24gram = 1.6604310-27 kg
the electron volt, eV Energy acquired by a particle with 1e charge when
accelerated through a potential difference of 1 volt.
1eV = 1.60210-19 Coul-volt = 1.60210-19 Joule
1u c 2 = 1.6604310-27 kg
1eV = 1.60210-19 Coul-volt = 1.60210-19 joule
c 2
1.60210-19 joule/eV
= 1.6604310-27 kg (2.998108 m/sec) 2
1.60210-19 joule/eV
= 9.315825108 eV = 931.58 MeV
Rest mass energy can be measure in terms of eV or MeV.
1u c 2
= 931.58 MeV/c 2 1u
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 tera (trillion)
• 109 giga (billion, “80 Gigabyte hard drive”)
• 106 mega (million, “128 Megabytes of RAM”)
• 103 kilo (thousand, 1 kilogram = 2.2 pounds)
• 10– 2 centi (hundredth, 1 in. = 2.54 cm)
• 10– 3 milli (thousandth, “milliVolt”)
• 10– 6 micro (millionth, micron=micrometer=10-6m)
• 10– 9 nano (billionth, “nanosecond”)
• 10– 12 pico (trillionth)
• 10– 15 femto
Particle symbol rest energy in MeV
electron e 0.5109989
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
or mass in MeV/c2
Rest energies
of a proton + neutron = 938.27200+939.56533 = 1877.83733 MeV
while of a deuteron H2 = 1875.61269 MeV
difference = 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.
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)
),()(2
ZAMmZAZmc
Bnucleusnp
),()(2
ZAMmZAZmc
BatomneutralnH
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”)
AuZAM D ),(
Notice with Au Zmp + (A-Z)mn
this is essentially the same as the Binding Energy
Binding Energy Per Nucleon The binding energy per nucleon of a nucleus is the binding
energy divided by the total number of nucleons in the nucleus
Binding energy per nucleon = total binding energy of nucleus _____________________ number of nucleons in the nucleus
In the case of
helium
28.3 7.1 MeV ________ = 4
The great of nuclides have a value of around 8MeV per
nucleon
Peaks at ~8.795 MeV near A=60
for A>50
~constant
8-9 MeV
Bin
din
g e
ne
rgy p
er
nu
cle
ar
pa
rtic
le (
nu
cle
on
) in
Me
V
The key to this release is the chain reaction
produced by the free neutrons realesed
during fission when the mass of U-235
exceeds a critical level (about 52kg- a
sphere17cm in diameter
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
2)1( AAA
then AA
B (not constant!)
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.
Nuclear Binding Energy Curve
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 Iron Limit
The Most Tightly Bound Nuclei 62Ni .
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.
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 Semi-emperical Mass Formula
is an approximate fit to all this data
np mZAZmZAM )(),(
The 1st correction to this (the “mass defect” or binding energy)
AaE vV
~proportional to volume
A
B/A
(M
eV)
true enough for
very small A !
Volume term
But the nuclei at the surface have fewer nearest-neighbor bonds!
by something the nucleus’ surface area
3/2AaE ss
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!
Surface effect
Coulomb Energy The work done in collecting and compacting
total charge Q into a sphere of volume . 3
3
4R
potential at the surface of a spherical concentration of charge
with charge density: 3)3/4( r
Q
3/1)1( AZZaE cc
3/13/2)1(
AZZaAaAaB
csv
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
A = 127
isobars
-88
-86
-84
-80
-88
-86
-84
-80
52 54 50 56
Atomic number, Z
Ma
ss
de
fec
t D
(M
eV
)
Like the “closed shell” electronic configurations of the noble gases
we see “magic numbers” recurring…marking tightly bound,
extraordinarily stable nuclear configurations.
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:
4/3
4/3
0)(
A
a
A
a
p
p
A
=4He (with Z=N=2) and 16O (Z=N=8) are doubly magic!
even Z, N
odd Z, N
Z = N = 0
)()2/(
)1(
2
3/13/2
AA
ZAa
AZZaAaAaB
sym
csv
volume term
grows as
nuclei added
surface term
corrects for
surface nuclei
Coulomb term
repulsion due
to protons
symmetry term
drives number of
protons ≈ neutrons
pairing energy term
increased stability
for paired nuclei
2),()(),( cZABmZAZmZAM np
),(/)2/(
)1(),(
2
3/13/2
ZAAZAa
AZZaAaAaZAB
sym
csv
av=15.5 MeV
as=16.8 MeV
ac=0.72 MeV
asym=23.0 MeV
ap=34.0 MeV
Thank You