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Nuclear Physics
Elements and atoms
Nuclear Composition
Atomic Number and Atomic Mass
Isotopes, Isotones, Isobars & Mirror Nuclei
Mass Defect
Binding Energy
Binding Energy per Nucleon
The atom consists of two parts:
1. The nucleus which contains:
2. Orbiting electrons.
protons
neutronsNucleons
The term “nucleon” refers to either a proton or a neutron in the
nucleus
The term “nuclide” refers to a nucleus with a specific number of
protons and neutrons.
All matter is made up of elements (e.g. carbon,
hydrogen, etc.).
Atom of different elements contain different numbers of
protons.
The mass of an atom is almost entirely due to the
number of protons and neutrons.
……
XA
Z
Mass number
Atomic number
Element symbol
= number of protons + number of neutrons
= number of protons
XA
Z
A = number of protons + number of neutrons
Z = number of protons = number of electron
A – Z = number of neutrons
Number of neutrons = Mass Number – Atomic Number
Inside Atoms: neutrons, protons, electrons
Carbon (C )
Gold (Au)
Atomic number Z=6 (number of protons)
Mass number A=12 (number of protons + neutrons)
# electrons = # protons(count them!)
(atom is electrically neutral)
Atomic number Z = 79
Mass number A = 197
#electrons = # protons(trust me!)
Further layers of substructure:
If each proton were 10 cm across, each quark would be
.1 mm in size and the whole atom would be 10 km wide.
www.cpepweb.org
u quark: electric
charge = 2/3
d quark:
electric charge = -1/3
Proton = uud
electric charge = 1
Neutron = udd
electric charge = 0
Atomic Mass• Atomic mass = 1/12 of the mass of a neutral
Carbon atom (126C) or
• The mass of 1 carbon atom =
Because 1 mole of carbon means that 12 g
of atomic carbon contain Avogadro’s
number (N) of atoms.
Therefore
1 a.m.u.=1.66 x 10-27 kg
a.m.u.1N
12g
12
1
Unit of EnergyThe unit in which nuclear energy is expressed is
called electron volt.
The amount of energy acquired by an electron when
accelerated through a potential difference of 1
volt is called an electron volt (eV)
1 electron volt (or eV) = charge on electron x 1 volt
=1.6 x 10-19C x 1V
1eV =1.6 x 10-19 J
The bigger unit of nuclear energy is mega electron volt (MeV)
1MeV= 106 eV
Relation between a.m.u. and Energy
• According to Einstein’s equation
E= mc2
Here, m = 1a.m.u.=1.66 x 10-27 kg
and c = 3 x 108 ms-1
E = 1.484 x 10-10 J
E= 931.49 MeV
Energy Equivalent:
E = mc2 E = 931.49 MeV
The energy corresponding to 1 a.m.u. is 931.49 MeV
Particle Mass (kg) Mass (a.m.u.) Mass (MeV/c2) Electric Charge
(e)
Proton 1.6726 x 10-27 1.007276 938.28 +1
Neutron 1.6750 x 10-27 1.008665 939.57 0
Electron 9.1095 x 10-31 5.486 x
10-4
0.511 -1
11H atom 1.6736 x 10-27 1.007825 938.79 0
Properties of nucleons with different mass scale used
• Units:– The electric charge of an electron is -1 in these units.
– Mass units are “million electron volts” where 1 eV is a typical
energy spacing of atomic electron energy levels.
Nuclear Size
Nuclear diameter ~10-15 m
Atomic diameter ~10-10 m
AR3
3
4
3
1
AR 3
1
0 ARR
Where R0 is constant for all nuclei
and has a value = 1.2 fm (where 1fm = 10-15m)
If nucleus is spherical then
Isotopes,
The number of protons (atomic number) determines which element it belongs to. Atoms with the same atomic number but different number of neutrons are called isotopes.
Examples:(1) 1H, 2H and 3H (2) 35Cl and 37Cl
(3) 235U and 238U.
U235
92U
238
92
There are many types of uranium:
A
Z
Number of protons
Number of neutrons
A
Z
Number of protons
Number of neutrons
U235
92U
238
92
There are many types of uranium:
Isotopes of any particular element contain the same
number of protons, but different numbers of neutrons.
A 235
Z 92
Number of protons 92
Number of neutrons 143
A 238
Z 92
Number of protons 92
Number of neutrons 146
Isobars,
The nuclei of an atom having the same atomic mass but different number of protons are called isobars.
Example:40Ar and 40Ca.
Ar40
18Ca
40
20A 40
Z 18
Number of protons 18
Number of neutrons 22
A 40
Z 20
Number of protons 20
Number of neutrons 20
Isobars of any particular element contain the same number of nucleon,
but different numbers of protons.
Isobars,
Isotones,
The nuclei having the same number of neutrons are called isotones. e.g., 15N and 16O.
Example:15N and 16O.
N15
7O
16
8A 15
Z 7
Number of protons 7
Number of neutrons 8
A 16
Z 8
Number of protons 8
Number of neutrons 8
Isotones of any particular contain the same number of neutons.
Isotones,
• The nuclei having the number of proton
in one nucleus is equal to number of
neutrons in other nucleus or vice versa.
Mirror Nuclei
Example:
73Li and 74Be
Mass Defect
The nucleus is formed by bringing protons and neutrons together,
the mass of the nucleus so formed is less than the sum of the
masses of the constituent protons and neutrons.
Mass of 11H atom = 1.007825 a.m.u.
+ mass of neutron = 1.008665 a.m.u.
Expected mass of 21H atom= 2.016490 a.m.u.
However,
The measured mass of 21H atom=2.014102 a.m.u
Difference ΔM = 0.002388 a.m.u.
This mass difference is called mass defect and is
denoted as ΔM
Mass Defect ΔM = [Zm(11H)+ Nm(n)]-m( A
ZX) in a.m.u.
Hydrogen
atommH = 1.007825 a.m.u.
neutron mn = 1.008665 a.m.u.
Deuterium
atommD = 2.014102 a.m.u.
= 2.016490
The mass of a deuterium atom is less than the sum of the masses of a
hydrogen atom and a neutron
Mass Defect Cont…….
ΔM = 0.002388 a.m.u.
0 50 100 150 200 250 300-20
0
20
40
60
80
100
16
8O
Pack
ing
Fra
cti
on
Mass Number A
Packing Fraction
It is observed that the atomic masses are very close to
whole numbers, but they invariably differ from integers
By a small amount
This deviation of the atomic
mass from whole number is
expressed in the form of a
quantity called Packing
Fraction
Definition: It is defined as
the mass defect per nucleon
Packing
fraction A
AX
A
mf
A
Z
Packing Fraction Cont…….Packing fraction measures the comparative stability of the nucleus.
The smaller the value of f, the larger is the stability of the nucleus.
From Figure:
1. For very light nuclei f is maximum and hence these are
unstable nuclei.
2. As the value of A increses, f goes on decreasing till it becomes zero
for A=16. Therefore the packing fraction of 168O is zero.
3. Beyond A=16 with the increase of A the value of f decreases
(-ve value) and becomes maximum with further increase of mass
number , the value of f starts increasing (becoming less –ve) till
it again becomes zero at A>240. Thus, one can conclude that
the nuclei of the middle range are stable whereas the heavy nuclei
(A>240) are unstable.
4. All the nuclieds have the tendency to move from region of higher
f to the lower f. This is possible for lighter nuclei by means of fusion
and for heavy nuclei by means of fission
Binding Energy
E = mc2
The energy equivalent to the mass defect is called the binding energy
of the nucleus.
Definition: It can be defined as the energy required to separate a
Nucleus into its constituents particles.
Mass Defect ΔM = [Zm(11H ) + Nm(n)]- m( A
ZX) in
a.m.u.
EBE = Δmc2
= [{Zm(11H )+ Nm(n)}-m( A
ZX)] c2
= [{Zmp+ (A-Z)mn}- AZX] a.m.u.
= [{Zmp+ (A-Z)mn}- AZX](931.49 MeV/a.m.u)
Einstein’ mass energy equation:
Note: All masses are in atomic mass units
Hydrogen
atommH = 1.007825 a.m.u.
neutron mn = 1.008665 a.m.u.
Deuterium
atommD = 2.014102 a.m.u.
= 2.016490
The energy equivalent of the missing mass is called the binding
Energy of the nucleus The greater is the binding energy of a nucleus
more is the energy required to break it
Binding Energy Cont…….
EBE = Δmc2
= 0.002388 x 931.49 MeV= 2.224 MeV
Deuterium nucleus
=
Proton
Neutron2.224-MeV
gamma ray
The binding energy of the deuterium nucleus is 2.224 MeV.
A gamma ray whose energy is 2.224 MeV or more can split
A deuterium nucleus into protons and neutrons
Note: A gamma ray whose energy is less than 2.224 MeV
cannot do this.
Range of BE from 2.224 MeV for 21H (deuterium)
To 1640 MeV for 20983Bi (an isotope of metal bismuth)
Numerical
.
Q1. The binding energy of the neon isotope 2010Ne is 160.647 MeV
find its atomic mass. [Ans=19.992 a.m.u.]
Q2. The mass number of 3517Cl is 34.9800 a.m.u. calculate its binding
energy. Given
Mass of =1.007825 a.m.u.
Mass of proton = 1.007276 a.m.u.
Mass of neutron = 1.008665 a.m.u.
Binding Energy per Nucleon
Definition: The amount of energy required to release a
nucleon from the nucleus is called binding energy per nucleon.
Binding energy per nucleon
(BE/A) = Binding Energy/ Total number of nucleons in the nucleus
BE/A = [{Zm(11H )+ Nm(n)}-m( A
ZX)] c2 /A
= [{Zmp+ (A-Z)mn}- AZX]/A a.m.u.
= [{Zmp+ (A-Z)mn}- AZX](931.49 MeV/a.m.u)/A
Binding energy per nucleon of a nucleus determines
The stability of the nucleus.
If BE/A of a nucleus is less, the nucleus is less stable
whereas the nucleus is more stable if its BE/A is higher.
The left figure shows the
binding energy per nucleon
as a function of mass
number.
Iron is the most stable
element.
Large amount of energy
(yield) is released by fusion
of light elements into
heavier elements or by
fission of heavy elements
into lighter elements.
The BE/A of 21H is 2.2
MeV/2 =1.1 MeV/nucleon
The BE/A of 20983Bi is
1640 MeV/209 =7.8
MeV/nucleon
Binding Energy per Nucleon Cont…….
Fusion
+ + Energy+ + Energy
Fission
1. BE of all stable nuclei
is positive
2. Peaks in BE at A=4n
for light nuclei
corresponds to 42He, 8
4Be, 126C, 16
8O, 2010Ne
and 2412Mg.
3. In the medium mass
region, i.e. between
A=25 to A=125 the
average BE/A is
constant.
4. There is slight decrease
in the BE/A for A>125
Binding Energy per Nucleon Cont…….
Fusion
+ + Energy+ + Energy
Fission
Iron (Fe) has most binding energy/nucleon. Lighter have too
few nucleons, heavier have too many.
BIN
DIN
G E
NE
RG
Y i
n M
eV/n
ucl
eon
92238U
10
Binding Energy Plot
Fission
Fusion = Combining small atoms into large
Fission = Breaking large atoms into small
9/26/2011 33
Binding Energy per Nucleon Cont…….
Two remarkable conclusions
1. If we can somehow split a heavy nucleus into two medium size one,
each of the new nuclei will have more binding energy per nucleon
than the original nucleus did. The extra energy will be given off
and it can be a lot.
For Example: If the uranium nucleus 23592U is broken into two smaller
Nuclei, the binding energy difference per nuclei is about 0.8MeV
Therefore, the total energy given off
(0.8 MeV/nucleon)(235 nucleons) = 188MeV.
This is the energy produced in a single atomic event.
Note: In ordinary chemical reaction involve the rearrangement of
the electrons in atoms and liberate only a few eV per reacting atom.
Here the splitting of heavy nucleus which is called Fission, involves
100 million times more energy per atom then say the burning of coal
and oil
Binding Energy per NucleonCont…….
2. If we join two light nuclei together to give a single nucleus of
medium size. It also means more binding energy per nucleon
in the new nucleus.
For Example: If two 21H deuterium nuclei combine to form a 4
2He helium nucleus over 23 MeV energy is released Such a process is
called nuclear fusion.
Note: Nuclear fusion is the main source of energy in sun and
other stars.
Numerical
Q1. The mass number of 3517Cl is 34.9800 a.m.u.at is the BE/A Given
Mass of =1.007825 a.m.u.
Mass of proton = 1.007276 a.m.u.
Mass of neutron = 1.008665 a.m.u.
Q2. (a) Find the energy needed to remove a neutron from the nucleus
of the calcium isotope 4220Ca.
(b) Find the energy needed to remove a proton from 4220Ca
nucleus. Why are these energies different?
Subatomic particles interact by exchanging integer-spin
“boson” particles. The varied interactions correspond
to exchange of bosons with different characteristics.
Force Strength Carrier Physical effect
Strong nuclear 1 Gluons Binds nuclei
Electromagnetic .001 Photon Light, electricity
Weak nuclear .00001 Z0,W+,W- Radioactivity
Gravity 10-38 Graviton? Gravitation
Nuclear ForcesNuclear Forces are attractive in nature. This is true as result
of positive BE for all nuclei. If the BE is –ve, then the nuclei
would not be stable and the repulsive forces would be existing in
the nucleus. The nuclear forces are very strong in nature
Meson Theory of Nuclear forces
In 1936, A Japanese Scientist Yukawa predicted that
The nuclear forces must be arising because of some
constant exchange of particles between the nucleons.
According to him, all nucleon i.e. protons and neutrons
could be held together if they share these particles.
The particles were named as mesons.
Later these particles were detected in1947, and were
found to have a mass 270 times mass of electron
According to this theory:
1. All nucleons consists of identical cores surrounded
by a cloud of one or more mesons.
2. Mesons may be neutral π0 or carry either charge (π+or π-).
3. The difference between proton and neutron is supposed to
be in the composition of their respective meson clouds.
4. The forces that act between one neutron and another neutron
and proton-proton are the result of the exchange of neutral
meson (π0) between them. The forces between neutron and
proton is due to the exchange of charged mesons (π+or π-)
between them.
Meson Theory of Nuclear Forces
Neutron Proton
π+ π- and π0
Meson
Cloud
-+
+
+
+
--
-+
--
-+
++
http://en.wikipedia.org/wiki/Nuclear_force
1. A neutron emits π- meson and convert into proton.
n p + π-
then, the proton absorb this π- meson and convert into neutron.
i.e. p + π- n
Similarly
p n + π+
and n + π+ p
Thus the protons and neutrons continuously exchange their
Nature by absorbing and emitting π mesons.
π0 is exchanged not between neutrons and protons but between
Proton-proton or neutron-neutron as explained below.
p p + π0
p + π0 and
n n + π0
n + π0
Analogy
Imagine two interacting nucleons as two dogs and mesons as
a piece of a bone which each of them is trying to snatch from
each other. The bone can be considered to be rapidly exchanged
between the two dogs as neither of them can part with it. Thus the
bone keeps the two dogs bound together.
Similarly, the exchange of π meson keep the two nucleon bound together in the nucleus.
Properties of nuclear forces
1. Nuclear Forces are short range.
2. Nuclear forces are charge independent.
3.Nuclear forces are strongest forces in nature.
4. Nuclear forces are spin dependent.
5. Nuclear forces are saturated forces.
6. Nuclear forces are non-central.
7.Exchange character.