Crackiitjee.in.Phy.ch26
Short Description
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Description
cr
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ki
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Table of Contents Nuclear Physics...................................................................................................................... 3 Properties of Nucleus ............................................................................................................. 4 Nucleus Density ..................................................................................................................... 5 Nuclear Quantum States.................................................................................................................5 Special Points .................................................................................................................................6
Nuclear Force ........................................................................................................................ 7 Four Fundamental Forces ...............................................................................................................7 Properties of Nuclear Forces ...........................................................................................................7
ee .in
Yukawa - Meson Exchange Theory ...................................................................................... 8 Different Types of Exchange ...........................................................................................................9 Dog Anology ..................................................................................................................................9
Nuclear Reactions................................................................................................................ 10
itj
Short Representation of Nuclear Reaction..................................................................................... 10
ki
Laws Governing Nuclear Reaction ................................................................................................. 10 Classification of Nuclear Reaction ................................................................................................. 11
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Q- Value of a Nuclear Reaction ..................................................................................................... 12
Radioactivity ....................................................................................................................... 13
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Facts about radioactivity .............................................................................................................. 13
Various modes of Delay ....................................................................................................... 15 Alpha Decay ................................................................................................................................. 15 Facts from analysis of energy spectrum of -particles ................................................................... 16 Beta () decay ............................................................................................................................. 17 Some Definitions .......................................................................................................................... 19 Electromagnetic Process (Gamma Ray) ......................................................................................... 21 Special Points ............................................................................................................................... 22
Neutrino/Anti-neutrino Hypothesis ...................................................................................... 27 Orbital-K-electron Capture ........................................................................................................... 28
1
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Nucleon Emission ................................................................................................................ 29 Proton Emission................................................................................................................... 30 Binding Energy .................................................................................................................... 30 Q - Equation ........................................................................................................................ 32 Special Points ...................................................................................................................... 34 Threshold Energy ................................................................................................................. 36 Successive Transformation................................................................................................... 37
cr
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Examples ............................................................................................................................. 39
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Nuclear Physics Nuclear physics deals with nucleus and nucleus related phenomenon. Nucleus is the central core of every atom in which entire positive charge and almost entire mass of the atom is concentrated. Volume of nucleus is very small as compared to atomic volume. Nucleus consists of neutrons and protons which are together referred as nucleons. Neutron is electrically neutral and proton is positively charged. Mass of neutron is slightly more than mass of proton. Representation of nucleus:
Z = Atomic number A ZX
symbol of element
In any atom,
ee .in
A = mass number
ki
Number of neutrons = A - Z
ac
∴
itj
Number of proton (Z) = Number of electrons
cr
Isotopes: Atoms having different mass numbers but belonging to the same chemical element and having the same atomic number are known as isotopes. Isobars: Atoms having the same mass number but different atomic numbers are known as isobars. Isotones: The having same number of neutrons are called isotones. Mirror Nuclei: Nuclei having same mass number A but with the proton number (Z) and neutron number (A - Z) interchanged are called mirror nuclei.
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Properties of Nucleus 1. Charge Qnucleus = Ze
2. Size (Radius) r = R0 A1/3 A = mass number
3. Density =
itj
If A is the mass number,
ee .in
R0 = 1.2 × 10-15 m = 1.2 fermi
ki
Mass of nucleus = 1.67470 × 10-11 A kg
ac
Volume of nucleus = r3 = (R0 A1/3)3 = 7.23 × 10-45 A m3
cr
Density of nucleus = 2.3 × 1017 kg/m3
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Nucleus Density
Nucleus Density
This is constant for all isotopes
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This is of order of density of black holes
Nuclear Quantum States
The various discrete energy levels in which nucleus is capable of existing, are called nuclear quantum states.
ki
itj
When nucleus makes transition from one state to another, the difference of energy of two levels is emitted in the form of electro-magnetic radiation which always lies in region irrespective of nucleus.
ac
Spin Motion
+
cr
Nuclear Spin:
Orbital Motion
This is the angular momentum arising out of the spin as well as the orbital motion of the nucleus.
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Special Points (i)
Total angular momentum = The energies of beta rays and gamma rays emitted from nuclei are typically of the order of 1 MeV. But these processes are transitions of nucleus from one state to another so their energies are differences between nucleon energies in two different states, the actual nucleon energy should be much larger. A crude approximation can be electrostatic energy E required to insert a proton into a nucleus.
itj
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(ii)
The nucleons also behave like tiny bar magnets having intrinsic magnetic moment. Therefore, the nucleus possesses a resultant magnetic moment. We know that if there is no external torque acting on a system, its angular momentum is conserved. Since an isolated nucleus is such a system, its angular momentum is one of its constant properties. In quantum physics, conserved quantities are represented by quantum numbers. The quantum number for total angular momentum of a nucleus is I and these two are related by
ki
E=
ac
For a medium weight nucleus (Z = 50, A = 120) E = 13 MeV
cr
This much energy would be released if the proton were allowed to come out of the nucleus, but still it does not ordinarily come out. This means that it is bound in the nucleus by even more energy. Roughly we can say that energies of nucleons in the nucleus are of the general order of 10 MeV. Since the velocity of a 10 MeV nucleon is only 15% of the speed of light, this mean that relativistic effects can be ignored in considering the motion of nucleon.
(iii)
The motion of nucleon in the nucleus is governed by the laws of quantum physics. This can be understood by calculating de-Broglie wavelength of nucleons with energy of about 10 MeV. λ=
= 6
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= 9.3 × 10-15 m This is clearly of the order of the size of a nucleus, so wave nature of matter is to be considered. Hence nucleus can’t be treated as classical system. It is to be treated as quantum system.
Nuclear Force Nuclear force is responsible for the stability of nucleus.
Four Fundamental Forces (i) Gravitational interaction
(iii) Nuclear interaction (iv) Weak interaction
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(ii) Electromagnetic interaction
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Nuclear force is the strongest force and gravitational force is the Strongest.
ac
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Gravitational force < Electromagnetic force < Nuclear Force
Properties of Nuclear Forces
(ii)
(iii) (iv)
(v)
Nuclear forces are attractive in nature. The magnitude which depends upon inter nucleon distance is of very high order. For small nucleon separation, it is repulsive. Nuclear forces are charge independent. Nature of force remains the same whether we consider force between two protons, between two neutrons or between a proton and a neutron. These are short range forces. Nuclear forces operate between two nucleons situated in close neighborhood. Nuclear forces decrease very quickly with distance between two nucleons. Their rate of decrease is much rapid than that of inverse square law forces. The forces become negligible when the nucleons are more than 10-12 cm apart. Nuclear forces are spin dependent. Nucleons having parallel spin are more strongly bound to each other than those having anti-parallel spin.
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(i)
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(vi) (vii)
Nuclear force does not obey principle of superposition. When distance between nucleons becomes less than 0.5 fermi, the nuclear forces become strongly repulsive. (viii) The nuclear forces show saturation properties i.e. each nucleon interacts with its immediate neighbors only, rather than with all the other nucleons in the nucleus. (ix) Nuclear forces are non-central forces. This shows that the distribution of nucleons in a nucleus is not spherically symmetric.
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Yukawa - Meson Exchange Theory According to this theory, a nucleon consists of a core surrounded by a cloud - mesons which may be charged or neutral.
-
ac
+
ki
itj
- Meson Particle
0
cr
There is continuous exchange of - meson particles among nucleus which is responsible for nuclear force acting between nucleuses.
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Different Types of Exchange P-P Force: It is the force between two neighbouring protons. It is due to the exchange of an 0 meson between them. It is represented in the form of a reaction as follows: P1 P2 + 0
P1* + 0 P 2*
Proton P1 emits 0 and gets converted into P1, a proton with different co-ordinates. This 0 is absorbed by P2 which also get converted into a new proton P2.
N1 N2 + 0
N1* + 0 N 2*
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N-N Force: It is the force between two neutrons. It is also due to exchange of 0 between them.
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Dog Anology
itj
P-N Force: It is the force between proton and neutron situated close to each other.
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This analogy helps us to understand the nature of nuclear forces. Consider the two interacting nucleons to be two dogs having bone clenched in between their teeth very firmly. Each one of these dog wants to take the bone. So we cannot separate them easily. They seem to be bound to each other with a strong attractive force. The meson plays the role of the common bone in between.
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Nuclear Reactions
+
Target
C
nucleus
+
O
Product
Outgoing
nucleus
nucleus
radiation
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Particle
P
Compound
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Incoming
T
itj
I
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There is no difference between chemical reaction and nuclear reaction. In chemical reaction, it is outermost electrons which take part where as in nuclear reaction; it is nucleus which takes part. There is no difference between nuclear energy and chemical energy. The difference is of magnitude only. A radioactive substance breaks up by emitting radiation. The daughter nucleus, left behind, has different physical and chemical properties and is assigned a new place in the periodic chart. Thus, radioactivity is the phenomenon by which a substance gets converted into another one. This change can be brought about by artificial method, by bombarding a given nucleus with some radiation. The particle constituting the incident radiation must possess sufficient kinetic energy so as to penetrate into the given nucleus. As they enter the given nucleus, a compound nucleus is formed which is generally unstable. The compound nucleus then breaks up to produce nucleus by emitting radiation. The process is schematically represented as
Short Representation of Nuclear Reaction When nitrogen is bombarded with - particle, following reaction takes place: 2He
4
+ 7N14 9F18 8O17 + 1H1
Laws Governing Nuclear Reaction (i) (ii)
Law of Conservation of Charge: The electric charge involved in a nuclear reaction must be same before and after the reaction. Law of Conservation of number of nucleus: The total number of nucleons involved in a nuclear reaction must be same before and after the reaction.
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(iv)
(v)
(vi)
Law of Conservation of energy: The total energy (rest + K.E.) of the reacting particles must be equal to the total energy of the product particles. Law of Conservation of Linear Momentum: The total linear momentum of the reacting particles must be equal to the total linear momentum of the product particles. Law of Conservation of Angular Momentum: The total angular momentum of the reacting particle must be equal to the total angular momentum of the product particles. Law of Conservation of Spin: The total spin of the reacting particles must be equal to the total spin of the product particles.
Classification of Nuclear Reaction I.
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(iii)
Elastic Scattering: The incident particle gets deflected without any change in its energy, i.e. 4
+
70Au
197
70Au
197
+ 2He4
itj
2He
Inelastic Scattering: If the bombarding particle passes close to target, it gets deflected. Due to strong repulsion, the target particle also acquires some energy. So, the energy left with the scattered particle is less than that it had initially,
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II.
ac
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The bombarding particle passes sufficiently large distance away from the target nucleus so as to get repulsion which changes its direction of motion without any change in its direction of motion without any change in its energy.
1 1H
+ 3Li7 3Li7 + 1H1
means existence of 3Li7 in one of its excited states. III.
Simple Capture: The incoming particle is captured by the target nucleus. The product nucleus which is generally in the form of excited state decays to the ground state by emitting - rays 1 1H
IV.
+ 6C12 7N13 7N13 + hf
Disintegration: The intermediate compound nucleus breaks up and results in a product nucleus and an outgoing particle. The product nucleus has different 11
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chemical properties as compare to the target particle. Majority of nuclear reactions belong to this category. (a) Disintegration by - particles 2He
+ 5B10 7N13 6C13 + 1H1
4
(b) Disintegration by protons 1 1H
+ 5B11 6C13 4Be8 + 2He4
(c) Disintegration by neutrons 1
+ 5B10 5B11 3Li7 + 2He4
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0n
(d) Disintegration by deutrons 2 1H
+ 5B10 6C12 2He4 + 2He4 + 2He4
(e) Photo disintegration
4Be
9
+ 0n 1
ki
itj
+ 4Be9
I
+
T
P
+
O
+
Q
I = Incoming particle/radiation
cr
Where,
ac
Q- Value of a Nuclear Reaction
T = target nucleus P = Product nucleus O = outgoing particle/radiation
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Radioactivity Facts about radioactivity Radioactivity is spontaneous disintegration of nucleus The phenomenon of radioactivity is statistical in nature Number U-238 nucleus in 1-gm 1 x6x1023 238
= 2.5 x 1021
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The phenomenon of radioactivity is unaffected by external conditions like pressure, temperature etc.
itj
The phenomenon of radioactivity is a nuclear phenomenon i.e. the seat of
ki
radioactivity lies in the nucleus.
ac
Law of conservation of charge holds in radioactivity
cr
Total change before disintegration = Total change after disintegration Radioactivity is a random process Each of the products of disintegration is a result of a new element having different physical and chemical properties as compared to that as parent nucleus Law of Radioactivity Rate of disintegration of radioactive substance at any instant is directly proportional to the number of different nucleus dN Rate of integration dt
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N = Number of nuclei at any instant of time dN N dt
dN N dt
= Disintegration constant Negative sign indicates that the population of radioactive nuclei is decreasing with time
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Let at t = 0, N = No dN dt N
Integrating both sides we get:
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N
ac
N = Noe-t
itj
No
R
dN dt N 0
ki
N
Population is decreasing exponentially
0
+ 14
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Various modes of Delay Radioactivity
Natural Radioactivity
Artificial Radioactivity
Alpha Decay
-decay
A
AZ—21D 4
ZX
+ 2He4 + Q -particle
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daughter nucleus
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It is the phenomenon of emission of an -particle from a radioactive nucleus. When a nucleus emits an alpha particle, its mass number decreasing by 4 and charge number decreases by 2.
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ac
ki
-particle is nothing but doubly ionised helium nucleus. It consists of two protons and two neutrons. The nucleus of a He is positively stable. Its binding energy is 28.3 MeV.
From Q value concept Q = KED + KE
From conservation of linear momentum PD = P Q=
PD2 KE 2MD
P2 KE 2MD 15
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2MKE KE 2MD
M KE KE D MD M
Or KE
Q
A 4 A Q
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A 200, KE = 0.96Q
Facts from analysis of energy spectrum of -particles
itj
ki
ac
iii)
Energy of -particles from a specific radioactive nucleus is definite Energy spectrum of R-particles has a fine spectrum i.e. many particles with a small difference of energies are emitted. The energy spectrum of -particles of some of the nuclei has a main group of energy together with some particles of very high energy.
cr
i) ii)
Classical physics rules out the possibility of emission of -particle from nucleus. -decay is explained by using Quantum Physics by invoking the concept of barrier penetration or tunnel effect.
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Beta () decay a) - decay b) + decay - decay Electron emission
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It is the phenomenon of emission of electrons from a radioactive substance These electrons are emitted when a neutron is converted into proton, an electron and anti-neutron Mass number remains unchanged but nuclear charge increases by one unit. Element moves from one place to the right in the periodic table. Anti-neutrino ( ) has zero mass and zero charge and it is postulated to balance nuclear spin
0n
1
0n
1
+ 0n 1 +
decay
+ -1e0 + + Q Q = Q-value of β- decay process Q = (Mx-My)C2 = ΔmC2 Δm = mass defect pr process - decay is caused by weak interaction For example: A ZX
A
ac
ki
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ZHY
cr
decay
6C
14
7N
14
+-1e0 +
decay 13Al
20
14Si
39
+-1e0 +
decay 56Ba
141
57La
141
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T1/2 = 18 min decay 57La
141
58Ce
141
T1/2 = 3.7 Hr decay 58Ce
141
59Pr
141
T1/2 = 28 days
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+ decay Positron emission
It is known as positron emission In + decay, mass number remains same but atomic number decreases by one unit and element moves one place to the left in the periodic table. 1P
1
0n
decay
1
+-1e0 +
itj
cr
ac
ki
A 0 ZXA Z-1Y +-1e + + Q Q = (Mx-My-2me)c2 Neutrino is postulated to balance spin Positron ejected from nucleus quickly collides with electron in the surroundings and the two particles annihilate to release two γ ray photons. For example:
decay 10Ne
19
9f
19
+-1e0 +
decay 6C
11
5B
11
+-1e0 +
18
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Some Definitions Life time of any radioactive nuclei can be anything from zero to infinity whereas radioactive sample will have infinite life time. Another form of Radioactive Law: Equal fractions decay in equal time Basic terms a) Half time (T1/2) It is defined as the time during which population of radioactive nuclei is reduced to half. N = Noe-t At t = T1/2 , N=No/2
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No/2 = Noe-t T1/2 = 2/ 2 = 0.6931
ki
itj
T1/2 = 0.6931/
ac
b) Average Life
cr
av =
1 Sum of life times of all nuclei = No 1 av = No
No
tdN 0
No
tdN 0
dN = -Ndt = -Noe-tdt av =
1
19
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c) Activity of radioactive substance This is defined as rate of disintegration (dps) A
dN N dt
A = -Noe-t Ao = Initial Activity
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A = Aoe-t
A
cr
ac
ki
itj
A0
0
t
S.I. Unit of Activity is Becquerel (Bq) 1 Bq = 1 dps
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Other units of radioactivity
Curie (Ci)
Rutherford (Rd)
1 Ci = 3.7 x 1010 dps This is also the activity of 1 gm of radium
1 Rd = 106 dps
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1 Ci = 3.7 x 104 Rd
Electromagnetic Process (Gamma Ray)
ZX
-decay
cr
A
ac
ki
itj
It is the phenomenon of emission of gamma ray photon from radioactive nucleus. This occurs when an excited nucleus makes a transition to a state of lower energy The range of energy is 0.1-1.5 MeV and it emitted as electromagnetic radiation of very short wavelength called γ-rays. A ZX
+ hf
daughter nucleus
-ray
In gamma ray, neither the proton number nor the neutron number changes. Only the quantum states of the nucleon changes. There is no difference between the γ-rays and X-rays, the difference lies in the origin only. γ-rays are of nuclear origin and X-rays are of atomic origin.
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Integration of -rays with matter
Compton Effect
Photoelectric Effect
This takes place when -rays strike with an electron bound to a nucleus
Pair Production
When -rays have high energy (>1.02MeV) an electron-positron pair is produced simultaneously due to the interaction between electric field of the nucleus and gamma rays.
ac
Special Points
ki
itj
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This takes place when -rays proton strikes a free or loosely bound electron
cr
1. An excited nucleus sometimes comes to ground state by emitting many gamma rays successively. 27CO
60
- decay
E1 = 1.17 MeV E2 = 1.33 MeV
Energy level diagram showing the emission of rays by 27C60 22
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cr
ac
ki
itj
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2. Gamma rays and X-rays have same wavelength. These two can't be distinguished. The difference lies in their origin only. 3. The nuclear processes which occur spontaneously are called decay. When a process occurs spontaneously, changing a system from one state to another, conservation of energy requires that the final state be the lower energy than the initial state and that the difference in these energies be sent out of the system in some way. In all cases, this is accomplished by energetic particles being emitted. 4. The graph shows possible decay of different nuclei:
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Properties of -rays, -rays and -rays
6.
Specific Activity
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5.
Useful Relation
cr
7.
ac
This is defined as activity of radioactive substance per unit mass
n
1 N No 2
n = number of half lives =
8.
t T1/2
Number of nuclei decayed upto limit N’ = No – N N’ = No (1 – e-t)
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N’
ee .in
N0
t
ki
9. Probability of survival (Ps)
itj
0
N et No
cr
ac
Probability of survival =
10. Probability of decay (Pd) Ps + Pd = 1 Pd = 1 - Ps = 1 - et
11. Radioactive substance is being produced at constant rate (P) dN Rate of accumulation dt 25
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= Rate of production – Rate of disintegration dN P N dt
N
P (1 et )
N
ac
ki
itj
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0
t
cr
It is clear from the graph that in producing radioactive isotopes, it clearly does not pay to extend the production period over more than a few half lives.
12. Relation between av and T1/2 T1/2 = 0.6931/ = 0.6931 x
1
or T1/2 = = 0.6931 x av
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Neutrino/Anti-neutrino Hypothesis Energy spectrum of -particle was found to be continuous Lead to the discovery of Neutrino/Anti-neutrino
N
Kinetic Energy of - Particles
Kmax
ac
0
ki
itj
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Relative Number of - Particles
cr
Nuclear reaction representing -decay was found to violate spin conservation From conservation of linear momentum P P P 0
Isolated proton can’t decay into neutron. Energy consideration does not allow it. Mass of neutron is more than the mass of proton. Neutron can decay into proton outside the nucleus. Detection of neutrino/anti-neutrino is very difficult as they interact weakly with matter can even penetrate without being absorbed.
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Orbital-K-electron Capture Nucleus may capture an orbital electron and convert a proton into a neutron with emission of a neutrino +-1e0 + Final product after decay is a nucleus whose charge is Z-1 1P
1
0n
1
K electron A
ZX
+e
A Z-1
-
+
Capture
ki
itj
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Orbital electron capture and β+ decay are competitive process. Q = (Mx – Mc)C2 This process increases the N/P ratio. Electron capture is not common. It occurs in nuclei where the N/P ratio is low and the nucleus has insufficient energy for positron emission (1.02 MeV) For example: For example: 7 0 7 4Be +-1e 3Li + 40 40 +-1e0 18Ar + 19K
cr
ac
In situation where the difference in mass of parent and daughter nuclei is equivalent to more than the required 1.02 MeV for positron emission, both positron emission and K capture occur,
48 23V
22Ti
48
K capture(42%)
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Nucleon Emission This form of decay is rare and only takes place with highly energetic nuclei This is because the binding energy of the neutron in the nucleus is high (about 8 MeV) Example neutron 37Kr
87
37Kr
87
+ 0n 1
37Rb
87
ee .in
emission
38Sr
87
cr
ac
ki
itj
The combination of two neutrons and two protons is particularly strong because of pairing effects. If the last two protons and two neutrons of a nucleus are bound by less than 28.3 MeV, then emission of an alpha particle is energetically possible. The Q-value of - decay is given by Q = [M(ZXA) - M(Z-2DA-4) - M(2He4)]c2 If Q>0 then alpha decay is possible. The disintegration energy Q appears in the form of kinetic energy in the daughter nucleus and the alpha particle. The fraction of the disintegration energy carried off by the alpha particle can be calculated by applying conservation of energy and momentum.
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Proton Emission Except for nuclei is a very high energy state, proton emission is unlikely as the energy needed to remove a proton is about 8 MeV
Binding Energy The energy corresponding to mass defect (Δm) of nucleus is known as binding energy.
+
cr
BE = Δm c
Nucleus
ac
Nucleons at infinite separation 2
ki
itj
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m = mass defect per nucleon = packing fraction (P) A
If the mass of the nucleus ZXA is M, then the mass defect is Δm = *Z mp + (A - Z) mn - M] where mp and mn are the masses of the proton and neutron respectively. And
Binding energy = Δm.c2 = [Z mp + (A - Z) mn - Mat] c2
Where Mat is the mass of the atom and mH is the mass of hydrogen atom. However, in this case, the small difference due to the binding energy of the electrons with the nucleus neglected.
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B.E.
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The binding energy per nucleon = B
(i)
ki
itj
ee .in
A
Elements having atomic weights between 40 to 120 have roughly a constant A
(iii)
A
of 56Fe. Binding energy per nucleon is an indication of stability. The more the binding energy per nucleon, the greater is the stability. Fe (Iron) is the most stable nuclei having binding energy per nucleon 8.89 MeV.
cr
(ii)
ac
value of B around 8.5 MeV. In fact the maximum value of B is 8.79 MeV in case
The average value of B of nuclei having atomic weight less than 40 and greater A
than 120 is smaller than 8.5 MeV. For heavier nuclei the value of B decreases to A
around 7.5 MeV. (iv)
There exists a cyclic occurrence of maximum value of B for elements having A
mass number multiple of four in the low mass number region. This is due to pairing effect.
31
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Q - Equation The analytical relationship between the kinetic energy of the projectile, outgoing particle and nuclear disintegration energy Q is called Q –equation. I+T
P+O+Q
I = Incoming particle T = Target nucleus P = Product nucleus
cr
ac
ki
itj
ee .in
O = Outgoing particle
I
=
O
+
32
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ee .in
The momentum triangle is indicated below:
We get that
PP2 = PI2 + PO2 = 2 PI PO cos We know that
itj
2 KE = P
ki
2m
2 MP KEP = 2 mI KEI + 2 mO KEO – 2.2 mIKEImOKEO cos
Or
2 mIKEImOKEO cos KEP = mIKEI + mOKE O MP MP MP
cr
ac
∴
Also from the definition of Q –value, we have Q = KEP + KEO – KEI Making substitution, we get Q = KEo 1 mO - KEI 1 mO - 2 mIKEImOKEO cos
MP
MP
MP
This is known as Q –equation.
33
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If outgoing particle is scattered at 90 with respect to the line of motion of incoming particle, then =
and Q –value equation reduces to 2
Q = KEo 1 mO - KEI 1 mO MP MP
Special Points
ee .in
itj
Q = (mI + mT) – (mO + mP)
ki
where the masses are the nuclear masses, and Q is in mass units. (a) Consider the reaction, 23 92
ac
(ii)
Since the Q –equation is based on mass energy conservation in a nuclear reaction, it holds for all type of nuclear reactions. The exact masses can be replaced by the corresponding mass numbers in many applications without significant error. For very accurate calculations, the neutral atomic masses are used. Let us see how we can use the isotopic masses (neutral atomic masses) to obtain Q value in alpha and beta decay reactions. For convenience, let the element symbol represent the isotopic mass. We recall that
U
234 90
Th + 42 He
Q = (U – 92e) – [(Th – 90e) + (He – 92e)]
cr
(i)
where e stands for electronic mass. i.e.
Q = (U – 92e) – [(Th + He – 92e)
= [U – (Th + He)] ∴ Isotopic masses can be used to evaluate Q. (b) Consider the reaction,
14 6
C
β- decay
14 7
N
0 1
e 00 v
34
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Q = (C – 6e) – [(N – 7e)] = (C – 6e) – (N – 6e) =C–N ∴
Isotopic masses can be used to evaluate Q.
(c) Consider the reaction,
C
63 29
Cu
Q = (Zn – 30e) – [(Cu – 29e) + e] = (Zn – 30e) – (Cu – 29e + 2e)
e 00 v
itj
= Zn – (Cu + 2e)
0 1
ee .in
63 30
Β+ decay
cr
ac
ki
Thus we see that the isotopic masses can be used to compute Q value in a positron decay reaction, provided two electron masses are included with that of the product particle.
35
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Threshold Energy The threshold energy is defined as the minimum kinetic energy of the incident particle which will initiate an indoergic reaction. This is expressed by KEth. We can calculate the threshold energy by using the centre of mass coordinate system, the linear momentum is always zero, before and after the system. Let KE’I be the kinetic energy of the incident particle in centre of mass coordinate system. An endoergic reaction is possible if KE’I | Q | Let Mred be the reduced mass of incident particle and the target nucleus, Then MIMT (MT MI )
KE’I =
1 1 MIMT Mred v12 = v 12 2 2 (MT MI )
1 2
or
1 M1 v12 2
∴
Threshold energy
v 12 | Q |
ki
MIMT (MT MI )
itj
∴
ee .in
Mred =
cr
ac
MIMT |Q| (M M ) T I
KEth = MIMT | Q | (MT MI )
36
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Successive Transformation λ1 A1
A2
λ2
A3
λ3
λn-1
……………….. An-1
An
λn
At t = 0 N1 = N1 (0) N2 (0) = N3 (0) = …………………… = Nn – 1 (0) = Nn (0) = 0
dN1 = - λ N1 dt
ee .in
The family of difference
dN2 = λ1 N1 – λ2 N2 dt
equation representing
ki
itj
successive transformation.
ac
dNn = λ N – λ N n-1 n-1 n n dt
cr
N1 = N1 (0) e 1t
1 1 N (0) 1t N1 (0) e 2t e 1 2 1 2 1
N2 =
=
1N1 (0) 1t e e 2t 2 1
N3 = C31 e 1t + C32 e 2t + C33 e 3t C31 =
12N1 (0) (2 1 )(3 1 )
37
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12N1 (0) (1 2 )( 3 2 )
C33 =
12N1 (0) (1 3 )( 2 3 )
cr
ac
ki
itj
ee .in
C32 =
38
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Examples Example 1: Calculate the activity of K40 in 100 kg man assuming that 0.35% of the body weight is potassium. The abundance of K10 is 0.012%, its half life is 1.31 x 109 years. Solution: Total mass of potassium is 100 kg man = 100 x 0.35 x 10-2 = 0.35 kg Mass of K40 = 0.350 x 0.012 x 10-2 = 4.20 x 10-5 kg
N
6.023x1023 x4.2x105 6.32425x1020 40
ki
6.023x6.32425x1020 1.31x109 x365x24x60x60
itj
Activity of K40 = N = (0.693/T1/2)N
ee .in
From Avogadro’s hypothesis, 1 kg atom of a substance consists of 6.023 x 10 23 atoms. Hence, the number of K40 atoms is given by
ac
= 1.061x104 disintegrations/sec
cr
=0.287 micro-curie Ans.
Example 2: If a radioactive material initially contains 3.0 milli grams of uranium (U234), how much it will contain after 150000 years? What will be its activity at the end of this time? (T1/2 = 2.50 x 105 years, = 8.80 x 104 per sec). Solution: Let m be the mass of uranium U234 remaining after 150000 years Number of atoms in mass m = N = m x 6.023 x 1023 / 234 No = 3 x 10-6 x (6.023 x 1023 / 234) Substituting the numerical values in equation N = Noe-t, we have 39
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mx6.023x1026 6.023x1026 3x106 x e (0.693 / 2.5x105 )x150000 234 234
or m = 3 x 10-6 e-0.416 0.416 = loge (3 x 10-6/m) or m = 1.98 x 10-6 kg Ans. Activity of uranium at the end of 150000 years =
dN N dt
= 8.8 x 10-14 x 1.98 x 10-6 x 6.023 x 1023/234
ee .in
= 4.5 x 105 disintegrations/sec We know that one mole contains Avogadro number of atoms.
N = N0 e-λt
ac
=
ki
Initial number of atoms = N0 =
itj
Number of atoms in mass (m) = N =
Ans.
cr
m = 1.98 × 10-6 kg
40
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Example 3: A small quantity of solution containing Na24 radionuclide (half life 15 hour) of activity 1.0 micro-curie is injected into the blood of a person. A sample of the blood of volume 1 cm3 taken after 5 hours shows an activity of 296 disintegrations per minute. Determine the total volume of blood in the body of the person. Assume that the radioactive solution mixes uniformly in the blood of the person. (1 curie = 3.7 × 10 10 disintegrations per second). Solution: We know that λ=
= 1.283 × 10-5 /sec
=
A=
= λ N0
∴
3.7 × 104 = 1.283 × 10-5 × N0
or
N0 =
ee .in
Also, activity is given by
= 2.883 × 109
ki
itj
Let the number of radioactive nuclei present after 5 hours be N1 in 1 cm3 sample of blood. = λ N1
or
N1
=
cr
or
ac
Then,
N1 = 3.844 × 105
Let N0’ be the number of radioactive nuclei in per cm3 of sample, then
= N’0 = (2)5/15 × N1 (2)1/3 × 3.844 × 105 = 4.878 × 105 Volume of blood V =
=
= 5.91 Liters
Ans.
41
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Example 4: A radionuclide with half life T, is produced in a reactor at a constant rate p nuclei per second. During each decay, energy E0 is released. If production of radionuclide is started at t = 0. Calculate (a) (b)
rate of release of energy as a function of time total energy released upto time t.
Solution: (a) Let at any instant, the number of nuclei in the radionuclide be N. Rate of decay = λ N At any instant t, net rate of accumulation of nuclei
∴
ee .in
=p-λN =
[1 – e-(t ln 2/T)]
cr
N=
ac
On solving, we get
ki
=
itj
On integrating both sides, we get
A = λ N = p [1 – e-(t ln 2/T)]
The rate of release of energy at any time t is A E0 = p E0 [1 – e-(t ln 2/T)]
Ans.
(b) Total number of nuclei produced upto time t = pt Nuclei decayed upto time t = (pt – N) Energy released upto this time = (pt – N) E0 [1 – e-(t ln 2/T)]
= pt E0 -
42
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Ans.
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Example 5: A radioactive source in the form of a metallic sphere of radius 10 -2 m emits β –particles at the rate of 5 × 1010 particles per second. The source is electrically insulated. How long it take for its potential to be raised by 2 volts, assuming that 40% of emitted β –particles escape the source. Solution: Let t be the time required to raise to potential by 2 V. Then number of β –particles emitted in t second is 5 × 1010 t. Now the number of β –particles escaping from sphere is 40% i.e. 2 × 1010 t. So, the charge developed
= (3.2 × 10-9 t) Coulomb But
Q = (40 R) V
ee .in
Q = (2 × 1010 t) (1.6 × 10-19) Coulomb
= 17.6 × 10-13
itj
= (8.85 × 1012) × (10-2) (2)
17.6 × 10-13 = 3.2 × 10-9 t
or
t=
ki
∴
Ans.
cr
ac
= 5.53 × 10-5 sec
Example 6: In an ore containing uranium, the ratio of U238 to Pb206 nuclei is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of U238. Take the half life of U238 to be 4.5 × 109 years. Solution: Given that U238 : Pb206 = 3:1 Let us assume that the number of Pb206 nuclei is x, then the number of U238 nuclei will be 3x (= N). Assuming that all the Pb206 present in the ore is due to U238, the initial number of U238 nuclei will be 3x + x = 4x (= N) We know that
N = N0 e-λt 43
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∴
3x = 4x e-λt
or
eλt =
or
λ t = ln
∴
t=
=
= 1.868 × 109 years
× ln
λ
Ans.
248 96Cm
And
2He
214
4
= 248.072220 u
= 244.064100 u
= 4.002603 u
itj
94Pu
ee .in
Example 7: The element curium 96Cm248 has a mean life of 1013 seconds. Its primary decay modes are spontaneous fission and a decay, the former with a probability of 8% and the latter with the probability of 92%. Each fission releases 200 MeV of energy. The masses involved in -decay are as follows:
ki
Calculate the power output from a sample of 1020 Cd atoms.
cr
ac
Solution: Number of atoms undergoing fission Nf =
× 1020
Number of atoms undergoing -decay N =
× 1020
Energy released in fission process Ef = Nf × 200 MeV =
× 200
= 16× 1020 MeV Energy released in -decay process E = Mass defect × 931 × N
44
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= 0.005517 × 931 ×
× 1020
= 4.725 × 1020 MeV Total energy released, E = Ef + E E = 16 × 1020 + 4.725 × 1020 = 20.725× 1020 MeV Power output P = Watt
ee .in
=
= 3.3 × 1020 Watt Ans.
ac
ki
itj
Example 8: A nucleus at rest undergoes a decay emitting an - particle of de- Broglie wavelength = 5.76 × 10-15 m. If the mass of the daughter nucleus is 223.610 a.m.u. and that of the - particle is 4.002 a.m.u., determine the total kinetic energy in the final state. Hence, obtain the mass of the parent nucleus in a.m.u. (1 a.m.u. = 931.470 MeV/c2). Solution: From de-Broglie relation, momentum of -particle is given by …………………………. (i)
cr
P =
λ
COLM gives Pd = P K.E. is given by
K = K + Kd =
+
……………… (ii)
Where d denotes daughter nucleus K=
45
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From equation (i) & (ii), we get K=
λ
Substituting the values, we get K = 6.25 MeV Ans. Further, (mp - m - md) c2 = K ∴
(mp – 223.61 – 4.002) c2 = 6.25
ee .in
mp – 223.61 – 4.002 = 6.25 MeV/c2 mp = 227.62 a.m.u Ans.
cr
Solution:
ac
ki
itj
Example 9: A thermonuclear device consists of a torus of diameter 3 m with a tube of diameter 1 m, containing deuterium gas at 10-2 mm mercury pressure and at room temperature. A bank of capacitors of 1200 F is discharged through the tube at 40 kV. If only 10% of the electrical energy is transformed to plasma kinetic energy, what is the maximum temperature attained? Assuming that the energy is equally shared between the deuterons and electrons in the plasma.
Cross-sectional area of the torus =
m2
Circumference = 3 m Volume of Torus = 32/4 = 7.4 m2 Pressure of the gas = 10-5 x 13.6 x 103 x 9.81 = 1.34 N/m2 From the equation PV = NkT We have
Nk = 1.34 x 7.4/293 = 0.0338
The energy obtained from the discharge
46
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= CV2 = x 1200 x 10-6 x (4 x 104) = 9.6 x 105 J
Energy transformed to plasma K.E. = 9.6 x 105 J We know that the average kinetic energy associated with the gas molecule is given as E = NkT Since, each deuterium molecule produces two ions and electrons, hence 4 NkT = 9.6 x 104
ee .in
or T = 4.75 x 105 K. Ans.
itj
Example 10: Calculate the energy released by the fission of 2 gm of 92U235 in kWh. Given that the energy released per fission is 200 MeV.
ki
Solution:
cr
ac
The number of atoms in 2 gm of
235 92U
atoms
=
Energy released per fission = 200 MeV
∴ Energy released by 2 gm of
= 200 × 1.6 ×
J = 3.2 ×
J
235 92U
=
× 3.2 ×
J
=
×
kWh
= 4.55×
kWh.
Ans.
47
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Example 11: Assuming that 200 MeV of energy is released per fission of uranium atom find the number of fission per second required to release one kilowatt power. Solution: Energy released per fission = 200 Mev = (200 ×
) × 1.6 ×
= 200 × 1.6 ×
J
J
The rate at which energy is to be released = 1 kilowatt = 1000 Joules per sec
∴
Number of fission per second =
Joules per sec.
ee .in
=
fission per second Ans.
itj
=3.125 ×
10
+ 0n1 3Li7 + 2He4
ac
5B
ki
Example 12: When thermal neutrons are used to induce the reaction
cr
alpha particles are emitted with an energy of 1.83 MeV. Given the masses of boron, neutron and 2He4 as 10.0167 amu ,1.0084 amu, and 4.00386 amu respectively. What is the mass of 3Li7? Solution: The given reaction is 5B
10
+ 0n1 3Li7 + 2He4 + Q
where ‘Q’ is the energy released in the reaction. This energy appears in the form of kinetic energy of the product particles. Let v1 and v2 be the velocities of 3Li7 and 2He4 after the reaction where m1 and m2 are masses.
48
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According to the law of conservation of momentum = Net kinetic energy , Q =
+
Q= = ∴
Or
Q= Q=
ee .in
or
=
Q=
+
ac
=
MeV
ki
Q = 1.83 ×
itj
= 1.83 MeV
= 3.08 ×
a.m.u.
cr
Substituting the given masses and value of Q in the given reaction 10.0167 + 1.00894 = 3Li7 + 4.00386 + 3.08 × 3Li
7
3Li
= (10.0167 + 1.00386) – (4.00386 – 0.00308)
7
= 7.0187 a.m.u. Ans.
49
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Example 13: A stationary He+ ion emitted a photon corresponding to the first line of the Lyman series. This photon liberates electron from a stationary hydrogen atom in the ground state. Find the velocity of the liberated electron. Solution: Energy of a photon emitted for transition n2 to n1 is given by eV …………….. (1)
Δ E = 13.6
This transition energy is shared between recoiling helium ion and photon. From conservation of energy,
Δ E = hf + K.E. …………….. (2)
ee .in
hf
ki
itj
V
ac
Where K.E. is kinetic energy of recoiling atom and hf is energy of photon. From conservation of momentum, =-
……………….. (2)
cr Photon
He
Here negative sign indicates that He+ ion recoils after emission of photons. Kinetic energy of recoiling of He+ ion =
=
=
where m is mass of He+ ion. From equation (2), Δ E = hf + or
h2f2 + (2mc2) hf – 2mc2 ΔE = 0 50
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On solving above quadratic for hf, we get Δ
hf =
We can neglect
Δ
Δ
in denominator as mc2 > ΔE
E = 40.8 eV
+ Electron
ee .in
+
H+
hf ΔE
ki
Hence,
itj
H Atom
ac
For first line of Lyman series
eV = 40.8 eV
cr
Δ E = 13.6
n2 = 2 and n1 = 1
Ionization energy of Hydrogen atom is 13.6 eV. Since this is less than energy of incident photons, the excess energy is transformed to kinetic energy to liberated electrons, mv2 = (40.8 – 13.6) eV = 37.2 eV v = 3.1 × 106 m/s
Ans.
51
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