Chapter 23 Nuclei (Nuclear Physics) free study material by TEACHING CARE online tuition and coaching classes
Chapter 23 Nuclei (Nuclear Physics) free study material by TEACHING CARE online tuition and coaching classes
Rutherford’s ascattering experiment established that the mass of atom is concentrated with small positively charged region at the centre which is called ‘nucleus’.
Nuclei are made up of proton and neutron. The number of protons in a nucleus (called the atomic number or proton number) is represented by the symbol Z. The number of neutrons (neutron number) is represented by N. The total number of neutrons and protons in a nucleus is called it’s mass number A so A = Z + N.
Neutrons and proton, when described collectively are called nucleons.
Nucleus contains two types of particles : Protons and neutrons
Nuclides are represented as
Neutron.
_{Z} X ^{A}; where X denotes the chemical symbol of the element.
Neutron is a fundamental particle which is essential constituent of all nuclei except that of hydrogen atom. It was discovered by Chadwick.
 The charge of neutron : It is neutral
 The mass of neutron : 6750 ´ 10^{–27} kg
 It’s spin angular momentum :
1 ´ æ h ö J – s
2 ç 2p ÷
è ø
 It’s magnetic moment : 9.57 ´ 10^{–27} J/Tesla
 It’s half life : 12 minutes
 Penetration power : High
 Types : Neutrons are of two types slow neutron and fast neutron, both are fully capable of penetrating a nucleus and causing artificial
Thermal neutrons
Fast neutrons can be converted into slow neutrons by certain materials called moderator’s (Paraffin wax, heavy water, graphite) when fast moving neutrons pass through a moderator, they collide with the molecules of the moderator, as a result of this, the energy of moving neutron decreases while that of the molecules of the moderator increases. After sometime they both attains same energy. The neutrons are then in thermal equilibrium with the molecules of the moderator and are called thermal neutrons.
Note : @Energy of thermal neutron is about 0.025 eV and speed is about 2.2 km/s.
Nucleus.
(1) Different types of nuclei
The nuclei have been classified on the basis of the number of protons (atomic number) or the total number of nucleons (mass number) as follows
 Isotopes : The atoms of element having same atomic number but different mass number are called All isotopes have the same chemical properties. The isotopes of some elements are the following
1 H 1 , 1 H 2 , 1 H 3
8 O16,
8O17,
8O18
_{2} He ^{3} ,
_{2} He ^{4}
17 Cl35 , 17Cl 37
92 U 235 ,
92U 238
 Isobars : The nuclei which have the same mass number (A) but different atomic number (Z) are called Isobars occupy different positions in periodic table so all isobars have different chemical properties. Some of the examples of isobars are
_{1} H ^{3} and
_{2} He ^{3} ,
_{6}C^{14} and 7 N ^{14} , 8 O^{17} and
9 F 17
 Isotones : The nuclei having equal number of neutrons are called For them both the atomic number (Z) and mass number (A) are different, but the value of (A – Z) is same. Some examples are
_{4} Be ^{9} and 5 B^{10} , 6 C^{13} and 7 N ^{14} , 8 O^{18} and 9 F ^{19} , 3 Li^{7} and 4 Be ^{8} , 1 H ^{3} and 2 He ^{4}
 Mirror nuclei : Nuclei having the same mass number A but with the proton number (Z) and neutron number (A – Z) interchanged (or whose atomic number differ by 1) are called mirror nuclei for
_{1} H ^{3} and 2 He ^{3} , 3 Li^{7} and 4 Be^{7}
(2) Size of nucleus
 Nuclear radius : Experimental results indicates that the nuclear radius is proportional to A^{1/3}, where A is the
mass number of nucleus i.e.
R µ A1 / 3
Þ R = R0 A^{1} ^{/} ^{3} , where R_{0} = 1.2 ´ 10^{–15} m = 1.2 fm.
Note : @Heavier nuclei are bigger in size than lighter nuclei.
 Nuclear volume : The volume of nucleus is given by V = 4p R^{3} = 4 p R^{3} A Þ V µ A
3 3 ^{0}
 Nuclear density : Mass per unit volume of a nucleus is called nuclear
Nuclear density(r ) = Mass of nucleus = mA
Volume of nucleus
4 p (R
3 0
A1 / 3 )3
where m = Average of mass of a nucleon (= mass of proton + mass of neutron = 1.66 ´ 10^{–27} kg) and mA = Mass of nucleus

Þ r =
3m 4pR ^{3}
= 2.38 ´ 10^{17} kg / m^{3}
Note : @r is independent of A, it means r is same of all atoms.
@ Density of a nucleus is maximum at it’s centre and decreases as we move outwards from the nucleus.
(3) Nuclear force
Forces that keep the nucleons bound in the nucleus are called nuclear forces.
 Nuclear forces are short range These do not exist at large distances greater than 10^{–15} m.
 Nuclear forces are the strongest forces in
 These are attractive force and causes stability of the
At low speeds, electromagnetic repulsion prevents the collision of nuclei
At high speeds, nuclei come close enough for the strong force to bind them together.
 These forces are charge
 Nuclear forces are noncentral
Nuclear forces are exchange forces
According to scientist Yukawa the nuclear force between the two nucleons is the result of the exchange of particles called mesons between the nucleons.
p – mesons are of three types – Positive p meson (p^{+}), negative p meson (p ^{–}), neutral p meson (p^{0}) The force between neutron and proton is due to exchange of charged meson between them i.e.
p ® p ^{+} + n, n ® p + p ^{–}
The forces between a pair of neutrons or a pair of protons are the result of the exchange of neutral meson (p^{o})
between them i.e.
p ® p‘+p ^{0}
and
n ® n‘+p ^{0}
Thus exchange of p meson between nucleons keeps the nucleons bound together. It is responsible for the nuclear forces.
DogBone analogy
The above interactions can be explained with the dog bone analogy according to which we consider the two interacting nucleons to be two dogs having a common bone clenched in between their teeth very firmly. Each one of these dogs wants to take the bone and hence they cannot be separated easily. They seem to be bound to each other with a strong attractive force (which is the bone) though the dogs themselves are strong enemies. The meson plays the same role of the common bone in between two nucleons.
(4) Atomic mass unit (amu)
The unit in which atomic and nuclear masses are measured is called atomic mass unit (amu)
1 amu (or 1u) = 1 th of mass of
12
_{6} C^{12} atom = 1.66 ´ 10^{–27} kg
Masses of electron, proton and neutrons
Mass of electron (m_{e}) = 9.1 ´ 10^{–31} kg = 0.0005486 amu, Mass of proton (m_{p}) = 1.6726 ´ 10^{–27} kg = 1.007276 amu
Mass of neutron (m_{n}) = 1.6750 ´ 10^{–27} kg = 1.00865 amu, Mass of hydrogen atom (m_{e} + m_{p}) = 1.6729 ´ 10^{–27} kg = 1.0078 amu
Massenergy equivalence
According to Einstein, mass and energy are inter convertible. The Einstein’s mass energy relationship is given by E = mc ^{2}
If m = 1 amu, c = 3 ´ 10^{8} m/sec then E = 931 MeV i.e. 1 amu is equivalent to 931 MeV or 1 amu (or 1 u) = 931 MeV
(5) Pair production and pairannihilation
When an energetic g–ray photon falls on a heavy substance. It is absorbed by some nucleus of the substance and an electron and a positron are produced. This phenomenon is called pair production and may be represented
by the following equation
hn
(g photon)
= 1 b ^{0} +
(Positron)
1 b 0 (Electron)
+1b0
The restmass energy of each of positron and electron is
E_{0} = m_{0}c^{2} = (9.1 ´ 10^{–31} kg) ´ (3.0 ´ 10^{8} m/s)^{2}
= 8.2 ´ 10^{–14} J = 0.51 MeV
hn
gphoton
+Ze
Nucleus
–1b0
Hence, for pairproduction it is essential that the energy of gphoton must be at least 2 ´ 0.51 = 1.02 MeV. If the energy of gphoton is less than this, it would cause photoelectric effect or Compton effect on striking the matter.
The converse phenomenon pairannihilation is also possible. Whenever an electron and a positron come very close to each other, they annihilate each other by combining together and two g–photons (energy) are produced. This phenomenon is called pair annihilation and is represented by the following equation.
+1 b 0 (Positron)
+ 1 b ^{0} (Electron)
= hn
(g photon)
+ hn
(g photon)
(6) Nuclear stability
Among about 1500 known nuclides, less than 260 are stable. The others are unstable that decay to form other nuclides by emitting a, bparticles and g – EM waves. (This process is called radioactivity). The stability of nucleus is determined by many factors. Few such factors are given below :
 Neutronproton ratio æN Ratioö

ç ÷
è ø
The chemical properties of an atom are governed entirely by the number of protons (Z) in the nucleus, the stability of an atom appears to depend on both the number of protons and the number of neutrons.
For lighter nuclei, the greatest stability is achieved when the number of protons and neutrons are
approximately equal (N » Z) i.e. N = 1
Z
Heavy nuclei are stable only when they have more neutrons than protons. Thus heavy nuclei are neutron rich compared to lighter nuclei (for heavy nuclei, more is the number of protons in the nucleus, greater is the electrical repulsive force between them. Therefore more neutrons are added to provide the strong attractive forces necessary to keep the nucleus stable.)
Figure shows a plot of N verses Z for the stable nuclei. For mass number upto about A = 40. For larger value of Z the nuclear force is unable to hold the nucleus together against the electrical repulsion of the protons unless the number of neutrons exceeds the number of protons. At Bi (Z = 83, A = 209), the neutron excess in N – Z = 43. There are no stable nuclides with Z > 83.
Note : @The nuclide 83 Bi ^{209} is the heaviest stable nucleus.
@ A nuclide above the line of stability i.e. having excess neutrons, decay through b ^{–} emission
(neutron changes into proton). Thus increasing atomic number Z and decreasing neutron number N.
In b ^{–}
emission,
N ratio decreases.
Z
A nuclide below the line of stability have excess number of protons. It
decays by b ^{+} emission, results in decreasing Z and increasing N. In b ^{+}
emission, the
N ratio increases.
Z
Proton number (Z)
 Even or odd numbers of Z or N : The stability of a nuclide is also determined by the consideration whether it contains an even or odd number of protons and
It is found that an eveneven nucleus (even Z and even N) is more stable (60% of stable nuclide have even Z
and even N).
An evenodd nucleus (even Z and odd N) or oddeven nuclide (odd Z and even N) is found to be lesser sable while the oddodd nucleus is found to be less stable.
Only five stable oddodd nuclides are known : 1 H ^{2} , 3 Li^{6} , 5 Be^{10} , 7 N ^{14} and 75 Ta^{180}
 Binding energy per nucleon : The stability of a nucleus is determined by value of it’s binding energy per In general higher the value of binding energy per nucleon, more stable the nucleus is
Mass Defect and Binding Energy.
(1) Mass defect (Dm)
It is found that the mass of a nucleus is always less than the sum of masses of it’s constituent nucleons in free state. This difference in masses is called mass defect. Hence mass defect
Dm = Sum of masses of nucleons – Mass of nucleus
= {Zmp + (A – Z)mn } M = {Zmp + Zme + (A – Z)mz } M‘
where m_{p} = Mass of proton, m_{n} = Mass of each neutron, m_{e} = Mass of each electron
M = Mass of nucleus, Z = Atomic number, A = Mass number, M¢ = Mass of atom as a whole.
Note : @The mass of a typical nucleus is about 1% less than the sum of masses of nucleons.
(2) Packing fraction
Mass defect per nucleon is called packing fraction
Packing fraction (f ) = Dm = M – A
where M = Mass of nucleus, A = Mass number
A A
Packing fraction measures the stability of a nucleus. Smaller the value of packing fraction, larger is the stability of the nucleus.
 Packing fraction may be of positive, negative or zero
(iii) At A = 16, f ® Zero
(3) Binding energy (B.E.)
The neutrons and protons in a stable nucleus are held together by nuclear forces and energy is needed to pull them infinitely apart (or the same energy is released during the formation of the nucleus). This energy is called the binding energy of the nucleus.
or
The binding energy of a nucleus may be defined as the energy equivalent to the mass defect of the nucleus. If Dm is mass defect then according to Einstein’s mass energy relation
Binding energy = Dm × c^{2} = [{m_{p}Z + m_{n}(A – Z)} – M]× c^{2}
(This binding energy is expressed in joule, because Dm is measured in kg)
If Dm is measured in amu then binding energy = Dm amu = [{m_{p}Z + m_{n}(A – Z)} – M] amu = Dm ´ 931 MeV
(4) Binding energy per nucleon
The average energy required to release a nucleon from the nucleus is called binding energy per nucleon. Binding energy per nucleon = Total binding energy = Dm ´ 931 MeV
Mass number (i.e. total number of nucleons)
Binding energy per nucleon µ Stability of nucleus
Binding Energy Curve.
A Nucleon
It is the graph between binding energy per nucleon and total number of nucleons (i.e. mass number A)
 Some nuclei with mass number A < 20 have large binding energy per nucleon than their neighbour
For example
_{2} He ^{4} , 4 Be ^{8} , 6 C^{12} , 8 O^{16} and 10 Ne ^{20} . These nuclei are more stable than their neighbours.
 The binding energy per nucleon is maximum for nuclei of mass number A = 56 (_{26} Fe ^{56} ). It’s value is 8
MeV per nucleon.
 For nuclei having A > 56, binding energy per nucleon gradually decreases for uranium (A = 238), the value of binding energy per nucleon drops to 5 MeV.
Note : @When a heavy nucleus splits up into lighter nuclei, then binding energy per nucleon of lighter nuclei is more than that of the original heavy nucleus. Thus a large amount of energy is liberated in this process (nuclear fission).
@ When two very light nuclei combines to form a relatively heavy nucleus, then binding energy per nucleon increases. Thus, energy is released in this process (nuclear fusion).
B.E. A
A
Nuclear Reactions.
The process by which the identity of a nucleus is changed when it is bombarded by an energetic particle is called nuclear reaction. The general expression for the nuclear reaction is as follows.
X + a
¾¾® C
¾¾®
Y + b + Q
(Parent nucleus)
(Incident particle)
(Compound nucleus)
(Compound nucleus)
(Product particles)
(Energy)
Here X and a are known as reactants and Y and b are known as products. This reaction is known as (a, b) reaction and can be represented as X(a, b) Y
(1) Q value or energy of nuclear reaction
The energy absorbed or released during nuclear reaction is known as Qvalue of nuclear reaction.
Qvalue = (Mass of reactants – mass of products)c^{2} Joules
= (Mass of reactants – mass of products) amu
If Q < 0, The nuclear reaction is known as endothermic. (The energy is absorbed in the reaction) If Q > 0, The nuclear reaction is known as exothermic (The energy is released in the reaction)
(2) Law of conservation in nuclear reactions
 Conservation of mass number and charge number : In the following nuclear reaction
2 He 4 + 7 N 14 ® 8 O17 + 1 H 1
Mass number (A) ®  Before the reaction  After the reaction 
Charge number (Z) ®  4 +14 = 18 2 + 7 = 9  17 + 1 = 18 8 + 1 = 9 
 Conservation of momentum : Linear momentum/angular momentum of particles before the reaction is equal to the linear/angular momentum of the particles after the That is Sp = 0
 Conservation of energy : Total energy before the reaction is equal to total energy after the Term
Q is added to balance the total energy of the reaction.
(3) Common nuclear reactions
The nuclear reactions lead to artificial transmutation of nuclei. Rutherford was the first to carry out artificial transmutation of nitrogen to oxygen in the year 1919.
2 He 4 + 7 N 14 ® 9 F 18 ® 8 O17 + 1 H 1
It is called (a, p) reaction. Some other nuclear reactions are given as follows.
(p, n) reaction Þ (p, a) reaction Þ (p, g) reaction Þ (n, p) reaction Þ (g, n) reaction Þ
1 H 1 + 5 B11 ® 6 C12 ® 6 C11 + 0 n1
_{1} H ^{1} + 3 Li^{11} ® 4 Be ^{8} ® 2 He ^{4} + 2 He ^{4}
1 H 1 + 6 C12 ® 7 N 13 ® 7 N 13 + g
0 n1 + 7 N 14 ® 7 N 15 ® 6 C14 + 1 H 1
g + 1 H 2 ® 1 H 1 + 0 n1
Nuclear Fission and Fusion.
Nuclear fission
The process of splitting of a heavy nucleus into two lighter nuclei of comparable masses (after bombardment with a energetic particle) with liberation of energy is called nuclear fission.
The phenomenon of nuclear fission was discovered by scientist Ottohann and F. Strassman and was explained by N. Bohr and J.A. Wheeler on the basis of liquid drop model of nucleus.
(1) Fission reaction of U^{235}
 Nuclear reaction :
92 U 235 + 0 n1 ®
92 U 236
(unstable nucleus)
® 56 Ba141 + 36 Kr 92 + 30 n1 + Q
 The energy released in U^{235} fission is about 200 MeV or 8 MeV per nucleon.
 By fission of
_{92}U ^{235} , on an average 2.5 neutrons are liberated. These neutrons are called fast neutrons
and their energy is about 2 MeV (for each). These fast neutrons can escape from the reaction so as to proceed the chain reaction they are need to slow down.
 Fission of U^{235} occurs by slow neutrons only (of energy about 1eV) or even by thermal neutrons (of energy about 025 eV).
 50 kg of U^{235} on fission will release » 4 × 10^{15} J of This is equivalence to 20,000 tones of TNT
explosion. The nuclear bomb dropped at Hiroshima had this much explosion power.
 The mass of the compound nucleus must be greater than the sum of masses of fission
 The
Binding energy A
of compound nucleus must be less than that of the fission products.
_{36} Kr
 It may be pointed out that it is not necessary that in each fission of uranium, the two fragments _{56} Ba
are formed but they may be any stable isotopes of middle weight atoms.
and
Same other U ^{235}
fission reactions are
92 U 235 + 0 n1 ® 54 Xe140 + 38 Sr 94 + 20 n1
® 57 La148 + 35 Br 85 + 3 0 n1
® Many more
 The neutrons released during the fission process are called prompt
 Most of energy released appears in the form of kinetic energy of fission
Slow Neutron
92U235 92U236
(2) Chain reaction
In nuclear fission, three neutrons are produced along with the release of large energy. Under favourable conditions, these neutrons can cause further fission of other nuclei, producing large number of neutrons. Thus a chain of nuclear fissions is established which continues until the whole of the uranium is consumed.
In the chain reaction, the number of nuclei undergoing fission increases very fast. So, the energy produced takes a tremendous magnitude very soon.
Difficulties in chain reaction
 Absorption of neutrons by U ^{238} , the major part in natural uranium is the isotope U^{238} (99.3%), the isotope
U 235
is very little (0.7%). It is found that U ^{238}
is fissionable with fast neutrons, whereas U ^{235} is fissionable with slow
neutrons. Due to the large percentage of
U 238 ,
there is more possibility of collision of neutrons with
U ^{238} . It is
found that the neutrons get slowed on coliding with
U 238 ,
as a result of it further fission of U^{238} is not possible
(Because they are slow and they are absorbed by U^{238}). This stops the chain reaction.
Removal : (i) To sustain chain reaction 92 U ^{235}
is separated from the ordinary uranium. Uranium so obtained
(92 U 235 )
is known as enriched uranium, which is fissionable with the fast and slow neutrons and hence chain
reaction can be sustained.
 If neutrons are slowed down by any method to an energy of about 3 eV, then the probability of their
absorption by U ^{238}
becomes very low, while the probability of their fissioning U ^{235}
becomes high. This job is done
by moderators. Which reduce the speed of neutron rapidly graphite and heavy water are the example of moderators.
 Critical size : The neutrons emitted during fission are very fast and they travel a large distance before being slowed down. If the size of the fissionable material is small, the neutrons emitted will escape the fissionable material before they are slowed Hence chain reaction cannot be sustained.
Removal : The size of the fissionable material should be large than a critical size.
The chain reaction once started will remain steady, accelerate or retard depending upon, a factor called neutron reproduction factor (k). It is defined as follows.
k = Rate of production of neutrons Rate of loss of neutrons
® If k = 1, the chain reaction will be steady. The size of the fissionable material used is said to be the critical size and it’s mass, the critical mass.
® If k > 1, the chain reaction accelerates, resulting in an explosion. The size of the material in this case is super critical. (Atom bomb)
® If k < 1, the chain reaction gradually comes to a halt. The size of the material used us said to be subcritical. Types of chain reaction : Chain reactions are of following two types
Controlled chain reaction  Uncontrolled chain reaction 
Controlled by artificial method  No control over this type of nuclear reaction 
All neurons are absorbed except one  More than one neutron takes part into reaction 
It’s rate is slow  Fast rate 
Reproduction factor k = 1  Reproduction factor k > 1 
Energy liberated in this type of reaction is always less  A large amount of energy is liberated in this type of 
than explosive energy  reaction 
Chain reaction is the principle of nuclear reactors  Uncontrolled chain reaction is the principle of atom bomb. 
Note : @The energy released in the explosion of an atom bomb is equal to the energy released by 2000 tonn
of TNT and the temperature at the place of explosion is of the order of 10^{7} ^{o}C.
Nuclear Reactor.
A nuclear reactor is a device in which nuclear fission can be carried out through a sustained and a controlled chain reaction. It is also called an atomic pile. It is thus a source of controlled energy which is utilised for many
useful purposes.
Coolant
Concrete wall
Moderator
Cadmium
rods Core
Coolant out Turbine
Water
Heat
To electric generator
Condenser
(1) Parts of nuclear reactor
Fuel elements
Coolant in exchanger
 Fissionable material (Fuel) : The fissionable material used in the reactor is called the fuel of the reactor. Uranium isotope (U^{235}) Thorium isotope (Th^{232}) and Plutonium isotopes (Pu^{239}, Pu^{240} and Pu^{241}) are the most commonly used fuels in the
 Moderator : Moderator is used to slow down the fast moving neutrons. Most commonly used moderators are graphite and heavy water (D_{2}O).
 Control Material : Control material is used to control the chain reaction and to maintain a stable rate of This material controls the number of neutrons available for the fission. For example, cadmium rods are inserted into the core of the reactor because they can absorb the neutrons. The neutrons available for fission are controlled by moving the cadmium rods in or out of the core of the reactor.
 Coolant : Coolant is a cooling material which removes the heat generated due to fission in the Commonly used coolants are water, CO_{2} nitrogen etc.
 Protective shield : A protective shield in the form a concrete thick wall surrounds the core of the reactor to save the persons working around the reactor from the hazardous
Note : @It may be noted that Plutonium is the best fuel as compared to other fissionable material. It is because fission in Plutonium can be initiated by both slow and fast neutrons. Moreover it can be obtained from U ^{238} .
@ Nuclear reactor is firstly devised by fermi.
@ Apsara was the first Indian nuclear reactor.
(2) Uses of nuclear reactor
 In electric power
 To produce radioactive isotopes for their use in medical science, agriculture and
 In manufacturing of
PU 239
which is used in atom bomb.
 They are used to produce neutron beam of high intensity which is used in the treatment of cancer and nuclear
Note : @A type of reactor that can produce more fissile fuel than it consumes is the breeder reactor.
Nuclear fusion
In nuclear fusion two or more than two lighter nuclei combine to form a single heavy nucleus. The mass of single nucleus so formed is less than the sum of the masses of parent nuclei. This difference in mass results in the release of tremendous amount of energy
_{1} H ^{2} + 1 H ^{2} ® 1 H ^{3} + 1 H ^{1} + 4 MeV
_{1} H ^{3} + 1 H ^{2} ® 2 He ^{4} + 0 n^{1} + 17.6 MeV
or 1 H ^{2} + 1 H ^{2} ® 2 He ^{4} + 24 MeV
For fusion high pressure (» 10^{6} atm) and high temperature (of the order of 10^{7} K to 10^{8} K) is required and so the reaction is called thermonuclear reaction.
Fusion energy is greater then fission energy fission of one uranium atom releases about 200 MeV of energy.
But the fusion of a deutron
(1 H ^{2} )
and triton
(1 H ^{3} )
releases about 17.6 MeV of energy. However the energy
released per nucleon in fission is about 0.85 MeV but that in fusion is 4.4 MeV. So for the same mass of the fuel, the energy released in fusion is much larger than in fission.
Plasma : The temperature of the order of 10^{8} K required for thermonuclear reactions leads to the complete ionisation of the atom of light elements. The combination of base nuclei and electron cloud is called plasma. The enormous gravitational field of the sun confines the plasma in the interior of the sun.
The main problem to carryout nuclear fusion in the laboratory is to contain the plasma at a temperature of 10^{8}K. No solid container can tolerate this much temperature. If this problem of containing plasma is solved, then the large quantity of deuterium present in sea water would be able to serve as inexhaustible source of energy.
Note : @To achieve fusion in laboratory a device is used to confine the plasma, called Tokamak. Stellar Energy
Stellar energy is the energy obtained continuously from the sun and the stars. Sun radiates energy at the rate of about 10^{26} joules per second.
Scientist Hans Bethe suggested that the fusion of hydrogen to form helium (thermo nuclear reaction) is continuously taking place in the sun (or in the other stars) and it is the source of sun’s (star’s) energy.
The stellar energy is explained by two cycles
About 90% of the mass of the sun consists of hydrogen and helium.
Nuclear Bomb.
Based on uncontrolled nuclear reactions.
Example: 1 A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio 8 : 1. The ratio of radii of the fragments is [EAMCET (Engg.) 2001]
(a) 1 : 2 (b) 1 : 4 (c) 4 : 1 (d) 2 : 1
Solution : (a)
v_{1}
By conservation of momentum m_{1}v_{1} = m_{2}v_{2}
Þ v1 = 8 = m2
…… (i)
v2 1 m1
r æ A
ö1 / 3
æ 1 ö^{1} ^{/} ^{3} 1
v_{2} Also from r µ A^{1} ^{/} ^{3}
Þ 1 = ç^{ } ^{1} ÷
= ç ÷ =
r2 è A2 ø
è 8 ø 2

Example: 2 The ratio of radii of nuclei ^{27} Al
and ^{125} Te
is approximately [J & K CET 2000]

(a) 6 : 10 (b) 13 : 52 (c) 40 : 177 (d) 14 : 7
r æ A
ö1 / 3
æ 27 ö^{1/} ^{3} 8 6
Solution : (a) By using r µ A^{1} ^{/} ^{3}
Þ 1 = ç 1 ÷ = ç
÷ = =
r2 è A2 ø
è 125 ø
5 10
Example: 3 If Avogadro’s number is 6 ´ 10^{23} then the number of protons, neutrons and electrons in 14 g of _{6}C^{14} are respectively
(a) 36 ´ 10^{23}, 48 ´ 10^{23}, 36 ´ 10^{23} (b) 36 ´ 10^{23}, 36 ´ 10^{23}, 36 ´ 10^{21}
(c) 48 ´ 10^{23}, 36 ´ 10^{23}, 48 ´ 10^{21} (d) 48 ´ 10^{23}, 48 ´ 10^{23}, 36 ´ 10^{21}
Solution : (a) Since the number of protons, neutrons and electrons in an atom of 6 C^{14}
are 6, 8 and 6 respectively. As 14 gm
of 6 C^{14} contains 6 ´ 10^{23} atoms, therefore the numbers of protons, neutrons and electrons in 14 gm of 6 C^{14}
are
6 ´ 6 ´ 10^{23} = 36 ´ 10^{23} ,
8 ´ 6 ´ 10^{23} = 48 ´ 10^{23} ,
6 ´ 6 ´ 10^{23} = 36 ´ 10^{23} .
Example: 4 Two Cu^{64} nuclei touch each other. The electrostatics repulsive energy of the system will be
(a) 0.788 MeV (b) 7.88 MeV (c) 126.15 MeV (d) 788 MeV
Solution : (c) Radius of each nucleus R = R0(A)^{1} ^{/} ^{3} = 1.2 (64)^{1} ^{/} ^{3} = 4.8 fm
Distance between two nuclei (r) = 2R
So potential energy U = k × q 2
r
= 9 ´ 10^{9} ´ (1.6 ´ 10 ^{–}^{19} ´ 29)^{2}
2 ´ 4.8 ´ 10 ^{–}^{15} ´ 1.6 ´ 10 ^{–}^{19}
= 126.15 MeV.
Example: 5 When
_{92} U ^{235} undergoes fission. 0.1% of its original mass is changed into energy. How much energy is
released if 1 kg of 92 U 235 undergoes fission [MP PET 1994; MP PMT/PET 1998; BHU 2001; BVP 2003]
(a) 9 ´ 10^{10} J (b) 9 ´ 10^{11} J (c) 9 ´ 10^{12} J (d) 9 ´ 10^{13} J
Solution : (d) By using
E = Dm × c^{2}
Þ E = æ 0.1 ´

è
ö
1÷ (3
ø
´ 10^{8})^{2} = 9 ´ 10^{13} J
Example: 6 1 g of hydrogen is converted into 0.993 g of helium in a thermonuclear reaction. The energy released is
[EAMCET (Med.) 1995; CPMT 1999]
(a) 63 ´ 10^{7} J (b) 63 ´ 10^{10} J (c) 63 ´ 10^{14} J (d) 63 ´ 10^{20} J
Solution : (b) Dm = 1 – 0.993 = 0.007 gm

\ E = Dmc^{2} = 0.007 ´ 10^{–3} ´ (3 ´ 10^{8})^{2} = 63 ´ 10^{10} J

Example: 7 The binding energy per nucleon of deuteron
(^{2} H)
and helium nucleus
(^{4} He)
is 1.1 MeV and 7 MeV
Solution : (c)
respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is
[MP PMT 1992; Roorkee 1994; IITJEE 1996; AIIMS 1997; Haryana PMT 2000; Pb PMT 2001; CPMT 2001; AIEEE 2004]
(a) 13.9 MeV (b) 26.9 MeV (c) 23.6 MeV (d) 19.2 MeV
_{1} H ^{2} + 1H ^{2} ® 2He^{4} + Q
Total binding energy of helium nucleus = 4 ´ 7 = 28 MeV Total binding energy of each deutron = 2 ´ 1.1 = 2.2 MeV Hence energy released = 28 – 2 ´ 2.2 = 23.6 MeV
Example: 8 The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu. The binding energy of the helium nucleus will be [1 amu= 931 MeV] [CPMT 1986; MP PMT 1995; CBSE 2003]
(a) 28.4 MeV (b) 20.8 MeV (c) 27.3 MeV (d) 14.2 MeV
Solution : (a) Helium nucleus consist of two neutrons and two protons.
So binding energy E = Dm amu = Dm ´ 931 MeV
Þ E = (2 ´ m_{p} + 2m_{n} – M) ´ 931 MeV = (2 ´ 1.0073 + 2 ´ 1.0087 – 4.0015) ´ 931 = 28.4 MeV
Example: 9 A atomic power reactor furnace can deliver 300 MW. The energy released due to fission of each of uranium atom U ^{238} is 170 MeV. The number of uranium atoms fissioned per hour will be [UPSEAT 2000] (a) 5 ´ 10^{15} (b) 10 ´ 10^{20} (c) 40 ´ 10^{21} (d) 30 ´ 10^{25}
Solution : (c) By using
P = W = n ´ E
where n = Number of uranium atom fissioned and E = Energy released due to
t
each fission so
t
300 ´ 106 = n ´ 170 ´ 10^{6} ´ 1.6 ´ 10^{–}^{19}
3600
Þ n = 40 ´ 10^{21}
Example: 10 The binding energy per nucleon of O^{16} is 7.97 MeV and that of O^{17} is 7.75 MeV. The energy (in MeV) required to remove a neutron from O^{17} is [IITJEE 1995]
(a) 3.52 (b) 3.64 (c) 4.23 (d) 7.86
Solution : (c)
O17 ® O16 + 0n1
\ Energy required = Binding of O^{17} – binding energy of O^{16} = 17 ´ 7.75 – 16 ´ 7.97 = 4.23 MeV
Example: 11 A gamma ray photon creates an electronpositron pair. If the rest mass energy of an electron is 0.5 MeV and the total kinetic energy of the electronpositron pair is 0.78 MeV, then the energy of the gamma ray photon must be [MP PMT 1991]
(a) 0.78 MeV (b) 1.78 MeV (c) 1.28 MeV (d) 0.28 MeV
Solution : (b) Energy of grays photon = 0.5 + 0.5 +0.78 = 1.78 MeV
Example: 12 What is the mass of one Curie of U^{234} [MNR 1985]
(a) 3.7 ´ 10^{10} gm (b) 2.348 ´ 10^{23} gm (c) 1.48 ´ 10^{–11} gm (d) 6.25 ´ 10^{–34} gm
Solution : (c) 1 curie = 3.71 ´ 10^{10} disintegration/sec and mass of 6.02 ´ 10^{23} atoms of U ^{234} = 234 gm
\ Mass of 3.71 ´ 10^{10} atoms = 234 ´ 3.71´ 1010 = 1.48 ´ 10^{–}^{11} gm
6.02 ´ 10^{23}
Example: 13 In the nuclear fusion reaction ^{2} H +^{3} H ®^{4} He + n,
given that the repulsive potential energy between the two
1 1 2
nuclei is
 7 ´ 10^{–}^{14} J , the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s constant k = 1.38 ´ 10^{–}^{23} J/K ] [AIEEE 2003]
(a)
10^{9} K
(b)
10^{7} K
(c)
10^{5} K
(d)
10^{3} K
Solution : (a) Kinetic energy of molecules of a gas at a temperature T is 3/2 kT
\ To initiate the reaction
3 kT = 7.7 ´ 10^{–}^{14} J
2
Þ T = 3.7 ´ 10^{9} K.
Example: 14 A nucleus with mass number 220 initially at rest emits an aparticle. If the Q value of the reaction is 5.5 MeV. Calculate the kinetic energy of the aparticle [IITJEE (Screening) 2003]
(a) 4.4 MeV (b) 5.4 MeV (c) 5.6 MeV (d) 6.5 MeV
Solution : (b)
M = 220 m1 = 216
k_{2} p_{2}
m_{2} = 4
Qvalue of the reaction is 5.5 eV i.e. k_{1} + k _{2} = 5.5 MeV
……(i)
By conservation of linear momentum
p1 = p2 Þ
= Þ k_{2} = 54 k_{1} ……(ii)

On solving equation (i) and (ii) we get k_{2} = 5.4 MeV.
Example: 15 Let mp
be the mass of a proton, mn
the mass of a neutron, M1 the mass of a 20
nucleus and
M _{2} the mass of a


40 Ca nucleus. Then [IIT 1998; DPMT 2000]
(a)
M2 = 2M1
(b)
M2 > 2M1
(c)
M2 < 2M1
(d)
M1 < 10(mn + mp )

Solution : (c, d) Due to mass defect (which is finally responsible for the binding energy of the nucleus), mass of a nucleus is

always less then the sum of masses of it’s constituent particles ^{20}
is made up of 10 protons plus 10

neutrons. Therefore, mass of ^{20} Ne nucleus M1 < 10 (mp + mn )
Also heavier the nucleus, more is he mass defect thus 20 (mn + mp ) – M2 > 10(mp + mn) – M1
or 10 (mp + mn ) > M 2 – M1
Þ M2 < M1 + 10 (mp + mn)
Þ M 2 < M1 + M1 Þ
M 2 < 2M1
Radioactivity.
The phenomenon of spontaneous emission of radiatons by heavy elements is called radioactivity. The elements which shows this phenomenon are called radioactive elements.
 Radioactivity was discovered by Henery Becquerel in uranium salt in the year
 After the discovery of radioactivity in uranium, Piere Curie and Madame Curie discovered a new radioactive element called radium (which is 10^{6} times more radioactive than uranium)
 Some examples of radio active substances are : Uranium, Radium, Thorium, Polonium, Neptunium
 Radioactivity of a sample cannot be controlled by any physical (pressure, temperature, electric or magnetic field) or chemical
 All the elements with atomic number (Z ) > 82 are naturally
 The conversion of lighter elements into radioactive elements by the bombardment of fast moving particles is called artificial or induced
 Radioactivity is a nuclear event and not atomic. Hence electronic configuration of atom don’t have any relationship with
Nuclear radiatons
According to Rutherford’s experiment when a sample of radioactive substance is put in a lead box and allow the emission of radiation through a small hole only. When the radiation enters into the external electric field, they splits into three parts
 Radiations which deflects towards negative plate are called arays (stream of positively charged particles)
 Radiations which deflects towards positive plate are called b particles (stream of negatively charged particles)
 Radiations which are undeflected called grays. (E.M. waves or photons)
Note : @Exactly same results were obtained when these radiations were subjected to magnetic field.
@ No radioactive substance emits both a and b particles simultaneously. Also g–rays are emitted after the emission of a or bparticles.
@ bparticles are not orbital electrons they come from nucleus. The neutron in the nucleus decays into proton and an electron. This electron is emitted out of the nucleus in the form of brays.
Properties of a, b and grays
Features  a– particles  b – particles  g – rays 
1. Identity  Helium nucleus or  Fast moving electron  Photons (E.M. waves) 
doubly ionised helium atom (_{2}He^{4})  (b ^{0} or b ^{–} )  
2. Charge  + 2e  – e  Zero 
3. Mass 4 m_{p} (m_{p} = mass of proton = 1.87 ´ 10^{–27}  4 m_{p}  m_{e}  Massless 
4. Speed  » 10^{7} m/s  1% to 99% of speed of light  Speed of light 
5. Range of kinetic energy  4 MeV to 9 MeV  All possible values between a  Between a minimum 
minimum certain value to  value to 2.23 MeV  
1.2 MeV  
6. Penetration power (g, b, a)  1  100  10,000 
(Stopped by a paper)  (100 times of a)  (100 times of b upto 30  
cm of iron (or Pb) sheet  
7. Ionisation power (a > b > g)  10,000  100  1 
8. Effect of electric or  Deflected  Deflected  Not deflected 
magnetic field  
9. Energy spectrum  Line and discrete  Continuous  Line and discrete 
Radioactive Disintegration.
(1) Law of radioactive disintegration
According to Rutherford and Soddy law for radioactive decay is as follows.
“At any instant the rate of decay of radioactive atoms is proportional to the number of atoms present at that
instant” i.e.
 dNµ N dt
Þ dN = –lN . It can be proved that N = N_{0}e^{–}^{l}^{t}
dt
This equation can also be written in terms of mass i.e. M = M_{0}e^{–}^{l}^{t}
where N = Number of atoms remains undecayed after time t, N_{0} = Number of atoms present initially (i.e. at t = 0), M = Mass of radioactive nuclei at time t, M_{0} = Mass of radioactive nuclei at time t = 0, N_{0} – N = Number of disintegrated nucleus in time t
dN = rate of decay, l = Decay constant or disintegration constant or radioactivity constant or Rutherford
dt
Soddy’s constant or the probability of decay per unit time of a nucleus.
Note : @l depends only on the nature of substance. It is independent of time and any physical or chemical changes.
N
t
(2) Activity
It is defined as the rate of disintegration (or count rate) of the substance (or the number of atoms of any
material decaying per second) i.e.
A = – dN = lN = lN
dt
0 e –lt
= A0 e –lt
where A_{0} = Activity of t = 0, A = Activity after time t
Units of activity (Radioactivity)
It’s units are Becqueral (Bq), Curie (Ci) and Rutherford (Rd)
1 Becquerel = 1 disintegration/sec, 1 Rutherford = 10^{6} dis/sec, 1 Curie = 3.7 ´ 10^{11} dis/sec
Note : @Activity per gm of a substance is known as specific activity. The specific activity of 1 gm of radium – 226 is 1 Curie.
@ 1 millicurie = 37 Rutherford
@ The activity of a radioactive substance decreases as the number of undecayed nuclei decreases with time.
@ Activity µ
1
Half life
(3) Half life (T_{1/2})
Time interval in which the mass of a radioactive substance or the number of it’s atom reduces to half of it’s

initial value is called the half life of the substance. N
i.e. if
N = N0
2
then t = T1 / 2
Hence from
N = N0 e ^{–l}^{t}
N_{0}/2 N_{0}/4
N –l (T )
log 2
0.693
N_{0}/8

0 = N0 e
2
1 / 2
Þ T1 / 2 = e = l
0 1 2T 3T t
Time (t)  Number of undecayed atoms (N) (N_{0} = Number of initial atoms)  Remaining fraction of active atoms (N/N_{0}) probability of survival  Fraction of atoms decayed (N_{0} – N) /N_{0} probability of decay  
t = 0  N_{0}  1 1 2 1 4 1 8
æ 1 ö^{10} ç 2 ÷ è ø æ 1 ö^{n} ç 2 ÷ è ø  (100%) (50%)
(25%)
(12.5%)
» 0.1%  0 1 (50%) 2 3 (75%) 4 7 (87.5%) 8 » 99.9%
ìï æ 1 ö^{n} üï í1 – ç ÷ ý ïî è 2 ø ïþ 
t = T_{1/2}  N 0  
2  
t = 2(T_{1/2})  1 ´ N _{0} = N _{0}  
2 2 (2)^{2}  
t = 3(T_{1/2})  1 ´ N_{0} = N_{0}  
2 (2) (2)3  
t = 10 (T_{1/2})  N 0  
(2)^{10}  
t = n (N_{1/2})  N  
(2)^{2} 
Useful relation
After n halflives, number of undecayed atoms
æ 1 ö^{n}
N = N_{0} ç 2 ÷
æ 1 öt / T1 / 2
= N_{0} ç 2 ÷
(4) Mean (or average) life (t)
è ø è ø
The time for which a radioactive material remains active is defined as mean (average) life of that material.
Other definitions
 It is defined as the sum of lives of all atoms divided by the total number of atoms
i.e. t = Sum of the lives of all the atoms = 1
Total number of atoms l
 From
N = N
0 e –lt Þ
ln N N0
t
= –l
slope of the line shown in the graph
i.e. the magnitude of inverse of slope of ln N
N0
vs t
curve is known as mean life (t).
N

ln
 From
N = N
0 e –lt 0
If t = 1 = t Þ
l
N = N
_{0}e ^{1} = N
æ 1 ö = 0.37 N

e 0
= 37%
of N_{0}.
t
è ø
i.e. mean life is the time interval in which number of undecayed atoms (N) becomes 37% of original number of atoms. or
1 times or 0.37 times or
e
It is the time in which number of decayed atoms (N
– N) becomes
æ1 – 1 ö
times or 0.63 times or 63% of
original number of atoms.
_{0} ç ÷

è ø
 From T1/ 2
= 0.693
l
Þ 1 = t =
l
1
0.693
.(t
1 / 2
) = 1.44 (T1 / 2 )
i.e. mean life is about 44% more than that of half life. Which gives us t > T_{(1/2)}
Note : @ Half life and mean life of a substance doesn’t change with time or with pressure, temperature etc.
Radioactive Series.
If the isotope that results from a radioactive decay is itself radioactive then it will also decay and so on.
The sequence of decays is known as radioactive decay series. Most of the radionuclides found in nature are members of four radioactive series. These are as follows
Mass number  Series (Nature)  Parent  Stable and product  Integer  n  Number of lost particles 
4n  Thorium (natural)  90 Th 232  82 Pb 208  52  a = 6, b = 4  
4n + 1  Neptunium  93 Np 237  83 Bi 209  52  a = 8, b = 5  
(Artificial)  
4n + 2  Uranium (Natural)  92 U 238  82 Pb 206  51  a = 8, b = 6  
4n + 3  Actinium (Natural)  89 Ac 227  82 Pb 207  51  a = 7, b = 4 
Note : @The 4n + 1 series starts from
94 PU 241
but commonly known as neptunium series because
neptunium is the longest lived member of the series.
@ The 4n + 3 series actually starts from _{92} U ^{235} .
Successive Disintegration and Radioactive Equilibrium.
Suppose a radioactive element A disintegrates to form another radioactive element B which intern disintegrates to still another element C; such decays are called successive disintegration.
Rate of disintegration of
A = dN1
dt
= –l_{1} N1
(which is also the rate of formation of B)
Rate of disintegration of
B = dN 2
dt
= –l_{2} N 2
\ Net rate of formation of B = Rate of disintegration of A – Rate of disintegration of B
= l_{1}N_{1} – l_{2}N_{2}
Equilibrium
In radioactive equilibrium, the rate of decay of any radioactive product is just equal to it’s rate of production from the previous member.
i.e. l N
= l N
Þ l_{1}
= N 2
= t 2 =
(T1 / 2 )
1 1 2 2
l N t (T )
2 2 1 1 / 2 1
Note : @ In successive disintegration if N_{0} is the initial number of nuclei of A at t = 0 then number of nuclei of
product B at time t is given by N =
l1 N0
(e ^{–l} ^{t} – e ^{–l} ^{t} ) where l l
– decay constant of A and B.
Uses of radioactive isotopes
 In medicine
^{2} (l
1 2


– l_{1})
 For testing bloodchromium – 51 (ii) For testing blood circulation – Na – 24
 For detecting brain tumor Radio mercury – 203
 For detecting fault in thyroid gland – Radio iodine – 131
 For cancer – cobalt – 60
 For blood – Gold – 189
 For skin diseases – Phospohorous – 31
(2) In Archaeology
 For determining age of archaeological sample (carbon dating) C^{14}
 For determining age of meteorites – K ^{40}
 For determining age of earthLead isotopes
(3) In agriculture
 For protecting potato crop from earthworm CO^{60}
 For artificial rains – AgI (iii) As fertilizers –
P 32
 As tracers – (Tracer) : Very small quantity of radioisotopes present in a mixture is known as tracer
 Tracer technique is used for studying biochemical reaction in tracer and animals.
(5) In industries
 For detecting leakage in oil or water pipe lines (ii) For determining the age of
21
1 l_{1}, T_{1}, t_{1}
l
Example: 16 When 90 Th^{228}
transforms to 83 Bi^{212} , then the number of the emitted a–and b–particles is, respectively
[MP PET 2002]
Solution : (d)
(a) 8a, 7b (b) 4a, 7b (c) 4a, 4b (d) 4a, 1b
Z =90 ThA= 228 ® Z‘=83 Bi A‘= 212
Number of aparticles emitted na = A – A‘ = 228 – 212 = 4
4 4
Number of b–particles emitted nb = 2na – Z + Z‘ = 2 ´ 4 – 90 + 83 = 1.
Example: 17 A radioactive substance decays to 1/16^{th} of its initial activity in 40 days. The halflife of the radioactive substance expressed in days is [AIEEE 2003]
(a) 2.5 (b) 5 (c) 10 (d) 20
æ 1 ö t / T1 / 2
N 1 æ 1 ö ^{40} ^{/} ^{T}1 / 2
Solution : (c) By using
N = N _{0} ç 2 ÷ Þ N = 16 = ç 2 ÷
Þ T1 / 2 = 10
days.
è ø _{0} è ø
Example: 18 A sample of radioactive element has a mass of 10 gm at an instant t = 0. The approximate mass of this element in the sample after two mean lives is [CBSE PMT 2003]
(a) 2.50 gm (b) 3.70 gm (c) 6.30 gm (d) 1.35 gm
–lt
–l(2t )
æ 2 ö

ç l ÷
æ 1 ö^{2}
Solution : (d) By using M = M_{0}e
Þ M = 10e = 10e ^{è} ^{ø} = 10 ç ÷ = 1.359 gm
Example: 19 The halflife of ^{215} At
è e ø
is 100 ms. The time taken for the radioactivity of a sample of ^{215} At
to decay to 1/16^{th} of
its initial value is [IITJEE (Screening) 2002]
(a) 400 ms (b) 6.3 ms (c) 40 ms (d) 300 ms
æ 1 ö^{n}
N æ 1 ö ^{t} ^{/} ^{T}1 / 2
1 æ 1 ö t / 100
Solution : (a) By using
N = N _{0} ç 2 ÷ Þ N = ç 2 ÷ Þ 16 = ç 2 ÷
Þ t = 400 m sec.
è ø _{0} è ø è ø
Example: 20 The mean lives of a radioactive substance for a and b emissions are 1620 years and 405 years respectively. After how much time will the activity be reduced to one fourth [RPET 1999]
 405 year (b) 1620 year (c) 449 year (d) None of these
Solution : (c)
l_{a} = 1 per year and l_{b}
= 1
per year and it is given that the fraction of the remained activity A = 1
1620
405
A0 4
Total decay constant l = l_{a} + l_{b}
= 1 + 1 = 1 per year
1620 405 324
We know that A = A e ^{–l}^{t} Þ t = 1 log A0 Þ t = 1 log
4 = 2 log
2 = 324 ´ 2 ´ 0.693 = 449 years.
^{0} l ^{e} A
l e l e
Example: 21 At any instant the ratio of the amount of radioactive substances is 2 : 1. If their half lives be respectively 12 and 16 hours, then after two days, what will be the ratio of the substances [RPMT 1996] (a) 1 : 1 (b) 2 : 1 (c) 1 : 2 (d) 1 : 4
2´24
æ 1 ö 12

æ 1 ö^{n} N (N ) (1 / 2)^{n}1 2 ç 2 ÷ 1
Solution : (a) By using
N = N _{0} ç ÷ Þ
1 = 0 1 ´
= ´ è ø =
è 2 ø
N2 (N0 )2
(1 / 2) 2
1 2´24 1
æ 1 ö
ç ÷ 16
è 2 ø
Example: 22 From a newly formed radioactive substance (Halflife 2 hours), the intensity of radiation is 64 times the permissible safe level. The minimum time after which work can be done safely from this source is
[IIT 1983; SCRA 1996]
 6 hours (b) 12 hours (c) 24 hours (d) 128 hours
æ 1 ö^{n}
A 1 æ 1 ö^{0}
æ 1 ö^{n}
Solution : (b) By using
A = A_{0} ç 2 ÷ Þ A = 64 = ç 2 ÷ = ç 2 ÷
Þ n = 6
è ø _{0} è ø è ø
Þ t = 6
T1 / 2
Þ t = 6 ´ 2 = 12 hours.
Example: 23 nucleus of mass number A, originally at rest, emits an aparticle with speed v. The daughter nucleus recoils
with a speed  [DCE 2000; AIIMS 2004]  
(a) 2v /(A + 4)  (b)  4v /(A + 4)  (c)  4v /(A – 4)  (d)  2v /(A – 4) 
Solution : (c) m
v
+
Rest
According to conservation of momentum 4v = (A – 4)v‘ Þ
v‘ = 4v .
A – 4
Example: 24 The counting rate observed from a radioactive source at t = 0 second was 1600 counts per second and at t = 8 seconds it was 100 counts per second. The counting rate observed as counts per second at t = 6 seconds will be [MP PET 1996; UPSEAT 2000]
(a) 400 (b) 300 (c) 200 (d) 150
æ 1 ö^{n}
æ 1 ö8 / T1 / 2
1 æ 1 ö^{8} ^{/} ^{T}1 / 2
Solution : (c) By using
A = A_{0} ç 2 ÷ Þ 100 = 1600 ç 2 ÷
Þ 16 = ç 2 ÷ Þ T_{1} _{/} _{2} = 2 sec
è ø è ø
è ø
æ 1 ö6 / 2
Again by using the same relation the count rate at t = 6 sec will be
A = 1600 ç 2 ÷
= 200 .
è ø
Example: 25 The kinetic energy of a neutron beam is 0.0837 eV. The halflife of neutrons is 693s and the mass of neutrons is 1.675 ´ 10^{–}^{27} kg. The fraction of decay in travelling a distance of 40m will be
(a)
103
104
3
105
106
Solution : (c) v = =
= 4 ´ 10
m/sec
\ Time taken by neutrons to travel a distance of 40 m
Dt‘ =
40
4 ´ 10^{3}
= 10 ^{–}^{2} sec
\ dN = l N dt
Þ dN = l dt
N
\ Fraction of neutrons decayed in Dt sec in
DN = l Dt = 0.693 Dt = 0.693 ´ 10 ^{–}^{2} = 10 ^{–}^{5}
N T 693
Example: 26 The fraction of atoms of radioactive element that decays in 6 days is 7/8. The fraction that decays in 10 days will be (a) 77/80 (b) 71/80 (c) 31/32 (d) 15/16
æ N _{0} ö
æ log
N_{0} ö
æ 1 ö t / T1 / 2
T_{1/} _{2} log _{e} ç N ÷
N t ç e N ÷
Solution : (c) By using
N = N _{0} ç ÷
Þ t = è ø
Þ t µ log e 0 Þ ^{ } ^{1} = è ø1
è 2 ø
log e (2)
N t 2 æ log
N _{0} ö
Hence 6 =
10
log e (8 / 1) log e (N _{0} / N)
Þ log N 0
e N
= 10 log e
6
 = log e
32 Þ
ç
è
N 0 = 32 .
N
_{e} ÷
N ø 2
So fraction that decays = 1 – 1
32
= 31 .
32