Physics: Modern Physics-II PDF Study Material

TG: @Chalnaayaaar PRE-MEDICAL PHYSICS ENTHUSIAST | LEADER | ACHIEVER STUDY MATERIAL Modern PhIysics-II ENGLISH MEDIUM  TG: @Chalnaayaaar All rights including trademark and copyrights and rights of...

TG: @Chalnaayaaar PRE-MEDICAL PHYSICS ENTHUSIAST | LEADER | ACHIEVER STUDY MATERIAL Modern PhIysics-II ENGLISH MEDIUM  TG: @Chalnaayaaar All rights including trademark and copyrights and rights of translation etc. reserved and vested exclusively with ALLEN Career Institute Private Limited. (ALLEN) No part of this work may be copied, reproduced, adapted, abridged or translated, transcribed, transmitted, stored or distributed in any form retrieval system, computer system, photographic or other system or transmitted in any form or by any means whether electronic, magnetic, chemical or manual, mechanical, digital, optical, photocopying, recording or otherwise, or stood in any retrieval system of any nature without the written permission of the Allen Career Institute Private Limited. Any breach will entail legal action and prosecution without further notice. This work is sold/distributed by Allen Career Institute Private Limited subject to the condition and undertaking given by the student that all proprietary rights (under the Trademark Act, 1999 and Copyright Act, 1957) of the work shall be exclusively belong to ALLEN Career Institute Private Limited. Neither the Study Materials and/or Test Series and/or the contents nor any part thereof i.e. work shall be reproduced, modify, re-publish, sub-license, upload on website, broadcast, post, transmit, disseminate, distribute, sell in market, stored in a retrieval system or transmitted in any form or by any means for reproducing or making multiple copies of it. Any person who does any unauthorised act in relation to this work may be liable to criminal prosecution and civil claims for damages. Any violation or infringement of the propriety rights of Allen shall be punishable under Section- 29 & 52 of the Trademark Act, 1999 and under Section- 51, 58 & 63 of the Copyright Act, 1957 and any other Act applicable in India. All disputes are subjected to the exclusive jurisdiction of courts, tribunals and forums at Kota, Rajasthan only. Note:- This publication is meant for educational and learning purposes. All reasonable care and diligence have been taken while editing and printing this publication. ALLEN Career Institute Private Limited shall not hold any responsibility for any error that may have inadvertently crept in. ALLEN Career Institute Private Limited is not responsible for the consequences of any action taken on the basis of this publication. ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical ERNST RUTHERFORD (1871 –1937) British physicist who did pioneering work on radioactive radiation. He discovered alpha-rays and beta-rays. Along with Federick Soddy, he created the modern theory of radioactivity. He studied the ‘emanation’ of thorium and discovered a new noble gas, an isotope of radon, now known as thoron. By ® scattering alpha-rays from the metal foils, he discovered the atomic nucleus and proposed the plenatery model of the atom. He also estimated the approximate size of the nucleus. MARIE SKLODOWSKA CURIE (1867-1934) Born in Poland. She is recognised both as a physicist and as a chemist. The discovery of radioactivity by Henri Becquerel in 1996 inspired Marie and her husband Pierre Curie in their researches and analyses which led to the isolation of radium and polonium elements. She was the first person to be awarded two Nobel Prizes - for Physics in 1903 and for chemistry in 1911. 156  Physics : Modern Physics-II TG: @Chalnaayaaar ® Pre-Medical NUCLEAR PHYSICS 1. NUCLEUS (i) Central core of every atom. (ii) Discovered by Rutherford in α-scattering experiment. (iii) The order of nuclear size = 10–15 m or fm while the order of atomic size = 10–10 m or Å (iv) Protons and neutrons, together referred as nucleons. A (v) A nuclide is represented by ZX Z = atomic number = p (no. of protons) A = mass number = total no. of nucleons = n + p (vi) Atomic masses are generally represented by atomic mass unit (u) mass of C12 atom 1u = = 1.66 × 10–27 kg ® 12 mp = 1.6726 × 10–27 kg = 1.00727 u –27 mn = 1.6749 × 10 kg = 1.00866 u me = 9.1 × 10–31 kg = 0.00055 u 1.1 Types of Nuclei (i) Isotope : same Z (ii) Isobar : same A (iii) Isotone : same (A – Z) 1.2 Properties of Nuclei Size of Nucleus : (Order is fermi) As the number of nucleons in nucleus increases its size also increases and relation between its radius and R∝A 1/3 mass number is 1/3 R = R0A Here R0 is a constant and its value R0 ≈1.2 fm. Volume of Nucleus Volume ∝ R (But R ∝ A ) volume ∝ A 3 1/3 or Mass of Nucleus Its mass is quite small compare to gm or kg. Therefore it is measured in another unit - amu (Atomic Mass Unit) Mass of an nucleus of mass number A is ~ Amp ~ A amu or mass of an nucleus, m ∝ A Density of Nucleus (ρ) mass Am p Am p 3m p ρ= ≅= =  2.3 × 1017 kg / m 3 volume 4 3 4 3 4 πR 0 3 πR πR 0 A 3 3 It means ρ is independent of A. Density of nuclei of all types of element is same and its order is 10 kg/m 17 3 14 3 or 10 g/cm 157 ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical Illustrations Illustration 1. Calculate mass no. of that nucleus whose radius is half of Ge72. Solution. r∝A 1/3 r 1/3 2  A 1 A =   ⇒ = ⇒A=9 r  72  8 72 Illustration 2. 12 Find the density of 6 C Solution. Mass of nucleus ≈ 12mP = 12 × 1.66 × 10 –27 kg {mP = mass of proton} ® M 12 × 1.66 × 10 –27 kg ρ= = = 2.4 × 1017 kg m–3 4 4 π [1.2 × 10 –15 (12)1/3 ] m 3 3 πR 3 3 3 ___________________________________________________________________________________________________ 1.3 Forces acting inside the nucleus There are three forces interacting between nucleons, these are (i) Gravitational force - weakest force of nature (ii) Electrostatic repulsive (coulombian) force → only works between proton proton. This is stronger than gravitational force. (iii) Nuclear force → strongest intraction that holds nucleons together to form nuclei and it is powerful enough to overcome the electric repulsion of proton and proton. 1.4 Features of Nuclear Force (Fn ) :– 1. The strongest force in the universe. 2. Works only between the nucleons. 3. Very short range : only upto size of nucleus (3 or 4 fermi). More than this distance, nuclear force is almost zero. 4. Very much depends upon distance :– Small variation in distance may cause of large change in nuclear force while electrostatic force remains almost unaffected. 5. Independent of charge :– Interacts between n–n as well as between p–p and also between n–p. 6. Spin dependent :– It is stronger between nucleons having same sense of spin than between nucleons having opposite sense of spin. 7. It is not a central force :– Definition of central force (Fc) : Whose line of action always passes through a fixed point and its magnitude depends only on distance, if medium is same. → K Fc = (± r̂ ) is central force. rn Electrostatic and gravitational forces are central forces. 8. Nature :– (i) Attractive – If distance is greater than 0.8 fm or above. (ii) Repulsive – If distance is lesser than 0.8 fm. 158  Physics : Modern Physics-II TG: @Chalnaayaaar ® Pre-Medical 2. EINSTEIN'S MASS ENERGY EQUIVALENCE According to Einstein, mass can be converted into energy and energy into mass. This relation is given by - E = mc2 Here E = total energy associated with mass m ; c2 = used as a conversion coefficient 2.1 Mass defect (i) Mass of a nucleus is always less than the sum of masses of its constituent nucleons. This difference is called mass defect. A (ii) If observed mass of nucleus ZX be M, mass of proton is Mp and mass of neutron is Mn then mass defect = ∆m = [ZMp + (A – Z)Mn] – M. (iii) If M is taken as mass of atom of ZXA instead of mass of nucleus then ® ∆m = [Z(Mp + Me) + (A – Z)Mn] – Matom 2.2 Binding energy (Eb) (i) Binding energy of a nucleus is the energy required to split it into its nucleons (free). (ii) ∆Eb = ∆m.c2 (iii) It is always positive and numerically equal to the energy equivalent of mass defect (or equal to the energy liberated when it was formed)  ∆Ε  2.3 Binding Energy per Nucleon  b   A  10 32 56 100 Mo Fe Binding energy per nucleon (MeV) 16 S 127 I C O 197 12 An 8 4He 18 O 184 W 238 14 N U 6 6 Li 4 3 H 2 2 H 0 0 50 100 150 200 250 Mass number (A) FIGURE : The binding energy per nucleon as a function of mass number. (i) The value of binding energy per nucleon decides the stability of a nucleus. It is obtained by dividing binding energy by the mass number of given nucleus. (ii) The following figure shows the binding energy per nucleon plotted against the mass number of various atoms nuclei Greater the binding energy per nucleon, the more stable the nucleus. 56 (iii) It is maximum for isotope of iron −26 Fe and is 8.8 MeV/nucleon. It is the most stable nucleus. (iv) For Uranium, binding energy per nucleon is about 7.7 MeV/nucleon and it is unstable. (v) The medium size nuclei are more stable than light or heavy nuclei. 159 ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical 3. NUCLEAR FISSION Splitting of a heavy nucleus (A > 230) into two or more lighter nuclei when struck by a neutron. In this process certain mass disapears which is obtained in the form of energy (enormous amount) A + n → excited nucleus → B + C + Q 235 Hahn and Strassmann done the first fission (fission of nucleus of U ). 235 When U is bombarded by a neutron it splits into two fragments and 2 or 3 secondary neutrons and releases about 200 MeV energy per fission (or from single nucleus) Fragments are uncertain but each time energy released is almost same. Possible reactions are - U 235 + 0n1 → Ba + Kr + 30n1 + 200 MeV or U235 + 0n1 → Xe + Sr + 20n1 + 200 MeV ® and many other reactions are possible. (i) The average number of secondary neutrons is 2.5. (ii) Nuclear fission can be explained by using " liquid drop model" also. (iii) The mass defect ∆m is about 0.1% of mass of fissioned nucleus (iv) About 93% of released energy (Q) is appear in the form of kinetic energies of products and about 7% part in the form of γ - rays. 4. NUCLEAR CHAIN REACTION The equation of fission of U 235 is 92 U235 + 0n1 → 56Ba144 + 36Kr89 + 30n1 + Q These three secondary neutrons produced in the reaction may cause of fission of three more U235 and give 9 235 neutrons, which in turn, may cause of nine more fission of U and so on. Thus a continuous 'Nuclear Chain reacion' would start. –6 If there is no control on chain reaction then in a short time (10 sec.) a huge amount of energy will be released. (This is the principle of 'Atom bomb') If chain reaction is controlled then produced energy can be used for peaceful purposes. For example nuclear reactor (Based on fission) generats electricity. 4.1 Natural Uranium 235 238 It is mixture of U (0.7%) and U (99.3%) 235 U is easily fissionable, by slow neutron (or thermal neutrons) having K.E. of the order of 0.03 eV. 235 To improve the quality, percentage of U is increased to 3%. The improved uranium is called 'Enriched 238 235 Uranium' (97% U and 3% U ) 4.2 Losses of Secondary Neutrons Leakage of neutrons from the system Due to their high K.E. some neutrons escape from the system. Absorption of neutrons by U238 238 U is not fissionable by these secondary fast neutrons. But U238 absorbs some fast neutrons. 160  Physics : Modern Physics-II TG: @Chalnaayaaar ® Pre-Medical 4.3 Critical Size (or mass) In order to sustain chain reaction in a sample of enriched uranium, it is required that the number of lost neutrons should be much smaller than the number of neutrons produced in a fission process. For it the size of uranium block should be equal or greater than a certain size called critical size. No. of neutrons in a stage 4.4 Multiplication factor (K) = No. of neutrons in just previous stage (i) If size of Uranium used is 'Critical' then K = 1 and the chain reaction will be steady or sustained (As in nuclear reaction) (ii) If size of Uranium used is 'Super critical' then K > 1 and chain reaction will accelerate resulting in a explosion (As in atom bomb) (iii) If size of Uranium used is 'Sub Critical' then K < 1 and chain reaction will retard and ultimately stop. 5. NUCLEAR REACTOR (K = 1) ® Its main constituents are - 5.1 Nuclear Fuel : Commonly used are U235 , Pu239. 239 Pu is the best. But Pu239 is not naturally available and U235 is used in most of the reactors. 5.2 Moderator Its function is to slow down the fast secondary neutrons. Because only slow neutrons are capable for the fission of U235. The moderator should be light and it should not absorb neutrons. Commonly, Heavy water (D2O, molecular weight 20 gm.) are used. 5.3 Control rods They have the ability to capture the slow neutrons and can control the rate of chain reaction at any stage. Boron and Cadmium are best absorber of neutrons. 5.4 Coolant A substance which absorb the produced heat and transfers it to water for further use. Generaly coolant is water at high pressure 6. FAST BREADER REACTORS 239 The atomic reactor in which fresh fissionable fuel (Pu ) is produced along with energy. Fuel : Natural Uranium. During fission of U235, energy and secondary neutrons are produced. These secondary neutrons are absorbed by U238 and U239 is formed. This U239 converts into Pu239 after two beta decay. This Pu239 can be separated, its half life is 2400 years. 2β − 92 U238 + 0n1 → 92U239  → Pu239 (best fuel of fission) 94 239 This Pu can be used in nuclear weapons because of its small critical size than U235. 7. NUCLEAR FUSION It is the phenomenon of fusing two or more lighter nuclei to form a single heavy nucleus. A + B → C + Q (Fusion) The product (C) is more stable then reactants (A and B). and mc < (ma + mb) 161 ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical and mass defect ∆m = [(ma + mb)- mc] amu Energy released is E = ∆m × 931 MeV/amu The total binding energy and binding energy per nucleon C both are more than of A and B. ∆E = Ec – (Ea + Eb) Fusion of four hydrogen nuclei into helium nucleus - 4(1H ) → 2He + 2 +β + 2 ν + 26 MeV 1 4 0 (i) Energy released per fission >> Energy released per fusion  200   26  (ii) = Energy per nucleon in fission − 0.85MeV  1.02 MeV, When electron and positron combines they when interact with a nucleus produces pair of annihilates to each other and only energy is – + electron (e ) and positron (e ). released in the form of two gama photons. The energy equivalent to rest mass of e– (or e+)=0.51 MeV. The energy equivalent to rest mass of pair – + (e + e ) = 1.02 MeV. For pair production Energy of photon1.02 MeV. If energy of photon is more than 1.02 MeV, the extra energy (E-1.02) MeV devides approximately in equal amount to each particle as the kinetic  E Ph − 1.02  energy. (K.E.)e− or e+ =   MeV  2  If E < 1.02 MeV, pair will not produce. 162  Physics : Modern Physics-II TG: @Chalnaayaaar ® Pre-Medical GOLDEN KEY POINTS  The density of nuclear matter is independent of the size of the nucleus. The mass density of the atom does not follow this rule.  The radius of a nucleus determined by electron scattering is found to be slightly different from that determined by alpha-particle scattering. This is because electron scattering senses the charge distribution of the nucleus, whereas alpha and similar particles sense the nuclear matter.  The nature of the binding energy (per nucleon) curve shows that exothermic nuclear reactions are possible, when two light nuclei fuse or when a heavy nucleus undergoes fission into nuclei with intermediate mass.  For fusion, the light nuclei must have sufficient initial energy to overcome the coulomb potential barrier. That is why fusion requires very high temperatures.  Although the binding energy (per nucleon) curve is smooth and slowly varying, it shows peaks at nuclides like 4 He, 16O etc. This is considered as evidence of atom-like shell structure in nuclei. ®  A free neutron is unstable (n → p + e– + v ). But a similar free proton decay is not possible, since a proton is (slightly) lighter than a neutron.  Gamma emission usually follows alpha or beta emission. A nucleus in an excited (higher) state goes to a lower state by emitting a gamma photon. A nucleus may be left in an excited state after alpha or beta emission. Successive emission of gamma rays from the same nucleus is a clear proof that nuclei also have discrete energy levels as do the atoms.  Hydrogen bomb is based on fusion. Illustrations Illustration 3. The mass defect in a nuclear fusion reaction is 0.05%. What amount of energy will be liberated in one kg fusion reaction ? Solution. 0.05 Mass defect = ∆m = 0.05% of 1 kg = kg = 5 × 10–4 kg 100 Energy liberated = (∆m)c2 = (5 × 10–4) (9 × 1016)J = 45 × 1012 J Illustration 4. 235 What is energy released by fission of 1 gm U ? Solution. 235 NA Number of atom in 1 gm of U = 235 NA 6.023 × 1023 Energy released = × 200 MeV = × 200 MeV = 5 × 1023 MeV 235 235 23 –13 10 = (5 × 10 ) (1.6 × 10 J) = 8 × 10 J 8 × 1010 4 = 6 kWH = 2.22 × 10 kWH 3.6 × 10 163 ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical Illustration 5. 235 What is the power output of 92 U reactor if it takes 30 days to use up 2 kg of fuel and if each fission gives 185 MeV of usable energy ? Solution. 235 6.02 × 1023 × 2 × 103 Number of atoms in 2 kg of 92 U = = 5.12 × 1024 235 24 14 Therefore, energy released in 30 day = 5.12 × 10 × 185 MeV = 1.51 × 10 J 1 ⋅ 51 × 1014 ∴ Energy released per second = = 58.4 MW 30 × 24 × 60 × 60 Illustration 6. Obtain the binding energy of a nitrogen nucleus ( 14 7 ) N in MeV from the following data. mH = 1·00783u, mn = 1·00867 u, mN = 14·00307 u ® Solution. Mass defect ∆m = 7mp + 7mn – mN = 7 × 1·00783 + 7 × 1·00867 – 14·00307 = 0·11243 amu Binding energy = ∆m × 931 MeV = 0·11243 × 931 MeV = 104·67 MeV Illustration 7. Explain nuclear fission & fusion on the basis of binding energy of nucleus. [AIPMT 2004] Solution. Binding energy per nucleon C A R Q P Mass number In fission : nucleus A breaks into B & C UV In fussion : P & Q fuse to result in nucleus R W In both cases the net B.E. increases resulting in energy release. Illustration 8. 238 238 237 Show that the nucleus 92 U emitting a proton through the decay process 92 U → 91 Pa + 11H can not proceed spontaneously. [AIPMT 2006] Mass of uranium = 238.05079 a.m.u. Mass of paladium = 237.05121 a.m.u. Mass of proton = 1.00783 a.m.u. Solution. 2 2 Here Q = (238.05079 – 237.05121 – 1.00783)c = (– 0.00825u)c As the Q for this process is negative, the decay can not proceed spontaneously 164  Physics : Modern Physics-II TG: @Chalnaayaaar ® Pre-Medical Illustration 9. Write three characteristic features which distinguish nuclear force from coulomb force. [AIPMT 2007] Solution. Nuclear force Coulomb force (Write any three) (i) Short range Long range (ii) Not a central force Central force (iii) Spin dependent Spin independent (iv) Charge independent Charge dependent (v) Strong force Comparitively weak force ® Illustration 10. C decays by β emission. Write symbolically this decay process. + The radionuclide 11 6 Given that m( 11 6 C ) = 11.011434 u [AIPMT 2008] m( 11 5 B ) = 11.009305 u me = 0.000548 u, 1u = 931.5 MeV/c2 Calculate the Q-value. Solution. Equation of β+-decay of 6C11 ; 6C11 → 5B11 + +1β0 + ν + Q Q-value of reaction = ∆mc2 = m ( 6 C11 ) − 6me − m ( 5 B11 ) + 5me − me  c2 = m ( 6 C11 ) − m ( 5 B11 ) − 2me  c2 = [11.011434 − 11.009305 − 2 × 0.000548] uc2 2 = [0.001033] uc = 0.001033 × 931.5 MeV = 0.962 MeV Illustration 11. Calculate the percent increase in mass of an electron accelerated by a potential difference of 500 kV. [AIPMT 2004] Solution. Kinetic energy of electron = 500 keV & Rest mass energy of electron = 511 keV 2 2 Total energy = mc = m0c + KE = (511 + 500) keV m − m0 500 Percent increase in mass = × 100 = × 100 = 97.8 % m0 511 165 ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical BEGINNER'S BOX-1 1. The Q value of a nuclear reaction A + b → C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. 1 3 2 2 (i) 1 H +1 H →1 H +1 H 12 (ii) 6 C +12 20 4 6 C →10 Ne + 2 He Atomic masses are given to be ® m ( 11 H ) = 1.007825 u m ( 12 H ) = 2.014102 u m ( 13 H ) = 3.016049 u m ( 12 6 C ) = 12.000000 u m ( 10 20 Ne ) = 19.992439 u mass of He atom is 4.0015 amu 2. 35 35 Calculate the binding energy of 17Cl if mass of 17Cl nucleus is 34.98 amu, mass of neutron is 1.008665 amu and mass of proton is 1.007277 amu. 3. 235 Calculate the energy released by the fission of 2 g of 92U in kWh. Given that the energy released per fission is 200 MeV. 4. –11 If the energy released in the fission of one nucleus is 3·2 × 10 J, then find number of nuclei required per second in a power plant of 16 kW. 5. 235 Find out the mass of Uranium required per day to generate 10 MW power from the fission of 92U. 6. The mass defect in a nuclear fusion reaction is 0.3 percent. What amount of energy will be liberated in one kg fusion reaction ? 7. Two nuclei have their mass numbers in ratio 1 : 3. What is the ratio of nuclear densities ?[AIPMT 2006] 166  Physics : Modern Physics-II TG: @Chalnaayaaar ® Pre-Medical RADIOACTIVITY 1. RADIOACTIVITY : (i) Spontaneous emission of radiations from the nucleus is known as radioactivity and substances showing this property are called radioactive substances. (ii) Only unstable nuclei exhibit this property. (iii) A particular nuclide (element) can radiate only a particular type of radiations at a time, according to its requirement of stability. (iv) This phenomenon (Radioactivity) was discovered by Becquerel therefore the radioactive radiations are also called Becquerel radiations. (v) Later on Curie couple (Merie Curie and Pierre Curie) discovered many other radioactive substances. ® 2. NATURE OF RADIOACTIVE RADIATIONS : 2.1 Rutherford's Experiment :– γ-rays – + – α- β-rays + rays – + – + – + radium Pb box (best absorber of nuclear radiations) He put a sample of radioactive substance in a lead box and allow the emission of radiations through a small hole only. When the radiation enter into the external electric field, they split into three parts. Radiations which deflect towards negative plate are called α–rays. Radiations which deflect towards positive plate are called β–rays. Radiations which are undeflected, called γ–rays. (i) Alpha rays :– These are stream of positive charged particles i.e. particle nature. (ii) Beta rays :– These are stream of negative charged particles i.e. particle nature. (iii) Gamma rays :– These are electromagnetic waves. 2.2 Properties of α, β and γ rays :– Features α–particles β–particles γ–rays 1. Identity Helium nucleus or Fast moving electrons Electromagnetic wave doubly ionised helium ( –β or β ) (photons) 0 – 4 ion ( 2He ) 2. Charge Twice of proton (+ 2e) Electronic (– e) Neutral 167 ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical 3. Mass ≈ 4mp , (rest mass of β) rest mass = 0 mp–mass of proton = (rest mass of ele.) 4. Speed ≈ 107 m/s ≈ 107 m/s Only c = 3 × 108 m/s Their speed depends on β-particles come out with γ-photons come out with nature of the nucleus. different speeds from the same speed from all So, it is a characteristic same type of nucleus. types of nucleus. speed. Therefore can not be a So, can not be a characteristic speed. characteristic speed. 5. K.E. ≈ MeV ≈ MeV ≈ MeV 6. Energy spectrum Line and discrete Continuous Line and discrete 7. Ionization power 10,000 times of γ-rays 100 times of γ-rays 1 (α>β>γ) 1 1 ® (or times of α) (or times of β) 100 100 8. Penetration power 1 1 1 times of γ-rays times of γ-rays (γ>β>α) 10000 100 (100 times of β) (100 times of α) 9. Effect of electric Deflection Deflection No deflection or magnetic field (More than α) 3. TYPE OF RADIOACTIVE DECAY 3.1 α–decay In this decay, mass number decreases by 4 and atomic number decreases by 2. Its decay equation is A A –4 Z X → Z –2 Y + 24 He (Parent ) (Daughter ) Alpha In this decay total mass of product is less than the mass of parent. This difference in mass appears as kinetic energy of the products. The disintegration energy or Q value for α-decay. 2 Q = (mX – mY – mHe)c 3.2 β–decay (A) The basic nuclear process underlying β–-decay is the conversion of neutron to proton. n → p + e + ν – (i) Its decay equation A A Z X →Z +1 Y + −1o β + ν (ii) After –β° decay, n/p ratio decreases. (iii) – β° always comes out from the nucleus along with antineutrino. 168  Physics : Modern Physics-II TG: @Chalnaayaaar ® Pre-Medical (B) The basic nuclear process underlying β -decay is the conversion of proton into neutron. + p → n + e + ν + (i) Its decay equation A A Z X →Z –1 Y + +1o β + ν (ii) After +β° decay, n/p ratio increases. (iii) +β° comes out from the nucleus along with neutrino. 3.3 γ–decay :– Similar to an atom, nucleus also have certain energy levels and nucleons occupy them. After α-decay (or β decay), daughter nucleus may be in excited state and return to ground state by emitting photons of high energy (MeV order) called γ - photons. Equation of γ–decay :– (ZX )∗ → ZXA + γ – photons (or hν) A ∗Shows excited nucleus ® (i) γ emission don't change the structure of nucleus (ii) No change in Z and A GOLDEN KEY POINTS  In β-decay either an electron or a positron is emitted by a nucleus, along with an antineutrino or a neutrino. The emitted particles share the available disintegration energy. The electrons and positrons emitted in β-decay have a continuous spectrum of energies from zero to a limit [Q = (∆m) c ] 2  Properties of neutrino & antineutrino (i) Both are chargeless (ii) Have almost zero rest mass (very light particles) (iii) Have spin quantum number ± 1/2 and spin angular momentum ± h/2π similar to electron. (iv) These are suggested by Pauli to explain the problems of energy conservation, linear momentum conservation, spin conservation and spin angular momentum conservation in β–decay.  In β-decay parent & daughter are isobar.  The K-electron capture : In K-capture a nucleus captures one of the inner orbital electrons and a proton transforms into a neutron. Hence K capture is like positron decay, in both n/p ratio increases. In this event a vacancy is created in K-shell to fill up the vacancy, electron transition takes place and X-rays are emitted. Z XA + –1e0 (K-capture) → Z–1YA + X-rays + υ Illustrations Illustration 1. β0 α Th234  → aX → cYd  (p) → 90Th b − 230 90 Find a, b, c and d and identify particle p Solution. 0 0 α −β P= (− β ) 90 Th234  → 88 X 230  → 89 Y 230  → 90 Th230 169 ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical Illustration 2. 0 90 Th234 → nα , n ' β − 83 Bi214, Find n and n' Solution. 234 − 214 20 Number of α particle, n = N= α = = 5 4 4 Number of –β particle, n′ = N–β = Zf – [Zi – 2 × No. of α] = 83 – (90 – 2 × 5) = 83 – 80 = 3 BEGINNER'S BOX-2 0 1. 92 U238 α +β → aX , find a & b. →  b 0 0 2. a Xb  β α →  − 215 → CY β d → 110Y Find a, b, c and d.− 0 3. 92 U238  n α,n '_ β → 82 Pb206. Find n & n' ® 4. Thorium isotope 90Th 232 emits some α–particles and some β–particles and gets transformed into lead isotope 82Pb 200. Find the number of α and β particles emitted. 5. A radioactive nucleus undergoes a series of decays according to the following scheme : − α α γ A  → A1  β → A2   → A3  → A4 If the mass number and atomic number of A are 180 and 72 respectively, what are these numbers for A4? 6. Write nuclear reaction equations for (i) α-decay of 226 88 Ra (ii) α-decay of 242 94 Pu (iii) β–-decay of 15 P 32 (iv) β–-decay of 210 83 Bi (v) β+-decay of 11 6 C (vi) β+-decay of 43 Tc 97 (vii) Electron capture of 54 Xe 120 4. NUCLEAR REACTIONS  It can be written as X (a, b) Y i. e. X + a → Y + b  All nuclear reactions follow conservation of number of nucleons and charge (i.e.Z) conservation as well as energy + mass conservations, linear and angular momentum. 5. MATHEMATICAL DERIVATION OF EXPONENTIAL DECAY Rutherford and Soddy's law At an instant rate of decay of active nuclei is directly proportional to the number of active nuclei at that instant dN − = rate of decay of nuclei at time t dt dN N = active nuclei at time t − ∝ N or − dN = λN...(i) dt dt 170  Physics : Modern Physics-II TG: @Chalnaayaaar ® Pre-Medical Here λ is the decay constant which depends only on the nature of substance. dN equation (i) can be written as = – λ dt N dN Integrate it ∫ N =–λ ∫ dt loge N = – λt + C...(ii) Let at t = 0 number of active nuclei were N0 (by putting t = 0 and N = N0 in equation (ii) ) loge N0 = C Now equation (ii) is logeN = – λt + logeN0 N N logeN – logeN0 = – λt  loge = – λt i.e. = e–λt N0 N0 N = N0e λ – t...(iii) ® equation (iii) gives number of active nuclei in a sample at desire instant t. GOLDEN KEY POINTS  Number of nuclei, which has been decayed in duration t ⇒ N' = N0 – N = N0 (1 - e–λt)  λ is independent of amount of active substance (N or m) and time and any physical or chemical changes.  λ is called decay constant or disintegration constant or radioactivity constant or Rutherford Soddy's constant or the probability of decay per unit time of a nucleus. Graph : Time versus N (or N') N0 N'=N0(1–e–λt) 0.63N0 N N0 2 N 0.37N0= e0 N=N0e λ – t (0,0 Th Ta time 5.1 Half life (Th) It is the time during which number of active nuclei reduce to half of initial value. N0 If at t = 0 no. of active nuclei N0 then at t = Th number of active nuclei will be 2 from decay equation N = N0e–λt N0 ln2 0.693 0.7 = N0e–λTh  Th = = ≈ 2 λ λ λ 171 ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical 5.2 Mean or Average Life (Ta) It is the average of age of all active nuclei i.e. sum of times of existance of all nuclei in a sample 1 Ta = = initial number of active nuclei in that sample λ (i) At t = 0, number of active nuclei = N0 then N0 number of active nuclei at t = Ta is = N N 0 e −λ= T a −1 N 0 e= = 0.37N = 0 37% of N 0 e (ii) Number of nuclei which have been disintegrated within duration Ta is N' = N0 – N = N0 – 0.37 N0 = 0.63 N0 = 63% of N0 ® 1 T Th (iii) Ta = = h = = 1.44 Th λ ln 2 0.693 (iv) Within duration Th  50% of N0 decayed and 50% of N0 remains active (v) Within duration Ta  63% of N0 decayed and 37% of N0 remains active 5.3 Activity of a sample (A or R) (or decay rate) dN (i) It is the rate of decay of a radioactive sample R= − = Nλ or R = R0e–λt dt (ii) Activity of a sample at any instant depends upon number of active nuclei at that instant. R ∝ N (or active mass) , R∝m (iii) R also decreases exponentially w.r.t. time same as the number of active nuclei decreases. (iv) R is not a constant with N, m and time while λ, Th and Ta are constant R0 R0 (v) At t = 0, R = R0 then at t = Th   R= and at t = Ta   R= or 0.37 R0 2 e (vi) Similarly active mass of radioactive sample decreases exponentially. m = m0e–λt 0.693 N (vii) Activity of m gm active sample (molecular weight Mw) is R = λN m AV Th MW 23 here NAV = Avogadro number = 6.023 × 10 SI UNIT of R : 1 becquerel (1 Bq)= 1 decay/sec Other Unit is curie : 1 Ci = 3.70 × 1010 decays/sec 172  Physics : Modern Physics-II TG: @Chalnaayaaar ® Pre-Medical GOLDEN KEY POINTS  In radioactive sample a radioactive substance (A) converts into rather stable substance (B), B converts into more stable substance (C). This process ends into stable element. A → B → C → D →.......  Stability order D > C > B > A  Radioactivity order A > B > C > D  For stable element λ = 0  For stable element Th = ∞  At equilibrium, rate of production of B = rate of decay of B (and also for C,D, etc.) Therefore number of active nuclei of B becomes constant (or activity Nλ also becomes constant) At equilibrium radio activities of all daughter nuclides are equal i.e. R1 = R2 = R3 = R4... at radioactive equilibrium N1λ1 = N2λ2 = N3λ3 λ1 Th m2 m1 Th  MW   Th2  At equilibrium N2 = N1 = N1 2 = NA = NA 2  m2= m1  2   λ2 Th1 M W2 M W1 Th1  M W1   Th1  ®  Radiation dozes is measured in sieverts (Sv) or Rontgen.  Uses of radioactive isotopes in human life (a) In medicine (i) Testing of blood circulation – Cr57 (ii) Brain tumer detecting – Hg203 131 (iii) Thyroid testing (cancer) – I (iv) Cancer cure – Co60 (v) Blood cancer cure – Au189 /Na24 (b) In Archaeology (i) For determining age of archaeological sample ( ≈ 30,000 yr old) – C14 (carbon dating) 40 (ii) For determining age of earth or meteorites (very old) – K and Uranium (c) In Agriculture (i) For protacting potato from earthworm – Co60 (ii) Artificial rains by – AgI 32 (iii) As fertilizers – P  Geiger – Muller counter is used for detecting (or counting) the α particles and β–particles. Illustrations Illustration 3. 9 In a old rock, ratio of nuclei of uranium and lead is 1 : 1. Half life of uranium is 4.5 × 10 yrs. Let initially it contains only uranium nuclei. How old is the rock ? Solution. Let present active nuclei of uranium is N then intial active nuclei is 2N. 1 1 1 Present active fraction of uranium =   = 2 2 2t / T1/2 t or = 1 or t = T1/2 = 4.5 × 109 yr T1/2 173 ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical Illustration 4. The mean lives of a radioactive substance are T1 and T2 for α–emission and β–emission respectively. If it is decaying by both α–emission and β–emission simultaneously then find its mean life and decay constant ? Solution. 1 1 T1 T2 1  λ = λ1 + λ2  λ= +  τ= [ τ = ] T1 T2 T1 + T2 λ Illustration 5. The half lives of X and Y are 3 minutes and 27 minutes respectively. At some instant activity of both are same, then the ratio of active nuclei of X and Y at that instant is ? Solution. A1 = λ1N1 and A2 = λ2N2   0.693 0.693 A1 = A2    N1 = N2 T1 T2 ® N1 N2 N1 3 1   =  = =   N1 : N2 = 1 : 9 T1 T2 N2 27 9 Illustration 6. Decay constant of two radioactive samples is λ and 3λ respectively. At t = 0, they have equal number of active nuclei. Calculate when will be the ratio of active nuclei becomes e : 1. Solution. Number of active nuclei of two radioactive sample is N1 e N 01e −λt N1 = N01e–λt and N2 = N02e–3λt ∴ = = = e2λt [ N01 = N02 ] N2 1 N 02 e −3λt 1 ∴ 1 = 2λt ⇒ t= 2λ Illustration 7. The fraction of a radioactive sample which remains active after time t is. What fraction remains active after t time. 2 Solution. N 9 Active fraction = = e–λt At time t, = e–λt N0 16 1 1  9 2 3 ( At time t/2 active fraction = x = e–λt/2 = e – λt ) 2 So x =    16  = 4 Illustration 8. 10 Calculate the radioactive disintegration constant if 3.7 × 10 alpha particles are emitted by 1 gram of radium 23 per second. Avogadro's number is 6.03 × 10 and the mass number of radium is 226. Solution.  NA   6.03 × 1023  Activity = Nλ =  × m  λ ⇒ 3.7 × 1010 =  × 1 λ  Mw   226  3.7 × 1010 × 226 λ= –11 = 1.38 × 10 per second 6.03 × 1023 174  Physics : Modern Physics-II TG: @Chalnaayaaar ® Pre-Medical Illustration 9. A free neutron is unstable against β–decay with a half life of about 600 seconds.– (i) Write the expression of this decay process. (ii) If there are 600 free neutrons initially, calculate the time by which 450 of them have decayed. Also determine the initial decay rate of the sample. [AIPMT 2005] Solution. n→p+e + ν – (i) The number of undecayed neutron would be 150 by using N = N0e λ – t (ii) 150 = 600e λ ⇒ t = 2T1/2 = 1200 sec – t Decay rate (initial) R = λN0 = 0.693 Bq Illustration 10. 210 Obtain the amount of polonium (84Po ) necessary to provide a radioactivity source of 5.0 mili curie strength. The half life of polonium is 138 days. (given : 1 curie = 3.7 × 1010 disintregration/sec., Avogadro ® 26 number = 6.02× 10 per k-mole). [AIPMT 2006] Solution. dN 1 dN = – λN ⇒ N=– dt λ dt dN Given : = 5 × 10–3 × 3.7 × 1010 disint./sec. & T1/2 = 138 × 24 × 3600 sec. dt 138 × 24 × 3600 × 5 × 3.7 × 107 ⇒ N= = 3.18 × 1015 atoms 0.693 210 210 But mass of one 84Po atoms = 6.02 × 1023 210 210 × 3.18 × 1015 Amount of 84Po in grams required = 23 = 1.11 × 10–6 6.02 × 10 BEGINNER'S BOX-3 1. The half lives of radioactive elements x and y are 3 minute and 27 minute respectively. If the activities of both are same, then calculate the ratio of number of atoms of x and y. 2. Carbon has two stable isotopes. Natural carbon has 98.9% carbon–12 and 1.1% carbon–13, calculate the average atomic weight of carbon. 3. A radioactive isotope has a half life of T. After how much time is its activity reduced to 6.25% of its original activity ? 2 4. fraction of a sample disintegrates in 7 days. How much fraction of it will decay in 21 days ? 3 5. The half life of radium is 1600 years. After how many years 25% of radium block remains undecayed ? 6. A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value? 7. Obtain the amount of 60 27 Co necessary to provide a radioactive source of 8.0 mCi strength. The half-life of 60 27 Co is 5.3 years. 175 ®  TG: @Chalnaayaaar Physics : Modern Physics-II Pre-Medical ANSWERS KEY BEGINNER BOX-1 − −β (iii) 32 15 P   → 16 32 S + 0−1β + υ 1. (i) –4.031 MeV, endothermic − −β (ii) 5.64 MeV, exothermic (iv) 210 83 Bi   → 84 210 Po + 0−1β + υ 2. 278.92 MeV 3. 4.55 × 104 kWh. −β C  + → 11 (v) 5 B + + 1β + υ 11 0 6 4. 5 × 1014 5. 10.5 g + −β (vi) 97 Tc  → 97 42 Mo + + 1β + υ 0 6. 2.7 × 10 J 7. 1 : 1 14 43 BEGINNER BOX-2 (vii) 120 54 Xe + −1e 0  electron capture → 120 53 I + X-Ray + υ 1. a = 89, b = 234 BEGINNER BOX-3 2. a = 110, b = 219, c = 109, d = 215 1 1. 2. 12.011 amu ® 3. n = 8, n' = 6 9 4. Nα = 8, Nβ = 8 26 3. 4T 4. 5. 3200 years 27 5. Mass number = 172 and Atomic number = 69 6. (a) 5T, (b) 6.64T −α 6. (i) 226 88 Ra → 222 86 Rn + He 4 2 7. 7 µg −α (ii) 242 94 Pu  → 238 92 U + 2 He 4 Elementary particles Bosons Fermions (s = 0, ±1, ±2....) (s = ±1/2, ±3/2....) Photon Gluons Graviton π-mesons Baryons Leptons Quarks (Hypothetical) ↓ ↓ ↓ n, n , p, p e, e , ν, ν + up and down Higgs boson W boson etc. etc. etc. (s = 0) & Z boson Note : Fermions obey pauli's exclusion principle but bosons do not obey it. 176