Radioactivity

NUCLEAR PHYSICS INTRODUCTION Among about 2500 known nuclides, fewer than 300 are stable. The others are unstable structures that decay to form other nuclides by spontaneously emitting particles and electromagnetic radiation, a process called radioactivity. The time scale of these decay processes ranges from a small fraction of a microsecond to billions of years. The substances which emit these radiations are called as radioactive substances. It was discovered by Henry Becquerel for atoms of Uranium. Later iit was discovered that many naturally occurring compounds of heavy elements like radium, thorium etc. also emit radiations. At present, it is known that all the naturally occurring elements having atomic number greater than 82 are radioactive. For example some of them are; radium, polonium, thorium, actinium, uranium, radon etc. Later on Rutherford found that emission of radiation always accompained by transformation of one element (transmutation) into another. In actual radioactivity is the result of disintegration of an unable nucleus. Rutherford studied the nature of thesed radiations and found that these mainly consist of  particles (rays). -Particles : (2He4) These carry a charge of +2e and mass equal to 4mp. These are nuclei of helium atoms. The energies of -particles very form 5 Me V to 9 Me V ; theiir velocities vary from 0.01 – 0..1 times of c (velocity of light). They can be deflected by electric and magnetic field and have lower penetrating power but high ionising power. -Particles : (1e0) These are fast moving electrons having charge equal to e and mass me = 9.1 × 10–31 kg. Their velocities very from 1% to 99% of the velocity of light (c). They can also be deflected by electric and magnetic fields. They have low ionising power but high penetrating power. -particles : (00 ) These are electro-magnetic waves of nuclear origin and of very short wavelength. They have no masss. They have maximum penetrating power and minimum ionising power. The energy released in a nuclear reaction is mainly emitted in from these -radiations. RADIOACTIVE DECAYS Generally, there are three types of radioactive decays (i)  decay (i) (ii)   and   decay (iii)  decay  decay In  decay, the unstable nucleus emits an  particle. By emitting  particle, the nucleus decreases it's mass energy number and move towards stability. Nucleus having A>210 shows  decay.By releasing  particle, it can attain higher stability and Q value is positive. B.E. A A A=210 112 (ii)  decay In beta decay (N/Z) ratio of nucleus is changed. This decay is shown by unstable nuclei. In beta decay, either a neutron is converted into proton or proton is converted into neutron. For better understanding we discuss N/Z graph. There are two type of unstable nuclides Region of stable nuclei N N=Z 45° Z  A type Move towards stability For A type nuclides (N/Z)A > (N/Z) stable To achieve stability, it increases Z by conversion of neutron into proton 0 n  1p + e +  , ZX  1 1 –1 A N 1 e  Y +  particle  Z+1 A A This decay is called  –1 decay. Z Kinetic energy available for e–1 and  is, Q  K  K  K.E. of  satisfies the condition 0 < K< Q  B type B For B type nuclides (N/Z)B  (N/Z) stable neutron. That is, p  n  e    positron  (iii)   neutrino  , Z X A  Z 1 Y A  e  particle   Move towards stability N To achieve stability it decreases Z by the conversion of a proton into  Z  decay : when an  or  decay takes place, the daughter nucleus is usually in higher energy state, such a nucleus comes to ground state by emitting a photon or photons. Co 67 Ni* 67 E =1.17MeV E =1.33MeV Order of energy of  photon is 100 KeV e.g. 67 27 Co  E =2.5MeV 67 28 Ni *  higher energy state      , 67 28 Ni* 26 87 Ni   photon 114 NUCLEAR PHYSICS Properties of ,  and  rays Features  –particles –rays Helium nucleus or doubly Fast moving electrons Electromagnetic wave ionised helium atom ( 2He ) ( – or  ) (photons) Charge Twice of proton (+2e)  4mp Electronic (– e) Neutral Mass (rest mass of  rest mass = 0 mp–mass of proton = (rest mass of electron) 1.4 × 10 m/s. to 1% of c to 99% of c Only c = 3 × 108 m/s 2.2 × 107 m/s. (All possible values –photons come out with (Only certain value between this range) same speed from all between this range). –particles come out with types of nucleus. Their speed depends on different speeds from the So, can not be a nature of the nucleus. same type of nucleus. characteristic speed. So that it is a So that it can not characteristic speed. be a characteristic speed. K.E.  MeV  MeV  MeV Energy Line and discrete Continuous Line and discrete spectrum (or linear) (or linear) Ionization 10,000 times power (>>) of –rays 100 times of –rays 1 (or times of ) 1 00 Penetration 1 times of –rays 1 00 00 Identity    –particles 4 Speed 7 power (>>) Effect of electric Deflection 0 – 1 times of –rays 1 00 (100 times of ) 1 (or 1 times of ) 1 00 1(100 times of ) Deflection (More than ) No deflection or magnetic field Explanation By Tunnel effect By weak nuclear With the help of energy of emission (or quantum mechanics) interactions levels in nucleus Laws of Radioactive Decay 1. The radioactive decay is a spontaneous process with the emission of ,  and  rays. It is not influenced by external conditions such as temperature, pressure, electric and magnetic fields. 2. The rate of disintegration is directly proportional to the number of radioactive atoms present at that time i.e., rate of decay  number of nuclei. Rate of decay =  (number of nuclei) i.e. dN  N dt 114 where  is called the decay constant. This equation may be expressed in the form N  N0 dN   N t  dt 0 dN  dt . N  N  l n    t  N0  where N0 is the number of parent nuclei at t=0. The number that survives at time t is therefore N=N0e–t and t = N  2 .303 log10  0  this function is plotted in figure.   Nt  N0 – t N'=N0(1–e ) 0.63N0 Graph : Time versus N (or N') N N0 2 N 0.37N 0= e 0 (0,0) (1) N=N0e – t Th Ta time Half life (Th) : It is the time during which number of active nuclei reduce to half of initial value. If at t = 0 no. of active nuclei N0 then at t = Th number of active nuclei will be N0 2 From decay equation N = N0e–t 0 .6 9 3 N0 l n 2  = N0e–Tn  Th = =  2  (2)  0 .7  Mean or Average Life (Ta) : It is the average of age of all active nuclei i.e. Ta = (i) sum of times of existance of allnucleiin a sample 1 =  initialnumber of active nuclei in that sample At t = 0, number of active nuclei = N0 then number of active nuclei at t = Ta is N  N 0 e Ta  N 0 e 1  (ii) N0  0 .3 7N 0  37 % of N 0 e Number nuclei which have been disintegrated within duration T a is N' = N0 – N = N0 – 0.37 N0 = 0.63 N0 = 63% of N0 (a) Ta = Th Th 1 = = = 1.44 Th   ln 2  0 .6 9 3 (b) Within duration Th  50% of N0 decayed and 50% of N0 remains active (c) Within duration Ta  63% of N0 decayed and 37% of N0 remains active 116 NUCLEAR PHYSICS Activity of a sample (or decay rate) It is the rate of decay of a radioactive sample R   (i) dN  N or R = R0e–t dt Activity of a sample at any instant depends upon number of active nuclei at that instant. R  N (or active mass) , R  m (ii) R also decreases exponentially w.r.t. time same as the number of active nuclei decreases. (iii) R is not a constant with N, m and time while , Th and Ta are constant (iv) At t = 0, R = R0 then at t = Th (v) R R0 and at t = Ta  R = 0 or 0.37 R0 e 2 Similarly active mass of radioactive sample decreases exponentially. m = m0e–t (vi) Activity of m gm active sample (molecular weight Mw) is R = N =  R= 0 .6 9 3  N AV   m Th  MW  here NAV = Avogadro number = 6.023 × 1023 SI unit of R : 1 becquerel (1 Bq)= 1 decay/sec Other unit is curie : 1 Ci = 3.70 × 1010 decays/sec 1 Rutherford : (1 Rd) =106 decays/s Specific activity : Activity of 1 gm sample of radioactive substance. Its unit is Ci/gm e.g. specific activity of radium (226) is 1 Ci/gm. Ex. The half–life of cobalt–60 is 5.25 yrs. After how long does its activity reduce to about one eight of its original value? Sol. The activity is proportional to the number of undecayed atoms: In each half–life, the remaining sample decays to 1 1 1 1 half of its initial value. Since          , therefore, three half–lives or 15.75 years are required for the 2 2 2 8 sample to decay to 1/8th its original strength. Ex. A count rate meter is used to measure the activity of a given sample. At one instant the meter shows 4750 counts per minute. Five minutes later it shows 2700 counts per minute. (i) Find the decay constant. Sol.  dN Initial activity Ai = dt t 0 Dividing (i) by (ii), we get (ii) Also, find the half–life of the sample.  N 0  4 75 0 ...(i)  dN Final activity Af = dt t 5  N  2 700 ...(ii) N0 4750 = N 2700 t The decay constant is given by   Half–life of the sample is T= 2 .3 03 N 2 .3 0 3 4750 lo g  0 .1 1 3 m in -1 log 0  5 2700 t Nt 0 .6 9 3 0 . 6 9 3 = = 6.14 min  0 .1 1 3 116 Parallel radioactive disintegration Let initial number of nuclei of A is N0 then at any time number of nuclei of A, B & C are given by N0 = NA + NB + NC  dN A  d  NB  NC  dt dt A disintegrates into B and C by emitting  particle. Now,  dN B dN C d  1 N A and   2 N A   N B  N C     1   2  N A dt dt dt t1 t 2 dN A    1   2  N A   eff  1   2  t eff  t1  t 2 dt Ex. The mean lives of a radioactive substances are 1620 and 405 years for –emission and –emission respectively. Find out the time during which three fourth of a sample will decay if it is decaying both by –emission and –emission simultaneously. Sol. When a substance decays by  and  emission simultaneously, the average rate of disintegration av is given by av= +when  = disintegration constant for –emission only  = disintegration constant for –emission only Mean life is given by Tm = av × t = 2.303 log Ex. 1 1 1 1 1 1 1      ,  =  +   av   Tm T T 16 20 405 3 24 1 100 N0 , t = 2.303 log  t = 2.303 × 324 log 4 = 449 years. 25 Nt 3 2 4 B A radioactive decay is given by A  t1/ 2 8 yrs Only A is present at t=0. Find the time at which if we are able to pick one atom out of the sample, then probability of getting B is 15 times of getting A. Sol. A  B at t  0 N0 0 at N N0  N tt N Probability of getting A, PA = N 0 Probability of getting B, PB = Remaining nuclei are N N0  N N N N  15 PB = 15 PA  0  N0 = 16N  N= 0 16 N0 N0 N0 1 th of initial nuclei, hence required time t = 4 half lives =32 years 16 118 NUCLEAR PHYSICS Radioactive Disintegration with Successive Production    B  rate of production   A  dN A    N A ....(i) dt when NA in maximum dNA  rate of production  0    NA  0 , NA max = = dt   t t dN A    dt, Number of nuclei is N A  1  e t By equation (i)     N  A 0 0 Ex. 21    1 1 0 per sec 30   A  B A shows radioactive disintegration and it is continuously produced at the rate of 1021 per sec. Find maximum number of nuclei of A. Sol. At maximum, rproduction = rdecay  1021 = 1 N  N=30 × 1021 30 Soddy and Fajan's Group Displacement Laws : (i) –decay : The emission of one –particle reduces the mass number by 4 units and atomic number by 2 units. If parent and daughter nuclei are represented by symbols X and Y respectively then, XA  Z–2YA–4 + 2He4() Z (ii)  –decay : Beta particles are said to be fast moving electrons coming from the nucleus of a radioactive substance. Does it mean that a nucleus contains electrons? No, it is an established fact that nucleus does not contain any electrons. When a nucleus emits a beta particle, one of its neutrons breaks into a proton, an electron (i.e., –particle) and an antineutrino n  p  e   where n= neutron p = proton e = –particle Thus emission of a beta particle is caused by the decay of a neutron into a proton. The daughter nucleus thus has an atomic number greater than one (due to one new proton in the nucleus) but same mass number as that of parent nucleus. Therefore, representing the parent and daughter nucleus by symbols X and Y respectively, we have ZXA  Z+1YA +   (iii) –decay : When parent atoms emit gamma rays, no charge is involved as these are neutral rays. Thus there is no effect on the atomic number and mass number of the parent nucleus. However the emission of –rays represents energy. Hence the emission of these rays changes the nucleus from an excited (high energy) state to a less excited (lower energy) state. 118