Radioactivity - PDF

136 Nirav Cullege Physies (03 6 RADIOACTIVITY 1. Introduction: You have studied adioactivity in 12h sta...

136 Nirav Cullege Physies (03 6 RADIOACTIVITY 1. Introduction: You have studied adioactivity in 12h standard. If N, and N are the number of in the element at tme 0 and respectively. nucli N= N, e.(A) This law is known as cxponential law of disintegration or law of radioactive A IS called decay constant or disintegration costant of the element. decay You have also studied the propertics of radioactive radiations as well as and Soddy's laws of radoactive disintegration. Relation between hali life tine and decaN Rutherford constant was obtaned as 0 693.(B) 2. Average life time t : Al the nuclei do not disintegrate simeltaneously in a given sample of radioactive element. Some of the nuclei disintegrate s0on after their birth while some of them live for very long time. Thus, the ife of individual nucleus of aradioactive element lies betven zero to infinity. Hence, the average life time is defined as follows Total lite time of all nuclei in a given sample Average life time =..() Total number of nuclei in that sample Let Na be the number of undisintegrated nuclei in a given sample at time = 0and Nbe this number at tine =1 Let dN be anumber of nucBei disintegrating in a very simall interval of time dt atter this. This dN nuclei are alive upto a time Hence their total life time = | N...2) Total fe of all nuclei =..(3) and total number of nucle: in the begining dN..4) Radioactivity 137 Average life time according to definition, Nå..(5) cN Ny Now N = N, e-r...(6) N dt = N, e- (...(7) dN = N e- (- ) dr..(8) Now dN = (- No)...(9) No From equ. (3), equ. (5) and equ. (6), T =...(10) (-N) In equ. (3) integration is over number N while in eqn. (7) the integration is over time 1.When N= N, then I= 0and when N=0 then = o, Thus limits of integration are change accordingly...(11) Integrating by parts, | di =..(12).(13) 138 Nirav College Physics (103) From equ. (1|) and equ. (13),...(14) Thus, life time of a radioactive element is an inverse of ils decay constant. 3. Radioactive series As the radioactive element is unstable it is converted into new element if it emils a particle or ß particle.The disinitigrated element is known as parent element and newly produced element is known as daughter element. If this daughter element is also unstable. it disintigrates and produces other new daughter element. This proceeds ahead. Process of disintegration stops when a stable element is produced. Group of radioactive elements obtained due to the disintegration of different elements arranged in chronological order form radioactive series. They have been classified into four series from the study of nuclides obtained in nature. They are shown in fig. (1). fig. (2). fig. (3) and fig. (4). 24Pt 32Th 237|| 2 Am 140 281P 140 22Ac Np -Z’ N=A-Z 233U A 229Th N= 25Ra B0 2 2122 130 2408T: -emjssion 130 -emission 213Bi, 217|4t -emjssior) B-emissior 209T 213Pþ 80 84 88 92 80 92 84 88 Fig. 1 Fig. 2 Radioactivity 139 238|| Pa 231TK 140 rtu 140 27 Alc N=A-Z’ NA-Z 231Th 223 Pa a19Rh spb 211 Z pb 2113i 130 -emjssio 130 207T 211Po -em]ssion 210Bi -emissior -emjssion 214Pd 206 PHL 80 84 88 92 80 84 88 92 Fig. 3 Fig. 4 Atomic mass number of the element changes by 4 due to emission of an a praticle. Due to this fact total 4 types of such series are obtained. The value of the atomic mass number of all members of onTh2 series can be represented by 4n. Therefore it is kinown as 4n series. The same applies to the other series. Series Parent element Half life time of last - stable Parent product A = 4n Thorium 90Th232 1.39 x 1010 years 82Pb208 93Np237 2.25 × 100 years 83Bj209 A = 4n + | Neptunium A = 4n + 2 Uranium 92U238 4.5| x 109 years 82Pb206 92U235 7.07 × 10* years 82P207 A 4n +3 Actinium Half life time of gzNp237 , Tu = T = 2.25 x 100 years which is very small in comparision with the estimated age of carth - 100 years. So the members of this series are not obtained at present in nature. But they can be produced in the laboratory by bombarding neutrons on heavy elements. Branching disintigration : Generally each element disintigrate in one way or other by emitting. - particle or B- particle. But look at gBj12 in 4n' series in the figure. It disintigrates in two ways. gBi212 and sPo212 by emitting B- particle with a probability of 66.3 % and forms gTI208 140 Nirav College Physics (103) element oy emitting a - particle with a probability of 33.7 %. They then form the same 82Pbus by the emission of a and ß particles respectively. Thus a radioactive element forms process is called e same new element by disintegrating in more than one ways. This branching disintegration. ß - particle Let Ag and a be the probabilities of emission of a - particle and respectively. Therefore, the total activity dN...(18) = - N =- ( t g) N dt and average life time T =.(19) From this,.20) Ta rg. is known as branching ratio. 4. Radioactive growth and decay (Successive radioactive transformations) In radioactive process the element which is unstable, disintigrates and is converted into other element. If newly produced element (daughter element) is also unstable, it also disintigrates further and produces other new element. In this phenomenon certain differential equations are solved to know the amount of each element as a function of time. Let us suppose that parent element A disintigrates and produces daughter element B. Which also disintigrates and produces stable element C. i.e. A ’B -’C(stable). At time = 0, let, No = Number of nuclei of element A. Numberof nuclei of element B is zero. (This is observed in most of the cases in practice). , and A are the decay constants of elements A, Band Crespectively (Aç = 0). Na, Ng and Nç are the number of nuclei of elements A, B and C respectively at time = Rate of disintegration of element A is.(2) dt Element B is produced from element A. Therefore, the rate of increase of elene B is the same as the rate NA Element B itself is radioactive so it can disintigrate the rate g Ng Therefore, the rate of increase of element B is the same i.e. A, N, Ne rate of increase of element B is A..(22) dt Radioactivity 141 Here element C is stable. Therefore it does not disintigrate but increase by the rate y Nå...23) d1 N has no negative signe in equ. (22) and (23). Therefore it shows the rate of dt increase. Negative sign shows decay. Number of nuclei of element A at time , is Ng = N, e- AA!..(24) From equation (22),..(25) d1 + Ag Ng = y No e a!...(26) dt Multiply both the sides of equ. (26) by e eAB' + p N¡ eB! = , N, elag - A ) !..(27) dt d (N eB) = N, elB - N!....28) d Integrating equation (28), N, elaB - )! + C...(.29) Cis a constant of integration. Now at time = 0, N, =0. From equation (26). 0 =..(30).(31) N =No....(32) Equation (32) shows the number of nuclei of element Bat time i. 142 Nirav College Physics (103) In this way number of nuclei of element Cat time is given by the following equalion. N,+...(33) Nc This is not useful practically. f the number of nuclei elements B and C are not zero at 0 time / = 0 but say N B and N B respectively. Then for their number of nuclei at time =1 the term N, N (e- AB! ) is added in equ. (32) and the term N+ N [| e- AB] is added in equ. (33). Graphs of N, ’ and Np ’ Fig. 5 are shown in fig. (5). to become 144 Nirav College Physics (103) From figure (5) it is clear that is positive between the interval I=0 to |=| d max which shows that activity of A is greater than that of Bin this interval. NB becomes d1 negative between time (= may to = , which shows that now the activity of daughter is greater than that of parent. 7. Transient Equilibrium When parent is longer lived than the daughter element (T, > T), but the value of TA IS not very much greater than Tp, then transient equilibrium is established between them. Na = N, e a...(48)} and No -..49) From this, Ng =...(50) and (51)..(52) For very large value of (therefore after a very long time) e (AB- AA) can be neglected...(53) 0.693 Now take Ty/2= = T Tg = Constant T - T¡...(54) In this way after a long time (/ > >) activities of A and B becomes constant (not equal). This type of equilibrium is know as transient their activities will be maintained constant at each time. equilibrium. Thenafter, the ratio of 8gRa228 89AC228 Example - 1 : (T:=6.7 Years) 90Th228 8gR¡224 (T=6hours) (T= 1.9 Years) In this process, the presence of Ac is neglected due to very small value of T and oansidering direct interaction between Ra and Thz0, the ratio of their activities at the time of transient equilibrium (for very large value of )  10. Radioactive isotopes of light elements In patural radioactive series Thallium is having the which is Bactive. Nuclei haVing atomic umber less than TI lowest atomic number Z = 51 and radioactive also (radioactive isotones) are found in nature of course in very low proportion. There proportion is very low and their periodic time Is very large. Hence their activity is felt much less. Proportion of dioactivity 147 elements in the atmosphere has increased due to nuctear veapon tests on large scale he world. (Atom bomb, Hydrogen bomb etc.). Sr is an example of nuclear pollution. dangerous because its half life time is 27.7 years. C# is produced continuously because of the bombardment of neutrons from cosmic ys coming from the space on the Nitrogen in the atmosphere. Because of this activity of does not decrease but goes on increasing. Sone of the radiactive isotopes of lighter ements are shown in the following table. Element Type of nuclei l'ype of radioactivity Half life time in ycars Hydrogen 12.26 Carbon 6Cl4 5730 B 19ko L3 x I0 Potassium B Strontium 27.7 Neodynium 2.4 x |015 Somarium 62Sml47 1.06 x I041 Platinum 7gP{204 6 x 1011 12.Artificial radioactivity: In 1919 Rutherford discovered that one element can be transformed in to other artificially. Line diagram of the apparatus used by him in the experiment is shown in fig. (8). T T Screen Silver foil M Microscope Fig. 8 A screen of flourecent material ZnS is pasted outside on the silver foil. Scintillations produced on the screen are observed by microscope M. Radioactive substance (Po?12) is placed on a stand D. Its distance can be changed from the silver foil. By using tubes T any gas can be filled inside the chamber or can be removed from the chamber. In the bcgining the thickness of the silver foil is taken such that it absorbs all the a particles emited from the source. When oxygen or carban dioxide gas is illed in the chamber at atmospheric pressure it is found that scintillations do not occur even in the  148 Nirav College Physics (103) absence of silver foil when the distance between the screen and source Is 7 Cm Or more. Tnis mneans that all a - particles are absorbed in the distance of 7 cm. When Nitrogen gas is filled in the chamber it is known that scintillations are observed up to the distance of 40 cm between the source and the sereen. It was known from other experiments that a - particles emitted from the source cannot ravel through adistance of 40 em. From this Rutherford concluded that scintillations are produced by newly emitted particles by the bombardment of a-particles on Nitrogen nuclei. It was known that they are prot ons when tested by a nethod of magnetic spectrograph. This interaction can be represented by the following equation. 2He+ + H'. 2. After the discovery of artificial transformation of nucleus by Rutherford. Chadwick discovered neutron in 1932. Then after many artificial nuclear transformations were carried out by the bombardment of neutrons. Many product nuclei formed in these studies were found to be radioactive. The activity of such nucleus is called artificial radioactivity. Same laws of natural radioactivity can be applied to it. D. I. Curie and Joliot, in l934, were studying the effects of bombardment of particles on light elements. They saw that emission of radiation from the target nucleus continues even though the bombardment of a - particles on the target nucleus is stopped. Emitted particles were found to be positron. (Positron is an antiparticle of electron. It has the same charge and mass as of electron. But its charge is positive.) The whole process takes place in two steps. For example, consider the following process. (i) BO + He ,N (compound nucleus) (ii) zNI3 ’C + ß + v (neutrino) One mnore example : 13A|27 () + zHe 14Sj30 + B' + v (neutrino) (ii) In this way light elements were made radioactive artificially. After this discovery, an intensive programne of obtaining new artificial radioactive elements came into existance in all the laboratories of the world by the bombardment of particles obtained from accelerating machines and nuclear reactors on light elements. From the study of such experiments informations regarding the constitution of thouSands of such isotopes were obtained. Because of this, we have a better practical knowledge regarding the energy states of the nuclei. Some more illustrations of artificial radioactivity are given below (i) B+ He [,N5) C + H' (i) + -,B + y (anti neutrino)  149 Radioactivity 13A/2? I|Na24 + zHe ) IjNa24 12Mg4 + (i) Most of the artificial radio isotopes have short half life times. Therefore, they can be identified by chenical processes. Curie - Joliot had used Boron nitride (BN) in their reaction on B. a - particles (NaOH). were bombarded on it for few minutes and then it was heated with costic soda BN + 3 NaOH Na, BO, + NH, ‘ Therefore, they belived that only In this process only NH, gas is found radioactive. N is produced in this process. 12. Determination of the age of the earth : earth or on its Radioactivity remains unaffacted by large changes occuring inside the fact. surface or by earthquakes. So it is possible to estimate the age of the earth by this For several methods, based on this fact are used in practice. First method : 82Pb206 () emitts 8 a - particles and forms 9zU2is 8zPb207 (i) emitts 7 a - particles and forms 82PB208 90Th232 emitts 6 a - particles and forms (ii) Uranium series starts from Uz38 and its half life time is 4.5 x 10 years (450 crores its decay constant can be considered years). Pb is the stable end product of this series, and So, after so many years (= 10 years) presence of Pb in such sample will be Io be zero. be in secular equilibrium with predominent, because all other elements except Pb will continously decreases. Therefore Uranium. Hence, Pb continuously increases while Uranium element transformations can be used not only for first and second equ. (32) of successive element of it. i.e. for Uranium and Pb. of the series but for the first and last From equation (32), (e- a - e- dg!) Ng = N our case, y = hys g= ph 0, No = Ny and N Npp In Nph = -'N, (e Aq! - 1)....(60) Nph = N, (1 - e qi ) which is produced by radioactivity. Pb is present in the ore of UZ35 N, =Nph t Ny..(61) Ny = Npb t Nå present. and Nph represent the numbers of nuclei of Uraninm and lead N, 150 Nirav College Physics (103) Substituting the value of N,. from equation (60) in the above equation (61) = (Np t N,) (| - e A) -.(62) Npb + N e Ae = |- Nph N: Nph t N Npy t N Nph t N N A = In Nph t N N Nph t Nt In| N Age i of the earth can be estimated from the above equation by knowing the proportion of Uand Pb in the sample of Ufrom spectrochemical analysis. Now one should ensure that no ,He and U has escaped from the sample rock. The oldest rock from the crust of the earth have been found to have an age of about 3 x 10 years (300 crores years). While the age of meteors are found to be 4.5 x 10 years. This value (4.5 x 10* years) (450 crores years) is taken as the age of the eartlh. It is different from the formation of rock on the surface of the earth. Second method P6206 Age of the earth can be estimated from the ratio series. PL207 produced in the radioactive PL206 Pb207 y235 where and A are the decay constants of 238 and y235 of yi and are known. respectively. The value In this method there is a possibility that Radon produced in the activity of Uranium may escape as it is in the gaseous state. Thus for these two methods, the rock samples have to be taken pitchblende deeply buried. In such a case the gas would get trapped in rÍckfrom massive cavities and Radon cannot be expected to diffuse very far from the dense rocks Rdioactivity 151 goird mebtod : Carbon dating Cl# is continuously produced due to the bombardment of neutron from the cosmic rays oming from the outerstaller on Nitrogen in the atmosphere. ,N4 + t H! 6CH is a neutron - rich isotope of carbon containing 6 protons and 8 neutrons which disintigrates by emitting ß - particle. 6CI4 -BO + 7 (anti neutrino) Half life time of Cl is 5730 years. C of the atmosphere combine with hydrogen and oxygen and is continuously 6onverted into organic matter in atmosphere. If it would not have been produced continuously in the atmosphere it would have been used up and exhausted and would not be present in the atmosphere. Half life time (5730 years) of C is very large so that its archaeology and anthropology is much useful also used for studying the age of civilizations. When any plant or animal dies it stops to breath carbon in any form. From that moment decay of C in its dead body is the only process which continues. Though coal shown that and petroleum are organic in nature, they are also so old that experiments have there is no fraction of Cl4 in them. as its activity in the In young plant or animal activity of C* is found to be same ofcH atmosphere. (i.e. |5.3 disintigrations per gm per min.). By measuring the activity its estimate the time between of fossil of dead animal or dead plant, it is possible to death till today. O, and forms Radio-carbon (cl4 ) being produced in the atmosphere combines with use it with water and sunlight in the CO,. This CO, absorbed by green plants and they animal eat this plants so they process of photosynthesis to produce carbohydrates.Now become radioactive. that the ratio of radio carbon C and ordinary carbon (C2) is the same It is found and animals. in the bodies of all living organisms. i.e. plants the death of plants and animals they stop to breath radio - carbon, but radio After continues to decay. Therefore after the death of plant -carbon which is there in their bodies and ,c2 decrease with increase of tine, By knowing the value of this ratio ratio of ClH the time interval between the death of Iiving body till tlhe above ratio is found can be determined. and activity R of CO, produced out of it In fact fossils of the dead body are burnt Is measured with the help of a sensitive B- counter. 152 Niray College Physics (103 If R is the activity of the living organism of the sanne mass, R = R e- t In RÍ R t is found from the above equation. The Q-Equation For Nuclear Process 155 THE Q-EQUATION FOR 7 NUCLEAR PROCESS Introduction To investigate the nuclear structure the particles like proton, a. - particles deutron etc are bombarded on a given nucleus. When these particles are in close vicinity of the target nucleus, they interact with it. As a result of this interaction, the projecticle particles are either scattered or eject out other particles from the nucleus. They can also be complelely absorbed and y-rays are emiitted from the resulting nucleus. If atomic mass number/atomic number of the target nucleus changes during the process, the process is known as nuclear process (or reaction). Typically a nuclear reaction may be represented as follows. x+ X = Y+ y... (1) X= target nucleus Where, x =incident or projecticle particle y =emitted particle Y = product nucleus Lord Rutherford, in 1919, found that when nitrogen is bombarded with a-particles from poloniums protons are produced and these protons are capable of travelling through distance of 28 cm in air at atmospheric pressure. This reaction can be written as N + He ’go" + H' (2) This was a breakthrough in the scientific world, as it made it possible to convert a non-radioactive nucleus into another nucleus. Symbolically, the above process can also be Fepresented as (3) N (a, p) 0" (p = proton) In 1930, Cockcroft and Walson used artificially accelerated protons to produced the following reaction. 3Li' + p ’ He +a (4) still Alarge Number of such reactions have been studied by now, Not only that, but Such studies are going on, ln the following, we shall assume that () The projecticle particle possesses non relativistic energy and (iü) The target nucleus is stationary. We shall apply the laws of conservation of energy and momentum and with that, try to understand the possibility for a nuclear reactiot to take place. 156 Niray College Physies (103) 2. Types of nuclear reactions : The nuclear reactions are classified according to the types of projecticle particles. the product nucleus and the emitted particles. () Seattering IT the target and projectile particle remain the same before and after the process, the reaction is called scattering. In this process no new particle Is emitted. H the quantum states of the larget before and after the process are the same. the Scattering is called elastic scattering, In an elastic collision the kinetic energy is conserved. IT quantunm states, before and after scattering, of the target are different, the scattering Is termed as inelastic scattering. In an inelastic collision, although total energy is conserved, the kinetic energy is not conserved. In both types of scattering the law of conservatIon of momentum does hold. (ii) Pick-up reactions : When the projectile gains nucleon/s from the target, the rcaction is called the pick up reaction. For example, ;Li' p, a), He', 0l6 (,, H) ols etc. (ii) Stripping reactions : In this type, the projectile loses nucleon/s to the target. e.g. gOl (GHe'. H) F, N (a, P) 0" In the pick-up and stripping processes the nucleon is absorbed or given up by the target, in such a way that no other nucleons are affected. Therefore, these processes are called direct reactions. (iv) Compound nuclear processes : This reaction differs from direct reactions. Bohr, in his model to fission, suggested that the projectile particle fuses in the target nucleus to explain the nuclear called compound nucleus. The compound nucleus is in excited states and form a new nucleus 10-l6 s. A projectile usually takes about 102 second to pass disintegrates in about the time interval of 10-10s is pretty large "through' the nucleus. Thus, compared to 10-22 s. Thus, it can be assumed that compound nucleus does exist for some time. The mode of disintegration of the compound formed, In other words, the compound nucleus has nucleus does not depend on how it was no memory of its birth. A compound nucleus can be formed in several ways. It disIntegrates also in several ways. The example will clarify this point. following 1. The compound nucleus of (Zn* can be formed in the following two ways. The star (*) here, denotes the excited state.  157 he Q-Equation For Nuclear Process 29Cu63 ’ (Zn4y* ’ 30Zn63... (5) ’ 30Zn62 (6) ’ 29Cu62+.. (7) He + 28Nj60 3oZn63 (8) 2. (oZn t (9) 29Cu62 + ’ (10)... Here, first the compound nucleus is formed and then, after sometime, the 'yields' are obtained. In this sense, the reaction differs from the direct reactions. The mode of disintegration of a compound nucleus depends on its total angular momentum, parity and its excited state energy (excitation energy). It does not depend, as mentioned above, on its mode of formations. The compound nucleus model can be applied sucessufully to nuclei with atomic mass Dumber, A< 10 and projectiles with energies < 15 MeV. The *direct reaction mode" works well for projectiles having energies gerater than 15 MeV. For projectiles of energies greater than 50 MeV, the "optical model" is more useful. Thus, in the nuclear reactions, the energy plays an important role. Y, + y, Kinetic energy X+ r ground state afler before X groundstate ollision collision fntermediate excited states of compound nucleus Fig. (1) 158 Nirav College Physics (103) 3. The balance of mass and energy in nuclear reactions In nuclear reaction total (mass + energy) is conserved. In the reaction, x + X ’ Y t y, (E, + m, C') + (O + M,C2) = Ey + M,C? + E, + m,C.... (12) E, = kinetic energy of projectile. m, = mass of projectile The target nucleus is assumed to be stationary and hence its kinetic energy has been taken zero. M, is mass of the target nucleus. The other notations are similar. The difference between total kinetic energy after and before reaction is defined as ) value of the reaction. Q= (Ey t E,) - E,... (13) From equations (12) and (13), Q= [(M7 t m,) - (My t m,)] C, (14) If Q is positive the reaction is called exoergic and if Q is negative the nuclear reaction is endoergic. 4. Q-equation : The analytical (algebraic) relationship of the nuclear disintegration energy Q with the kinetic energies of the projectile and outgoing particles is called the Q-equation. A nuclear reaction has been displayed schematically in the laboratory frame of reference in fig. (2) rest mass of x. = m, rest mass of Y = M speed of x = V, speed of Y = V, kinetic energy of x = E Y kinetic energy of Y - E, momentum of x=2m, E, momentum of Y- J2M,E, X rest mass of X = M. rest mass of y= m, speed of X = 0 kinetic energy of X = 0 speed of y = v, kinetic energy of y = E, momentum of X = 0 momentum of y =J2m, E Fig. 2 The 0-Equation For Nuclear Process 159 Using the definition in terms of kinetic energy, Q = Ey + E, - E, (15) According to the law of conservation of linear momentum, P, = pP, cos t + py Cos (in X-direction) Using V2mE, /2mE, cOs + 2mE, cOS (16) In Y-direction, O =J2M, Ey sin - J2m,E, sin.. (17) Adding the square of equations (16) and (17). MyEy = m,E, + m,E, - 2 Jmm, E, E, cOs.. (18) Substituting for Ey in equation (15). 2/m,m, E,E,. (19) Q= E, - E, + M E, + M M; cos 2m,m, E, E, - E,1+ M E, My cOS (20) This is the standard from of the Q-equation. The same equation can also be obtained from momentum-triangle method as follows. Px Py Py Fig. 3 From the triangle shown in fig. (3), py = p + p, 2p, p, cos... (21) 160 Niray College Physics (103) Using p = 2mE 2MEy = 2m,E, + 2m,E, - 2 Jm, n, E, Ey COS (22) Ey = E, + m,E - 2 COs (23) M; My My Finally, from Q= Ey + E, - E we get, Q = E, | + 2m,m,E,E,.cose M, E,1+ M, My Natural atomic masses : The masses appearing in equation (20) are the masses of nuclei. Now, in the literature, nuclear masses are not available. Therefore, we should convert the equation for nuclear reaction or the equation determining Q value in terms of atomic masses. To understand this, consider the following examples. 1. In the process of emission of a-partieles from the radio-active onU238 the product nucleus is 9oTh234, 92U238 9Ths4 + He..(24) If we consider the nuclear masses, the equation for Q is Q= My -(Mh t MHe) C?... (25) Now, Mu mass of neutral atom of U- mass of 92 electrons. my - 92 m, (26).. Using such results in equation (25), Q = (må 92 m,) - ((mh -90 m,) - (MHe - 2m)] (Mh (27) Eqaution (27) shows that in the equation for which are available in the literature. Q-value we can use the atomic masses 2. Equation (27) has to be modified for the Look at the following reaction. reactions in which positrons are emitted. 3oZn63 29Cu63 + (29) (position) (neutrino) Q = (Mn - (Mcu t m,.] c2 (30) The Q-Equation For Nuclear Process 161 M.. and M, are nuclear masses of Zn and Cu respectively. Writing them in terms of atomic masses, Q= [(nz, -30 m) - {(mcy - 291,) + m,+] C? (31) Mcu (32) Mçu 2m (" m+ = m) (33) (mcu + 2 m)... (34) Thus, in the case of positron emission also we can use the but twice the mass of an electron has to be masses of neutral atoms, subtracted. 5. Solution of Q-equation We have obtained the following equation for Q. Jn,£, E,.cos Q= E,1+ My E, My... (35) Since, Q is constant, this equation can be rergarded as a quadratic equation in E,, Its general solution is, E,... (36) Where, Jmym,E,.cose , + M, (37) and W = MyQ+ E, (My n,) m, + My.. (38) Only when JE, is positive real number, the reaction is possible. Emission of m, becomes energetically impossible when Q value is negative, (My - m,) is negative (i.e. heavy projectile) and a large angle of observation 0, making cos 0 negative. Equation (36) tells us about the types of nuclear reactions for different energies (E.) of the bombarding particle. 1. Exoergic reactions : (i) When the energy of the projectile is low (e.g. thermal neutrons), E, can be taken as zero. Using this approximation in equation (37) and (38), =0 and W = M, (39) , + M 162 Nirav College Physics (103) QM (40) My t , Note that E, is independent of 0. (ii) Projecticle particles of finite enrgies : much less than that of product In most of thereactions, mass of projectile particle is E, Also, E is sinole nucleus, m, < M,, Therefore. W> 0for all the values of trom the equation valued for all the values of E,. In this case E,can be found (4) JE, The illustrative example of this case is sBl0 (a, p) C3 (Q = 4 MeV) (iüi) Double values of E, : In some nuclear reactions, E, is not single valued. It can have more than one value. The reason for this can be understood with the following example. NS (d, n) Q = 10 MeV. If we regard as product nucleus and 0l6 a product (outgoing) particle ('). m, = 16 and My = I. Therefore, from equation (38). + E,( - 2) (42) W= (: for d, m, = 2).. 16 + | 10 - E,.(43) 17 If the energy of incident deutron is greater than 10 MeV, W will be negative values. Thus, for E, > 10 MeV, in the = 0 direction, JE, will have two real positive This means that we shall have two groups of (monoenergetic) ol6 nuclei. 2. Endoergic reactions : inverse In such reactions Q