SECTION C: NUCLEAR RADIATION AND NUCLEAR ENERGY LOSS PROCESSES. " N & = '!t and so N = N 0. implying ln! N $

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1 SECTO C: UCLEAR RADATO AD UCLEAR EERGY LOSS PROCESSES n this section we discuss decay and transmutation processes in nuclei (including α, β, and γ decay, as well as fission and fusion processes), using the previous models where appropriate. Also we discuss the interaction of beams of nuclei with matter, focusing on the processes that govern energy loss and penetration distances. C.1 Radioactive decay equations Before considering specific mechanisms of decay, it is useful to consider some general properties. We start with a simple one-stage decay process, where the product of decay of the radioactive nucleus is stable, e.g., a 14 C 6 nucleus can decay into 14 7 (through beta decay) over several thousands of years. f denotes the number of radioactive nuclei at time t, and λ is the probabity of decay per unit time, then the rate of decrease (known as the activity) is! d!, where λ is the probabity of decay constant dt The unit of activity is the Bequerel (1 Bq 1 disintegration per second). Rearranging and integrating the above equation gives d!! t dt 0 0 implying ln! $ & '!t and so 0 e!!t. % 0 The decay constant can also be expressed in terms of the half-life (denoted by t 1/ ), which is the time for half of the nuclei to decay. Therefore exp( λt 1/ ) 0.5, which gives t 1/ ln 0.693!! Also the mean life (denoted by τ) can be found as ( 0 ) 0 e!t dt 0 ( )!! t 0 e!t dt 1!1 More generally, we might have a chain of decays if the initial product of a decay is itself radioactive and can undergo a further decay (usually with a different decay constants). An example is 79 Sr 38 (t 1/ ~ min) 79 Rb 37 (t 1/ ~ 3 min) 79 Kr 36 (t 1/ ~ 35 hr) 79 Br 35 For a -chain decay of the form A B C with decay constants λ A and λ B and C being stable, the rate of decay equations are! d A! A A,! d B! B B!! A A dt dt The solutions when λ A λ B are then easy found to be (e.g., can be verified by substitution):! A (t) 0 exp(!! A t), B (t) A 0 [ exp(!! A t)! exp(!! B t) ]! B!! A The results for A, B, and C ( 0 A B ) wl be discussed further in class. C. Alpha decay Energy release in α decay The α decay process is ( A, Z)! ( A 4, Z ) + 4 He ( 4 He α particle) n terms of the energy balance, where Q is the energy release and B is related to the binding energy, we have Zm p c + (A! Z)m n c! B(A, Z) Zm p c + (A! Z)m n c! B(A! 4, Z! )! B(4, )+Q After cancelling terms and solving for Q we get Q B(A! 4, Z! )! B(A, Z)+ 8.3 MeV, where the 8.3 MeV represents the experimental value of B for the α particle. 19

2 The decay process is favoured if Q > 0. We can substitute for the other binding energies using the liquid drop model (semi-empirical mass formula) to calculate Q. This leads to the general conclusion that α decay wl be favoured for large nuclei (i.e., large A and Z). Specifically, it is predicted to occur if Z is greater than about 66 (assuming nuclei that are near the bottom of the mass parabola). The experimental data broadly confirm this prediction, but there are several exceptions. Also there are some anomalies with regard to the rate of decay (sometimes it is several orders of magnitude small than expected by the above the above simple theory). To understand this, we need a QM theory. Quantum theory of α decay This would involve solving Schrödinger s equation using spherical polar coordinates and assuming a spherically symmetric potential V(r), like in the case of the Shell model. However, the potential wl be different in the present case. t wl be made up of:- short-range nuclear interactions (potential well) + Coulomb interaction term V C (r). When the α particle has formed and moves apart from the parent nucleus, we have nuclei with charges (Z )e and e. The Coulomb repulsion wl give (Z! )e V C (r) where r separation 4! 0 r The minimum value of r (denoted by r s ) wl be when the two spheres are just touching: r s r 0 (A! 4) 1/3 + r 0 4 1/3 with r fm The maximum value of V C (r) wl occur at this separation. For example, taking the case of α decay in 09 Bi 83, we get r s 8 fm and V C (r s ) 8 MeV For comparison, the experimental value for the energy release in this case is Q 3 MeV. The overall form of the total potential is something like: V(r) V C (r s ) Q r Clearly, if Q < V C (r s ) (as in the quoted example of 09 Bi 83 ), then classically the α particle which is formed inside the well cannot escape no α decay! However, according to QM the α particle has a non-zero probabity of tunneling through the barrier α decay can occur (if Q > 0), but with very low probabity in some cases, depending on the height and width of the barrier. 0 rs rc Width of tunneling barrier ~ r c r s 75 8 fm 67 fm for decay of 09 Bi 83 to occur. This value is many times the nuclear radius and it makes the tunneling probabity (and hence α decay) almost negligibly small in this case. A proper calculation would involve solving Schrödinger s equation this is complicated unless V(r) is further simplified (to be discussed in class). 0

3 C.3 Beta and gamma decay Beta decay We have already discussed the different forms of beta decay (β - decay, β + decay, and electron capture) in section B, where we mainly used the Liquid Drop model. This was done in terms of the mass parabola and conditions for the different types of decay to be energetically possible were deduced involving the nuclear masses. For example, we had the diagrams on pages 9 and 10 for A 135 and 140 respectively. Going further, there are some estimates to calculate the transition rates in beta decay (using an analog of the Fermi golden rule in QM, but doing a relativistic version of the theory). Also the role of the neutrino and antineutrino in the decay wl be reconsidered later in discussing conservation laws (specifically, for parity). Gamma decay When nuclei are produced in reactions or decay processes, they are often found to be initially in excited states. These can then emit energy to decay to the ground state (or lowest-energy state) of the same nucleus. Typically the energy spacing is of order ~ 50 kev (<< binding energy per nucleon ~ 8 MeV), and this amount of energy can be emitted as a single energetic photon (a γ particle of electromagnetic radiation). The conservation of energy for this γ decay gives M * c M c + E! + E reco where M* and M are the masses for the excited and ground state of the nucleus, Eγ is the energy of the emitted γ, and there is a reco energy term for the nucleus (so that momentum is conserved). We know, by conservation of linear momentum, that the momentum of the recoing nucleus is of magnitude p reco E! c and so E reco p reco M E! Mc This leads to a quadratic equation for the γ-ray energy of the form E! + Mc E!! M(M *! M )c 4 0 and the solution may then be found. (where the non-relativistic approximation is good) A more complete theory of gamma decay (to find transition rates and decay times, etc) can based on either using classical electromagnetism or (much better) using nonrelativistic QM. C.4 Fission and fusion Fission This is a process whereby a nucleus rapidly divides into approximately equal masses. (Technically, this is binary fission; fission decay into 3 or more heavy masses can occur but is extremely rare). We can work out a condition for fission to occur spontaneously by using the liquid drop model. For simplicity, we focus on fission into equal masses:- (A, Z)! ( 1 A, 1 Z) + ( 1 A, 1 Z) taking A and Z to be even Energy before (Zm p + m n )c! a 1 A! a A /3! a Z 3 A! a 4 (A! Z) % $ ' 1/3 A & Energy after (Zm p + m n )c! a 1 ( 1 A)! a ( 1 A)/3! a 3( 1 Z) (! a 4( 1 A! % $ Z) 1 A)1/3 ( 1 A) ' & 1

4 So the energy release is Q! 0.6a A / a 3Z > 0 for fission to occur A 1/3 Z This condition becomes A > 0.6a! 16 after substituting the standard values for a and a a 3 An improved version of the theory (due to Bohr and Wheeler) leads to the modified condition that Z A > 50 The main two factors playing a role in the improved theory are: (i) The shape of the stabity curve. As discussed previously the stabity curve for nuclei looks like: Z Z line Coordinates for nucleus undergoing fission Coordinates for fission into equal parts f the nucleus undergoing fission lies on the stabity line, it is clear that because of the curvature the products of fission into equal fragments wl not be on the line and wl have an excess of neutrons. This leads to a likelihood of further decay (e.g., by β - decay), changing the energy balance equation. Also fission into equal fragments is no longer necessary the optimum. (ii) The effect of a Coulomb barrier. There wl be a Coulomb barrier for the same general reasons as in the case of α decay, BUT the detas are different because the different nuclei are highly distorted in shape just before and after fission. These distortions modify the surface area term (proportional to a ) and the Coulomb energy term (proportional to a 3 ) in the liquid drop model. This changes the condition for Z /A. All the previous discussion was for spontaneous fission, i.e., it occurs because the energetics of the decay is favourable. However, we can also have induced fission, which refers to fission induced by supplying the nucleus with extra energy (such as by bombardment with neutrons as in a nuclear reactor for power generation). For example, consider the fission process for 35 U 9. We have Z /A (9) /35 36, which is less than the usual threshold of 50 and so fission does not occur spontaneously. However, if the 35 U 9 is bombarded with a slow neutron having KE ~ 1 MeV, then induced fission can occur: Barrier energy V C neutron capture energy + 1 MeV ~ 6 or 7 MeV. When the neutron has been captured, we have an unstable nucleus of 36 U 9 (often called a compound nucleus), which then decays into fragments in a time of ~ s. *oo*ge n O*O-ca) + Stage 1 Stage Stage 3 Stage 4 Stage 5 At t 0 we have Stage 1 consisting of the n and the 35 U At t ~ s we have stages, 3, and 4 where the compound 36 U forms, splits (with prompt n and γ emission) At t > s we have stage 5 where unstable fragments move apart (with further β, γ and n emission). n some nuclei (A, Z) that are less massive (therefore slightly more stable) than the above example, we might have Stages 1,, and 3 occur, to temporary form (A+1, Z) in an excited state. Then, instead of Stages 4 and 5,

5 we would have γ emission to give (A+1, Z) in its stable ground state. This process is called radiative capture of the n. c).9 (g -o c,o a.9. TL t \ +, T t i : \ z t\ + + O 6 + O o \ + \ t\ e V 3 + The figure on the left shows the calculated potential barrier (plotted versus A) for nuclear fission to take place. t lustrates why we only get fission at large enough A. The smooth-looking solid curve is found using a simple theory based on the Liquid Drop model, whereas the dotted curve is based on the Shell model and includes the additionalstabity effects associated with the magic numbers. 0 Fusion A Fusion is effectively the inverse process to fission, since it corresponds mainly to two nuclei combining to form a larger nucleus. The energy conditions for this to occur are rather restrictive, and it is of interest only for the smaller nuclei. Recall the earlier figure (in Section A) for the binding energy per nucleon: This has a peak at around A ~ 55, so the process of combining two much lighter nuclei to form a larger nucleus (with A less than about 55) wl increase the binding energy per nucleon. The consequent release of energy if this occurs can in principle be used as an energy source. The main problem is how to achieve this with sufficient efficiency, since there is a Coulomb potential barrier to overcome (by analogy with the case of α decay considered earlier. Generalizing the previous expression for the potential barrier height, we now have e Z V C (max) 1 Z which occurs when the separation of the nuclei is R 1 +R 4! 0 R 1 + R Here Z i and R i denote the atomic number and radius for nucleus i ( 1, ). Using 1/3 R i r s A i! r s (Z i ) 1/3 and then taking A 1 ~ A ~ A and putting in numerical values for the constants, we eventually estimate V C (max) ~ A 5/3 / 8 MeV Therefore, fusion is most likely to occur for very small A, and in such cases the barrier height is less than about 1 MeV. 3

6 How should this amount of energy be supplied so that fusion can occur? One way would be high-energy collisions (this turns out to be very inefficient because they are mainly elastic collisions and so there is not much energy transfer). Another way is to have very high thermal energies k B T, as in stellar systems. For 1 MeV of energy this corresponds to T ~ K, which is much larger than in stellar interiors (typically 10 7 K). Actually, 10 7 K is high enough, because there is a Maxwellian distribution of velocities (and energies) and there wl be enough energetic particles in the Maxwellian ta of this distribution. The main ongoing practical problem in the use of fusion technology is the containment of the high-temperature plasma combined with achieving the high-enough temperatures. Some of the simplest fusion reactions involve deuterons d ( H 1 ), e.g., d + p 3 He + γ (Q 5.5 MeV) d + d 3 H 1 + p (Q 4.0 MeV) C.5 Energy loss for charged particles in matter The energy range of the particles is typically in a range of ~ few kev up to ~ 10MeV (sometimes more). As a charged particle passes through matter, it wl generally interact and lose energy. The processes by which this occurs wl depend on the particle kinetic energy T, as well as on the type of particle and the target material. We divide the discussion into two cases: Heavy particles, e.g, protons, α particles Light particles, e.g., electrons, positrons Applications are in medical physics (e.g., ionizing effects of heavy particles in cell tissue for cancer studies) and in solid state physics. Energy loss by collisions for heavy particles Consider the process of a proton p passing through a thin sheet of metal (such as Al) and the probabity of transmission: The main process of energy loss for p is collision with the Al sheet electrons in the Al (via the Coulomb interaction). f the electrons are assumed to behave as if free, they wl acquire reco energy during the collision and this wl represent energy lost by p. p We use a non-qm treatment, and also initially we ignore relativistic effects. The general collision process is approximately as depicted below:- Charged heavy particle The incident heavy particle has charge ze, mass M, speed v, and kinetic energy T. Also b impact parameter. Magnitude of Coulomb force ~ ze 4! 0 b, which acts for a time (the collision time ) ~ b v b Electron, mass me charge e Coulomb force x 4

7 ze So, momentum imparted to electron ~ force time ~ 4! 0 bv For a better calculation, we would write down the instantaneous force and integrate with respect to time. Then the result becomes ze (only the numerical factor is changed)! 0 bv Therefore the reco KE imparted to the electron is 1 ze % z e 4!T(b) $ ' m e! 0 bv & 8! 0 b v m e ow we need to integrate over b to get the total energy loss. Consider a cylindrical shell between b and b + db : v dx Let there be n atoms per unit volume in the target metal, so there are nz electrons per unit volume. o. of electrons in the cylindrical shell nz πb db dx Total energy loss ( ΔT ) in length dx nz!t(b)! b db dx So,! dt dx! dt dx b max b min z e 4 nz b max db and the final result is 4! 0 v b m min e b z e 4 nz L with L ln b % max $ ' 4! 0 v m e & This quantity dt/dx is known as the specific energy loss or the absolute stopping power, and is often denoted by S(T). We stl need to specify b min and b max to determine the L factor: For b min, we use a value deduced from the Uncertainty Principle. n the centre-of-mass frame of the electron plus heavy particle, speed of electron ~ v, and so Δb m e v ~ ħ or Δb ~ ħ/m e v for the uncertainty in b. Therefore, we take b min ~ ħ/m e v. For b max, we use the fact that the electrons are not really free, but mostly are bound to a particular atom (with an ionization energy ). We can use this and the Uncertainty principle to define a time τ by τ ~ ħ/ (This is just ~ the period of orbital motion of the electron in a Bohr atom). This time can be compared with the collision time ~ b / v, so that if b / v << ħ /, this is a fast collision electron behaves same as if free ( able to reco) if b / v >> ħ /, this is a slow collision electron behaves same as if bound to atom (cannot reco). We choose b max such that b / v ħ / giving b max v ħ/.! Therefore, L ln b $! max & ln m $ e v & b min % av % where av is an average value of for all electrons. Comments: (1) S(T) is independent of M (mass of incident particle), but it does depend on its charge factor z and speed v. b min 5

8 eglecting the weak logarithmic dependence, S(T )! 1 v! 1 T at non- relativistic speeds (v << c). The general behavior (showing also the high-energy relativistic correction) is like:- S(T) ncrease due to relativistic terms () The range R of penetration of the heavy charged particle into the metal target can be worked out from! dt z e 4 nz L(v)! Mv dv dx 4! 0 v m e dx (using T ½Mv dt Mv dv) R Range is R! dx 4! 0m e M v v 3 i dv! 0 z e 4 v nz f L(v) The usefulness of this expression is limited by its breakdown at very low velocities (because of difficulties in fixing the lower limit of integration for the speed), but it is good for comparing R for different cases. Also, there are other loss mechanisms that become significant at low v, adding to the uncertainty in the above expression for R (see the text-book). (3) At very high incident energies, apart from the relativistic corrections for the collision formula, there is the additional mechanism of Cerenkov radiation. This is an electromagnetic process that can occur at high enough energies when a charged particle passes through a dense medium of refractive index n 0 (n 0 > 1). f the particle speed v > c/n 0 (the speed of light in the medium), the particle can radiate energy by forming a coherent wave front ( shock wave ) T angle θ A c/n 0 C Consider the particle moving along the straight line AB with speed v. A wave front spreads out from A with speed c/n 0 and from the later instantaneous positions of the particle along AB. A coherent wavefront (Huygen s construction) is formed as BC. Since we must have cosθ < 1 and v < c, a condition is that v B The energy loss is typically of order kev/cm (for a proton), which is usually much smaller than for collision loss. Energy loss for light particles (electrons and positrons) 6

9 Energy loss wl stl occur by the previous process of collision loss (via the Coulomb interactions), but there are other mechanisms that can become important and may dominate. One of these mechanisms is radiation loss of energy when a charge is accelerated (in the strong electromagnetic fields close to other atomic nuclei. From Maxwell s equations it can be shown that the radiation loss per unit time for an accelerated charge q is proportional to q a, where a is the acceleration. ow, for a given electromagnetic field, a 1/m where m is the mass. Comparing this mechanism for electrons and protons, Radn loss for e! m ~ p % $ ' ~ 10 6 Radn loss for p m e & Thus radiation losses are much more important for lighter particles, especially at higher values of the kinetic energy T and for target materials with higher Z. n practice, the radiation losses for electrons and positrons require special conditions and usually occur by (a) Bremsstrahlung (radiation involving strong electric field close to a nucleus) OR (b) Radiation in a high magnetic field (e.g., in a synchrotron) OR (c) Cerenkov radiation (high v in a medium, as discussed) The special conditions arise as a consequence of satisfying conservation of both energy and momentum for the emission of radiation: photon pph, Eph electron p 1, E 1 θ φ before after electron p, E We now show that it is impossible for this process to occur in isolation. We have: p 1 p cos! + p ph cos and p sin! p ph sin from momentum conservation and E 1 E + E ph from (relativistic) total energy conservation n addition, we have E 1 p 1 c + m e c 4, E p c + m e c 4 for each electron state and E ph p ph c for the photon Eliminating angle φ from the above equations eventually leads to E cos! 1 1+ m e c 1/! $ & > 1 p 1 c p 1 % This gives a contradiction The conservation laws cannot be satisfied in isolation. Therefore, we need an electric or magnetic field [as in cases (a) and (b) above] or a dense medium [as in (c)]. 7

Chapter VIII: Nuclear fission

Chapter VIII: Nuclear fission 1 Summary 1. General remarks 2. Spontaneous and induced fissions 3. Nucleus deformation 4. Mass distribution of fragments 5. Number of emitted electrons 6. Radioactive decay