The Physics of Energy sources Nuclear Fission


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1 The Physics of Energy sources Nuclear Fission B. Maffei Nuclear Fission 1
2 Introduction! We saw previously from the Binding energy vs A curve that heavy nuclei (above A~120) will gain stability by splitting into 2 fragments of A~60 each! This reaction will release energy (Q) due to difference in binding energy between the parent nucleus and the products! The reaction will also produce neutrons! This has been observed in early fission experiments in the early 40s! The reaction will quite probably produce! rays A simple energy release model B/A energy for heavy nuclei (A~200) is about 7.6MeV B/A energy for more stable nuclei (fragments) ~ 8.5MeV The change of B/A is about 0.9MeV per nucleon If we have 235 U fission we have a release of energy Q=235x0.9MeV= 211.5MeV A typical nuclear fission releases an energy of the order of 200MeV Nuclear Fission 2
3 Mass partition 235 An typical example of fission is 236 * U + n! U! La+ Br + 2n It is not unique and many different mass partitions are possible as shown in figure. For 235 U fragment mass numbers vary between 70 and 160 with most probable values at about 96 and 135 Ref 2 Distribution of fission fragment masses from the fission of 235 U The energy released in process will be transmitted to the fragments as Kinetic energy Nuclear Fission 3
4 Neutron production Ref 2! The line of stability of the nuclei is not totally a straight line! There is a slight curvature leaning towards neutron rich nuclei when A increases! Ratio N/Z for A>200 is ~ 1.5 Uranium, Plutonium..! Ratio N/Z for 70<A<160 is These are the fragments When the fission of the heavy nucleus occurs, the fragments end up having the same neutron to proton ratio (~1.5) than the original nucleus. So compared to the line of stability for nuclei of their mass, these fragments are neutron rich. They will decay towards the line of stability mainly by "  emissions (long) but also by rapid shedding of several neutrons. These will be crucial to initiate other fission reactions! chain reaction Several fission reactions for the same original element! Average number of neutrons produced Nuclear Fission 4
5 Imagine a typical fission reaction Induced fission reaction Pu! Pd + Cd 3n The calculation of the Q factor will give 188MeV (in agreement with the rough estimate of 200MeV we saw earlier). From this figure we might think that fission is quite a probable process due to the energy released! so could it be spontaneous? Actually no: do not forget the Coulomb interaction! Coulomb barrier Let s assume that the 2 fragments are uniformly charged spheres of radius r pd and r cd We have the energy due to the Coulomb potential between the 2 fragments E coulomb = e2 Z Pd Z Cd 4"# 0 r r being the distance between them! Let s try to bring these 2 nuclei as close as possible to each other r = r Cd + r Pd = r 0 ( A 1/ 3 + A 1/ 3) Cd Pd Reminder: nucleus radius 1/3 r = r0 A, where r0 = 1.2fm Nuclear Fission 5
6 Induced fission reaction (2) Replacing all the parameters by their value we find (strongly recommended to do that as an exercise) c Bc = Ecoulomb = 270MeV B c : Coulomb barrier Coulomb Potential energy (1/r variation) distance We see that in order to be able to glue the 2 fragments together, we need to cross this energy barrier due to the Coulomb repulsion Only when the 2 nuclei will be close enough (basically coming in contact) the strong force will be able to overcome the repulsion and keep them together as one nucleus Attractive strong nuclear interaction Based on Ref 2 So working the other way round, in order to have a high probability of spontaneous fission of the parent nucleus, the energy Q that can be released during the fission should be comparable to the Coulomb barrier. It is not the case here Nuclear Fission 6
7 Induced fission reaction (3) Note: Spontaneous fission does exist Even with a low probability, it will occur statistically Occurs quantum mechanically by tunnelling effect Some very heavy isotopes are so instable that fission overcomes the Coulomb barrier We would need A large (~300) and roughly Z 2 /A>47 However for some isotopes spontaneous fission is a non negligible decay mode Example: 252 Cf is used as a radioisotope (spontaneous fission branch ratio~3%) More generally for fission reaction to happen, it needs to be induced (triggered) We need to excite the nucleus to above the Coulomb barrier That will happen through absorption of a neutron Having no charge, the neutron will not have to go through a Coulomb barrier in order to be absorbed by the nucleus The addition of the neutron will leave the resulting nucleus in an excited state 235 U + n! 236 U Nuclear Fission 7 *
8 Activation energy! The previous crude model was mainly to explain why fission needs to be induced.! However, in reality the necessary extra energy amount to make fission happen is not that large. This is what we call activation energy. Energy plot from our simple model of Pu fission 270 MeV Coulomb barrier 188 MeV Fission energy released Reference: Energy of 2 fragments far apart Equivalent to Energy state of Pu nucleus Graphs based on Ref 3 So to make sure that the reaction will occur (probability=1), the neutron needs to provide an energy at least equal to the activation energy Nuclear Fission 8
9 Activation energy: crude model to more realistic one! The actual activation energy is not as high as we did calculate! Mainly for heavy nuclei! Part of that can be explained with our simple liquid drop model n U 236 U * +n(s) Due to nucleus stretching, the overall binding energy of the nucleus has changed Consider the semiempirical mass formula from the LDM B(Z, A) = a v.a " a s.a 2 / 3 " a c.z(z "1) " a sym (A " 2Z)2 A 1/ 3 A +#(A,Z) When the nucleus is stretched, the volume is constant but the surface increases:! Volume term is! the same! Surface and Coulomb terms will be modified! Binding energy decreased (for more details see ref 3) Nuclear Fission 9
10 Variation of activation energy with mass number! In order to get a better value of the activation energies more detailed models are used taking into account more sophisticated effects than the LDM (i.e. shell model). Around Uranium 5MeV typically Z Dark curve: LDM Thin curve: shell structure model A~280 Spontaneous fission Figs based on Ref 3 Activation energy for heavy nuclei. A and Z dependence Nuclear Fission 10
11 235 U + n! 236 U * Application to uranium When 235 U captures a neutron to form a compound state 236 U * the excitation energy is: [ ] # c 2 Q ex = m( 236 U * ) " m( 236 U) Assuming the kinetic energy of the neutron small, the energy of the compound state 236 U * can be found directly by the mass energies of 235 U and the neutron:! 236 * 235 m( U ) = m( U ) + mn = ( u u) = u Q ex = ( u " )! 931.5MeV / u = 6. 5MeV The activation energy of 236 U is 6.2MeV! 235 U can be fissioned with zero energy neutron. Similar calculation for 238 U! 239 * + n U gives Q ex =4.8MeV Activation energy for 239 U is 6.6MeV. We need a neutron of at least ~2MeV to get fission. The difference between the excitation energies (6.5 and 4.8MeV) is one of the explanations for the extreme difference in the fissionability of 235 U and 238 U Nuclear Fission 11
12 Fission energy budget Example on thermalneutron induced fission of 235 U (all figures from ref 2) For this reaction, there is an average of 2.4 neutrons per fission! About 87% of the total energy is released promptly! 90% of this in fragments KE.! ~13% in radioactive decay! Electrons,! rays energy converted into heat! Neutrino energy not recoverable Mean energy of each released neutron ~ 2MeV Energy spectrum of emitted neutrons Nuclear Fission 12
13 Neutrons released We have seen that there is a prompt release of a few neutrons per reaction These prompt neutrons have a range of kinetic energy By convention we have the following designation:! Thermal: E ~ ev Delayed neutrons! Epithermal: E ~ 1 ev Timescale ~ 6sec! Slow: E ~ 1 kev! Fast: E ~ 100 kev to 10 MeV About 1% of released neutrons Other neutrons are emitted during some of the fission decay chains (from heavy fragments) These are released within about sec after fission About 99% of released neutrons Ex: sec Rb "! Sr! Sr + n 56 " Decay branch "  followed by neutron em. 1.4% probability Nuclear Fission 13
14 Cross section A brief explanation For a more precise development see references at the end Intensity I=#A Flux # Consider an incident flux of particles on a surface A of width dx (Number of particles/m 2 /sec) A Intensity (number of If N target nuclei are exposed to the beam (number of atoms in volume A.dx) incident particles per second) ="! A N = " N! dx! A dx $ N being the atoms volume density (Number/m 3 ) Let s suppose that we have a reaction (i.e. fission) between the incident particles and the atoms in the volume A.dx. If the nuclei in the target act independently, the reaction rate will be proportional to the number of incident particles (! # or I), and the number of atoms N.! = Rate R $ R " N $ N #! # dx # A#! event rateper nucleus incident flux The constant of proportionality is called the cross section % % is the reaction rate per target atom per unit of flux. It is equivalent to a probability of the process to occur. and R = " # N #! Note: % as units of area (cross section). R number of reaction per unit of time Nuclear Fission 14
15 A classical approach of the crosssection (but wrong!) R 1 R R 2 Consider the reaction 135 Xe + n! 136 Xe The neutron (radius R 1 ) can be captured by the nucleus of Xeon (radius R 2 ) when the strong interaction act! distance between their centres is * r! R = R 1 + R 2 Neglecting R 1 in comparison with R 2, R=R 2 Then a simple picture of the cross section would give: $ 2 2 2/3 = # " R = # " r0 " A!120fm 2 However, the cross section for this specific reaction is actually %~10 6 fm 2 The classic approach is not the right one. In order to get the real values of the crosssections a proper quantum mechanical treatment has to be perform and other physical processes have to be taken into account. Units: we use a more appropriate one for crosssection 1 barn=100 fm 2 = m 2 Nuclear Fission 15
16 Example Given 10g of natural uranium (mass u) and a neutron flux of cm 2 s 1, a thermal fission cross section of 235 U of 584b, find the fission rate. Fission rate = N%# with N being the number of 235 U atoms. Abundance of 235 U in natural uranium= 0.72% N = 10g! 6.023! ! = 1.82! atoms Fission rate = 1.82! 10 20! 10 13! 584! 10 " 24 = 1.06! reactions / sec Note: if several reactions are possible, each with a cross section % i, total cross section: " =! total " i i Nuclear Fission 16
17 Cross section data Thermal neutrons Fast neutrons Cross sections of neutron induced fission of 235 U and 238 U as a function of the incident neutron KE Note the poor cross section of 238 U Nuclear Fission Figures from Ref 3 17
18 What happens to the emitted neutron?! There is an average of & neutrons emitted per reaction! Not all emitted neutrons will be able to trigger another fission.! There are other reactions that enter in competition! Not all neutrons are useful in a chain reaction! Each competitive reaction has a probability to occur! Reaction rate 4 possible reactions: 3 competitors to fission 235 U + n# 236 U * elasticscattering """"" # 235 U + n Back to square one 236 U * inelastic scattering """"" # 235 U * + n Reemitted neutron as lower KE leaving 235 U in an excited state Scattering 236 U * radiative capture """"" # 236 U +! Heavy nucleus + low energy incident neutron: most probable decay Nuclear Fission 18
19 Fission vs capture! Each of the previous reactions has a corresponding cross section.! The cross sections are normalised: we can sum them!! fission elastic with +! +!! capture inelastic total =! =! a f =! +! =! e +! +! s c i =! absorption scattering =! =! In order to get a good fission rate, we want % fission to be the dominant cross section From previous plot we saw that the fission cross section of 235 U is strongly dependent of the incident neutron KE. % s is relatively independent of the energy ~ 10b % fission = 584b for thermal neutrons (~0.025eV) % fission ~ 1.5b for fast neutrons (~2MeV), ~1/6 of % s Fast neutrons (mean of 2MeV) are released during fission We need to slow them down to thermal energies to sustain chain reaction! Moderation Nuclear Fission 19 s a
20 Number of useful neutrons! Each 235 U fission reaction, initiated by a thermal neutron, produces an average of &=2.4 neutrons.! Some neutrons are lost through radiative capture both with 235 U and 238 U. Suppose that we have pure 235 U Let s call the number of released neutrons available to induce fission & a (also called! )! f " a = "! +! f c = 2.06 Suppose that we have natural uranium which contains 0.72% 235 U 235 & 0.72' ( f ( U) ) a = ) $ 235 $ % 0.72' ( a( U) ' ( c( 238 #! U)!" = % f % c % a % s 235 U U We can vary & a by changing the enrichment (proportion of 235 U in natural uranium) Note: an enrichment of 1.6% leads to & a = Nuclear Fission 20
21 Reactor kinetics! We introduce the neutron reproduction (or multiplication) factor k.! This is the ratio in number of neutron from one generation to the next.! Neutrons are characterised by a time constant ', mean time before absorption occurs! Includes the time necessary to moderate the neutron (106 s)! Includes a diffusion time at thermal energies before absorption (103 s) If there are N neutrons at time t, then there will be on average kn neutrons at time t+', k 2 N at time t+2' and so on. Then in a short time interval dt, the increase will be: dn = (kn " N) dt # Giving dn(t) N(t) dt = (k "1) #! $ N(t) = N 0 e (k "1)t # Nuclear Fission 21
22 Energy rate! The rate of energy released during fission will be: Q value per fission x number of absorbed neutrons leading to fission in time dt The integration gives de = Q " N(t) #! We then have 3 cases % dt = QN 0 exp ' & E = Q N 0" (k #1) (k $1)t # $ (k #1)t ' exp & ) + C st % " ( making ( * dt ) % E " exp ' & (k #1)t $! k<1 the number of neutrons decreases with time. The energy produced also decreases. Subcritical: the chain reaction will stop after a while! k~1 The number of neutron! remains constant, so is the produced energy. Critical: the chain reaction will be steadily sustained.! k>1 The number of neutrons and the energy will increase exponentially with time. Supercritical: we get a bomb ( * ) Nuclear Fission 22
23 References Ref 2: Lilley, J. Nuclear Physics Principles and Applications (Wiley 2006) Ref 3: Kenneth S. Krane, Introductory Nuclear Physics (Wiley 1988) Nuclear Fission 26
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