uclear Physics ( th lecture) Cross sections of special neutron-induced reactions UCLR FISSIO Mechanism and characteristics of nuclear fission. o The fission process o Mass distribution of the fragments o nergy balance o Fission barrier o Fission neutrons: prompt and delayed neutrons uclear chain reaction o Time behaviour of the nuclear chain reaction, criticality o Methods for achieving self-sustaining chain reaction Cross sections of special neutron-induced reactions Capture of low-energy neutrons Usually it is an exothermic process, because of the binding energy of the neutron. The neutron is neutral, therefore there is no activation energy also neutrons with very small kinetic energy can induce this reaction. The reaction probability is proportional to the time what the neutron and the nucleus spend close together R radius of the nucleus This is the /v cross section ~ t v velocity of the neutron How does it look in a log() log( n ) graph?. example: neutron capture cross section of Cd n,g : /v region resonance Cd(n,g) Cadmium is an important n-absorber and shielding material!!. xample Fission cross-section for uranium isotopes /v region resonance region energy threshold for 8 U(n,f)
Reminder: Binding energy of nuclei (Weizsäcker) B The = const. cuts are parables! uclear energy B b b V F Z Z bc b bp (average energy of a nucleon) (energy = binding energy) The energy valley :, Z The place of Z min on the (,Z) map is between the blue and purple region This way the radioactive decays could be explained! Z 5 min in function of mass number () This helps to understand the nuclear energy production! Two ways of producing nuclear energy The fission process Fusion (make larger nuclei from small ones) Fission (make smaller nuclei from large ones) For example H+ H He + n (Q = 7, MeV) For the fusion: large gain in energy/nucleon (~ 5 MeV/nucleon) few nucleons (~... 5), Total ~ 8 MeV For the fission: small gain in energy/nucleon (~,85 MeV/nucleon) many nucleons (~5), Total ~ MeV 7 ) The nucleus gets in an excited state (for example by capturing a neutron) ) The shape gets deformed, a thinner neck forms ) Scissions into two parts, a few neutrons get emitted Heavy transuranium elements may fission spontaneously, no need to excite them. Then the first step may be missing. 8
The fission reaction: 5 U + n Fragment + Fragment +, n The composition (Z,) of the fragments cannot be given more precisely, since the fission process is stochastic also in this respect!! (This is why the average number of emitted neutrons is not an integer) Mass distribution of the fragments: If the fission was symmetric: 8 9U Pd 8 Pd The production of ~ fragments ~ less probable The fission is most likely asymmetric! ( 5 U +n therm ) This distribution depends on the fissile material and also on the energy of the neutrons! ~9 ~ 9 Fission on the (,Z) map /Z = /Z =,5 UTRORICH nuclei U 9 /Z = /9 =,5 Starting point (/Z ~,5) ( stable nucleus T~,5 9 y) nd points if the ratio remains /Z ~,5 (asymmetric fission) Highly radioactive( b decays) fission products eutron emission necessary (v ~,) nergy balance of the fission 5 U(n,f) The total energy will be released through several different processes. This influences the place and the time of heat production The kinetic energy of the fragments 8 MeV (8, %) The energy of the b-particles from the fragments 8 MeV (,9 %) nergy of the neutrons emitted in the fission 5 MeV (, %) nergy of the prompt γ-photons 7 MeV (, %) nergy of the γ radiation of the fragments 7 MeV (, %) nergy of the antineutrinos emitted by the b-decays of the fragments MeV (,9 %) TOTL 5 MeV (%) Short range (in the fuel, or close to it) Medium range (coolant, reactor vessel, biological shield) Very long range (leaves the reactor) Prompt (in time of the fission) xample: suppose that the U nucleus fissions the following way: U 9 Kr + Ba + n Determine, how far are the fragments at the scission point, when only Coulomb-forces act. ssume, that their total kinetic energy is 8 MeV. Solution: t the scission point they have only Coulomb potential energy, this will turn into their kinetic energy: ZZe 8, J Here Z d = (Kr), Z =5 (Ba) From this we get: d ~ 7, fm The radius of both nuclei using R r R Kr = 5, fm, R Ba =, fm The geometry of the scission :
Fission barrier The fission process begins with a small deformation. (deformation parameter: ) Z B bv bf During the deformation volume = constant, asymmetry, even-odd = constant How much energy is needed for that? The needed energy will be determined using the Weizsäcker-formula: Z bc b bp surface increasing bf bf af Z Z Coulomb-energy decreasing bc bc ac (protons get farther apart) The factors can be calculated, the result: a F ~,5 and a C ~,. The change in the energy: Z,5bF,bC If <, then the nucleus fissions spontaneously for any small deformation! Z,5b Z F,5, bf bc follows:,b C (since b F =,85 - J, b C =, - J) For large nuclei Z/ ~,9, therefore at Z > there is no activation energy for fission. (In fact, already around Z~, the activation energy is so small, that the nuclei undergo fission spontaneously quite quickly by tunnel effect.) The Periodical System ends because of the fission!!! 5 For Z < small deformations need some energy ( > ) this way a fission barrier is formed How can a slow (low energy) neutron make the 5 U nucleus fission? The height of the fission barrier around the uranium is ~ 7-8 MeV Importance of the pairing energy: From 5 U + n even-even nucleus ( U) larger Q From 8 U + n even-odd nucleus ( 9 U) smaller Q therefore 5 U + n fissions by low energy neutron, for 8 U +n there is an energy threshold (~,8 MeV) The binding energy of the neutron covers the activation energy of the fission! 5
The fission neutrons We saw: the fragments have too many neutrons after the fission neutron emission is inevitable However, the number of the emitted neutrons is not constant, it fluctuates around a mean value. It can be well approximated by a Gaussian. The mean value depends on the fissile nucleus, and on the energy of the fissioninducing neutron. FWHM: ~,5 (does not depend on the nucleus) Timely appearance of neutrons (relative to the time of fission) a) Prompt neutrons The largest part of the neutrons is emitted promptly just at the time of the fission. These are the prompt neutrons (t < - s). Their mean energy is ~ MeV nergy-spectrum: Watt-spectrum n n ~,8e sinh ( = MeV) n Mean value: v =, ( 5 U+n th fission) 7 8 b) Delayed neutrons We saw: the fragments have too many neutrons after the fission These all are b -decaying. In some cases after the b - decay the daughter nucleus is in such highly excited state, that neutron-emission is energetically possible delayed neutron possible example: (the neutron is emitted with,78 s half life!) There are many such kind of decays with different half lives, different yields The naming convention: precursor 9 The delayed neutrons are grouped into groups (according to half lives) 5 n (MeV),5,5,,,,5 T i (s) 5,,,, b i (%),,,8,55,7, Total: b =,5 % The delayed neutron ratio: Typical precursors 87 Br, Cs 88 Br, 7 I 89 Br, 8 I 9 Kr, 9 I, Cs I, 5 Cs 87 s, Xe The appearance of the delayed neutrons after the fission: ( t) i b e i t ln T i (delayed n) (delayed n) b = (total n) ~ (prompt n) Their role is very important in the control of the chain reaction!! 5
Chain reaction with neutrons eutron-balance What can happen with a neutron? scapes from the reactor Gets absorbed Induces new fission eutron generations,,, i, i+, i is the number of neutrons inducing fission in the i th generation eutron multiplication factor: If i keff i (definition) k eff <, the chain reaction is decreasing ( sub critical ) k eff =, the chain reaction is stationary ( critical ) k eff >, the chain reaction is increasing ( supercritical ) Time behaviour of the chain reaction i keff i i i i Some algebra: keff i i i The (average) time between two generations: t (generation time) Multiplying the two equations: t k eff keff We get: t d keff Using t we get t dt The solution of this differential equation is: t Starting point: e k eff t Obviously for k eff = we get (t)= = constant, For k eff > the (t) increases exponential in time, For k eff < the (t) decreases exponential in time. The changing rate is determined by For prompt neutrons the generation time is ~ s xample: Suppose that k eff =, The number of neutrons (and also the power of the reactor) changes in one second by e e,, Uncontrollable! (prompt-critical) k eff t e k eff t The generation time of the delayed neutrons will be extended by the half-life of the precursor it can reach even several seconds! The role of the delayed neutrons: increase the effective generation time! The system can be controlled, if k eff < without the delayed neutrons! Therefore, k eff < +b,5 should always be fulfilled! Reactivity: k eff k eff (definition) For a prompt-critical system: k eff =+b, therefore its reactivity: b b (since +b =,5 ~ ) b Commonly used unit of the reactivity is the $ (dollar), which is the reactivity in delayed neutron units. The reactivity is $, if b
Methods for achieving self-sustaining chain reaction (occurs on the surface) ( 8 U only absorbs, no fission) i keff The fraction of new fission should be increased. There are several methods for that i Slow down the neutrons (fission cross section increases) Decrease the fraction of the escape Decrease the fraction of absorption increase the ratio 5 U/ 8 U (enrichment) Large size (surface/volume) decreases Critical mass 5 7