We’ve noted a collision reaction that produces free neutrons: Total ,4n Cross Section ,3n ,2n ,n Alpha particle energy One of the most practical applications of nuclear reactions occurs with the compound nucleus resulting from A>230 nuclei absorbing neutrons. Often split into two medium mass nuclear fragments plus additional neutrons. NUCLEAR FISSION

Studying -rays bombarding beryllium 1930 Bothe & Becker Studying -rays bombarding beryllium produced a very penetrating non-ionizing form of radiation -rays? Irène and Frédéric Joliot-Curie knocked protons free from paraffin targets the proton energy range revealed the uncharged radiation from Be to carry 5.3 MeV 1932 James Chadwick in discussions with Rutherford became convinced could not be s since assuming Compton Scattering to be the mechanism, E>52 MeV!

Replacing the paraffin with other light substances, ionization (cloud) chamber Neutron chamber Replacing the paraffin with other light substances, even beryllium, the protons were still produced. Nature, February 27, 1932

Chadwick developed the theory explaining the phenomena as due to a 5.3 MeV neutral particle with mass identical to the proton undergoing head-on collisions with nucleons in the target. 1935 Nobel Prize in Physics 9Be has a loosely bound neutron (1.7 MeV binding energy) above a closed shell: 5-6 MeV  from some other decay Q=5.7 MeV Neutrons produced by many nuclear reactions (but can’t be steered, focused or accelerated!)

Natural sources of neutrons Mixtures of 226Ra ( source) and 9Be  ~constant rate of neutron production also strong  source so often replaced by 210Po, 230Pu or 241Am Spontaneous fission, e.g. 252Cf ( ½ = 2.65 yr) only 3% of its decays are through fission 97% -decays Yield is still 2.31012 neutrons/gramsec !

Gains ~1 MeV per nucleon! 2119 MeV = 238 MeV released by splitting A possible (and observed) spontaneous fission reaction 8.5 MeV/A 7.5 MeV/A Gains ~1 MeV per nucleon! 2119 MeV = 238 MeV released by splitting 119Pd 238U

Atomic (chemical) processes ~few eV Fission involves 108 as much energy as chemical reactions! Yet  ½ = 1016 yr is a rare decay: not as probable as the much more common -decay  ½ = 4.510 9 yr why?

From the curve of binding energy per nucleon the most stable form of nuclear matter is as medium mass nuclei.

The Q value (energy release) of this process is Consider: The Q value (energy release) of this process is The mass differences cancel since the total number of constituents remains unchanged. For simplicity, if we assume the protons and neutrons divide in the same ratio as the total nucleons:

The difference in binding energy comes from the surface and coulomb terms so the energy released can then be expressed in terms of the surface energy Es and the coulomb energy Ec of the original nucleus (A,Z).

Expressing the energy released in terms of the surface energy Es and the coulomb energy Ec of the original nucleus (A,Z). maximum Q is found by setting dQ/dy1 = 0 Note: maximum occurs when y1 = y2 = 1/2.

>0.35 Fission into two equal nuclei (symmetric fission) produces the largest energy output or Q value The process is exothermic (Q > 0) if Ec/Es > 0.7. in terms of the fission parameter, x >0.35 Suggesting all nuclei with (Z2/A) > 18 (i.e. heavier than 90Zr) should spontaneously release energy by undergoing symmetric fission. However

Half-life of spontaneous fission as a function of x where and R.Vandenbosch and J.R.Huizenga. Nuclear Fusion, Academic Press, New York, 1973.

binding the nucleus together There is a competition between the nuclear force binding the nucleus together and the coulomb repulsion trying to tear it apart

Induced fission as nuclear reaction suggests the absorption of the neutron (and its energy) may induce such distortions/vibrations in the nucleus.

The surface if any arbitrary figure can be expanded as If lm time-independent: permanent deformation of the nucleus If lm time-dependent: an oscillation of the nucleus

The Spherical Harmonics Yℓ,m(,) ℓ = 0 ℓ = 3 ℓ = 1 ℓ = 2

ℓ = 0 Nuclear Charge Density  z ℓ = 1

quadrupole deformation Lowest order to be considered: ℓ = 2 quadrupole deformation For which we write the nuclear radius The l=2, m=0 mode:

Z 

Nuclei do show spectra for such vibrational modes Example of a vibrational spectrum (levels denoted by the number of phonons, N) O.Nathan and S.G.Nilsson, Alpha- Beta- and Gamma-Ray Spectroscopy, Vol.1, (K. Siegbahn, ed.) North Holland, Amsterdam, 1965.

semi-major axis semi-minor axis We can approximate any small elongation from a spherical shape by semi-major axis semi-minor axis The semi-empirical mass formula surface of spheroid From which:

holding spherical shape With the surface energy (strong nuclear binding force) proportional to area Coulomb force deforming nucleus surface tension holding spherical shape which we can write in the form where Notice  > 0 (so the Coulomb force wins out) for: Same fission parameter introduced when estimating available Q in symmetric fission

comes from considering small perturbations from a sphere. As long as these disturbances are slight, the Separation, r, of distinct fragments linearly follows  r V(r) for small r At zero separation the potential just equals the release energy Q For Z2/A<49, is negative. separation r

While for large r, after the fragments have been scissioned  r r r V(r) for small r for large r separation r

distortions the figure shows the energy of deformation (as a factor For such quadrupole distortions the figure shows the energy of deformation (as a factor of the original sphere’s surface energy Es) plotted against  for different values of the fission parameter x. When x > 1 (Z2/A>49) the nuclei are completely unstable to such distortions.

Z2/A=49 Z2/A=36 The potential energy V(r) = constant-B such unstable states decay in characteristic nuclear times ~10-22 sec Z2/A=36 Tunneling does allow spontaneous fission, but it must compete with other decay mechanisms (-decay) The potential energy V(r) = constant-B as a function of the separation, r, between fragments.

No stable states with Z2/A>49! Tunneling probability drops as Z2/A drops (half-life increases).

m for the masses of the nuclear fragments we’re talking about, At smaller values of x, fission by barrier penetration can occur, However recall that the transmission factor (e.g., for -decay) is where m while for  particles (m~4u) this gave reasonable, observable probabilities for tunneling/decay for the masses of the nuclear fragments we’re talking about,  can become huge and X negligible.

. . . Neutron absorption by heavy nuclei can create a compound nucleus in an excited state above the activation energy barrier. As we have seen, compound nuclei have many final states into which they can decay: . . . in general: where Z1+Z2=92, A1+A2+=236 PROMPT NEUTRONS Experimentally find the average A1/A2 peaks at 3/2

Thermal neutrons E< 1 eV Slow neutrons E ~ 1 keV Fast neutrons E ~ 100 keV – 10 MeV Typical of decay Products & nuclear reactions The incident neutron itself need not be of high energy. “Thermal neutrons” (slowed by interactions with any material they pass through) have been demonstrated to be particularly effective. Cross section  incident particle velocity, v This merely reflects the general ~1/v behavior we have noted for all cross sections!

At such low excitation there may be barely enough available energy to drive the two fragments of the nucleus apart. Division can only proceed if as much binding energy as possible is transformed into the kinetic energy separating them out. (so MOST of the available Q goes into the kinetic energy of the fragments!) Thus the individual nucleons settle into the lowest possible energy configurations involving the most tightly bound final states.

There is a strong tendency to produce a heavy fragment of A ~ 140 (with double magic numbers N = 82 and Z = 50).

Recall Gains ~1 MeV per nucleon! 2119 MeV = 238 MeV A possible (and observed) spontaneous fission reaction 8.5 MeV/A 7.5 MeV/A Gains ~1 MeV per nucleon! 2119 MeV = 238 MeV released by splitting 119Pd 238U

Fragment kinetic energy 165 MeV Prompt neutrons 5 MeV 238 MeV represented an estimate of the maximum available energy for symmetric fission. For the observed distribution of final states the typical average is ~200 MeV per fission. This 200 MeV is distributed approximately as: Fragment kinetic energy 165 MeV Prompt neutrons 5 MeV Prompt gamma rays 7 MeV Radioactive decay fragments 25 MeV

235U

Recall Isobars off the valley of stability (dark squares on preceding slide) b-decay to a more stable state.

Recall a and b decays can leave a daughter in an excited nuclear state 1/2- 2- b- b- 187W 198Au 0.68610 1.088 MeV 0.61890 b- b- 0.20625 0.412 MeV 0.13425 5/2+ 0+ 187Re 198Hg

With the fission fragments radioactive, a decay sequence to stable nuclei must follow

With the fission fragments radioactive, a decay sequence to stable nuclei must follow 25 sec b,g 18 min b,g 4 hr b,g 33 d b,g 0.03% 65 sec b,g 13 d b,g 40 h b,g 6 sec b,g 7 min b,g 10 hr b,g 106 yr b,g 1.40% 5 sec b,g 3 hr b,g 4 h b,g sometimes or

For 235U fission, average number of prompt neutrons ~ 2.5 with a small number of additional delayed neutrons.

This avalanche is the chain reaction. with every neutron freed comes the possibility of additional fission events This avalanche is the chain reaction.

at all energies of the absorbed neutron. It is a FISSILE material. 235U will fission (n,f) at all energies of the absorbed neutron. It is a FISSILE material. However such a reaction cannot occur in natural uranium (0.7% 235U, 99.3% 238U)

Total (t) and fission (f) cross sections of 235U. 1 b = 10-24 cm2