Nuclear Forces • Opposite charges attract, like charges repel • Hydrogen has one proton and no neutrons • Everything else has more than one proton • Why don’t the protons repel each other? • If r is the spacing between two protons, the respective electrostatic repulsion force F between the two protons and the stored electrostatic energy U are 2 F e 2 4 0 r e 2 U 4 0 r What is r ?
• It is difficult to determine nuclear size using scattering by charged particles because long range Coloumb force dominates (like trying to determine diameter of sun using orbit of a comet) • Need some head-on collisions • Rutherford shot alpha particles at aluminum and obtained some 1800 scattering • At instant of direction reversal, the alpha particle must be stationary and all kinetic energy is converted into potential energy • This gives an upper bound for nuclear radius Estimate upper bound on aluminum nucleus radius based on backscatter of 7.7 MeV alpha particle
2 Z Z Ale U 4 0 r U 7. 7 MeV Z 2 Z Al 13 19 Z Z e 2 2 13 1. 6 10 r Al 4. 9 10 15 m 12 6 4 0 U 4 8. 8 10 7. 7 10 Determining size of nucleus using neutron scattering • Much easier, since no Coulomb interaction (no electrostatic repulsion) • Simply use R 1 R0 nd
• Neutrons either hit or miss (no grazing) • These give a nuclear radius scaling as 1/3 r r0A 15 where r0 1.37 10 m 1/3 Implications of r r0A
• Volume of a nucleus scales as A • Densities of all nuclei are about the same • Nuclear mass density is Am 27 3 1u 1.66043 10 1. 5 1017 kg m 4 r 3 4 15 3 3 3 1.37 10
Or more picturesquely, the nuclear density is about 150 thousand metric tons per cubic mm like the mass of an ocean liner crammed into the head of a pin Nuclear force
• Attracts protons and neutrons together • Balances proton electrostatic repulsion • Short range, dies out very quickly beyond nuclear radius • Protons and neutrons in nucleus described by quantum mechanics in manner similar to electrons in an atom • e.g. shell structure, Pauli exclusion principle hold • alpha particle is like a completed K shell (2 protons, 2 neutrons, one each spin up, spin down) and so is stable Nuclear force as a generic potential well
Nucleon trapped In potential well
L Nominal dimension Classical energy 2 2 E 1 mv 2 m v p 2 2m 2m Quantum mechanics: quantize momentum p k n where k n /nL
Quantized energy for L 2r0
p 2 2 2 2 E n n 2m 2 8mr0 342 2 2 1. 05 10 n 8 1. 6 10 27 2 1. 37 10 15 2 1. 1 10 12 n 2 Joules 7n 2 MeV Extremely small size of nucleus implies quantized energy levels are in MeV !!! Coulomb Barrier
• Potential well seen by protons and neutrons differs • Neutrons just see nuclear force (strong force) • Protons also see electrostatic repulsive force due to other protons • Nuclear force is short range, electrostatic force is long range Coulomb Barrier, cont’d
Potential seen by a neutron Neutron outside well feels no force, can wander in
Neutron trapped In potential well
L
Well is due to attractive nuclear force from all other neutrons, protons (short range) Coulomb Barrier, cont’d Proton outside well does not feel short range nuclear force Ridge (lip) due to repulsive but does feel repulsive electrostatic force is called Coulomb barrier force due to protons inside nucleus
Proton trapped In potential well
L Nominal dimension Well is due to attractive nuclear force from all other neutrons, protons (short range) Magnitude of nuclear energy
• If nuclear force balances proton repulsion, then stored energy U per pair of protons must be of the order of the electrostatic energy of two
protons separated by r0
2 e 13 U 1.78 10 J 1 MeV 4 0r0
• Any substantial change in configuration of protons, neutrons in nucleus will generally involve releasing or absorbing energies of the order of MeV Compare to energy of an electron orbiting in an atom • The distance is of the order of the Bohr radius so the order of magnitude of the energy for an electron orbiting an atom is e2 Uatom 30 eV 4 0rBohr
• Chemical processes involve rearranging electron orbitals and so are an order of magnitude smaller, a few eV or less Cross-section can also be used to characterize rate of a process • e. g. can define a cross-section for fission, or a cross-section for neutron capture • Cross-section is usually given in “barns” where 1 barn =10-24 cm2 • “like hitting a barn wall” Reaction-rate cross section
• Let R0 be incident number of particles per second that can instigate a process in target particles having density n in a sheet of thickness d
• Let Rprocess be the rate of the process in question • Define the cross-section for the process as Rprocess 1 process R0 nd Number of protons and neutrons as function of atomic mass • Protons repel each other • Nuclear force is “glue” between protons and neutrons that holds the nucleus together • Use “glue” analogy because glue is a short range force Electrostatic potential energy
In a sphere of charge, Gauss’s law shows that the charge behaves as if it is all at the center.
The radial electric field of a sphere of radius r with Z charges is Ze Er 2 4 0r The repulsive radial force on each of these charges is Ze2 Fr 2 4 0r Here e 1. 6 1019 is the charge on a proton 12 and 0 8. 854 10 is the permittivity of vacuum The work done per charge on bringing all the charges in from infinity to surface of a sphere of radius r is r U Frdr per charge r 2 Ze dr 2 4 0r Ze2 r 4 0r Ze2 4 0r The work done on all the charges for doing this is Z2 e2 U 4 0r This is the electrostatic potential energy of Z protons grouped together on surface of a sphere of radius r. If charges are uniformly distributed in the volume of the sphere, then find (HW) that 3 Z2 e2 U 5 4 0r Protons on surface of a sphere
Z2e 2 U 4 0r
Protons filling up volume of a sphere
3 Z2e 2 U 5 4 0r 3 Z2 e2 U 5 4 0r The nuclear radius scaled as 1/3 r r0A and since Z A/2 the electrostatic potential energy of nuclei scales with A as 2 2 A/22 e2 U 3 Z e 3 A5/3 1/3 5 4 0r 5 4 0r 0A and so the electrostatic energy per nucleon scales as U A2/3 A Thus, proportionally more neutrons are required to hold together nuclei with large A General structure of nuclide chart
• Average nuclear force between neutrons and protons is about twice as much as force between two neutrons or two protons
• Implies binding energy ought to be largest if there are equal numbers of neutrons and protons
• But, when there are lots of protons their mutual electrostatic repulsion becomes important, so need extra neutrons
• Large stable nuclides are neutron rich Nuclide Chart Trends
Proton rich Neutron rich
Heavy elements are progressively more neutron rich
Z=N Light nucleii have approximately the same number of neutrons as protons Heavy nuclei have about 1. 5 the number of neutrons as protons Breaking up a heavy nucleus results in neutron-rich light nuclei Thermal fission
A low energy neutron becomes attached to a heavy nuclide having an odd value of A.
The nuclide resonates like a jiggled water droplet. It develops a non-spherical shape.
The long-range electrostatic repulsion force is only slightly reduced by the non-spherical shape.
The short-range attractive nuclear force is greatly reduced by the non-spherical shape.
The repulsive electrostatic force wins and the nucleus splits in two (fission).
neutron
Nucleus with odd A Nucleus with even A + + + + + + + + + + + + + + + + + + Vibrating nucleus + + + + + + + +
+++++ +++++ +++++ +++++ +++++ +++++
Electrostatic repulsion +++++ +++++ +++++ +++++ +++++ +++++ Uranium fission example First neutron hits U-235 and excites it
235 236 92 U n 92 U where the asterisk means excited state (can vibrate)
144 89 Suppose the U-236 splits into 56 Ba, 36Kr and 3 neutrons Using 1/3 r r0A 15 where r0 1. 37 10 m the radii of the Barium and Krypton are 15 1/3 15 rBa 1. 37 10 144 7. 18 10 m 15 1/3 15 rKr 1. 37 10 89 6. 12 10 m If we consider the Barium and Krypton as two spheres just touching each other, the distance between the centers of the spheres is r rBa rKr and so the potential energy of the spheres is 2 ZBaZKre UBa,Kr 4 0r 56 36 1. 6 10 19 4 8. 8 1012 7. 18 1015 6. 12 1015 220 MeV This gets converted into kinetic energy as the two positive nuclides fly away from each other The numerical value of this estimate is about 25% more than the true value Shows that fission energy is mainly kinetic energy of fragments which comes from electrostatic repulsion
+++++ +++++ +++++ +++++ +++++ +++++ Nuclear Fission: A slow neutron collides with U-235, excites it, and then it breaks up into fragments, including a few neutrons Light nucleii have approximately the same number of neutrons as protons Heavy nuclei have about 1. 5 the number of neutrons as protons Breaking up a heavy nucleus results in neutron-rich light nuclei
Fission products will also have ~1.5 x number of neutrons as protons i.e., will lie below line of stability for their atomic mass number and so will be radioactive 235 Typical 235U fission U 235U -> 147La + 87Br +n Z=92, N=143
147La Z=57, N=90 ~8 excess neutrons compared to stable 87Br Z=35, N=52 ~7 excess neutrons compared to stable Quantifying fission process
• Thus, if we shoot neutrons at a sheet of U-235
with rate R0 and measure the number of reactions per second, we can determine the fission cross-section
Rfission 1 fission R0 nd Properties of cross-sections
• Cross-sections can depend on velocity of incident particle • There can be different cross-sections for different competing processes • Comparing cross-sections tells what will happen • Cross-sections can be measured in lab tests, can be computed from first principles for simple situations Fission cross-sections
• Find that fission cross-sections for nuclei with odd values of A are much higher than for even values of A • This is because adding a neutron to an odd-A nucleus is akin to completing an atomic shell (like adding an electron to chlorine, for example) Comparison of U-235 and U-238 fission cross-sections Natural uranium is 99.3% U-238 and 0.7% U-235
1 eV 1 MeV
5 1 0
4 Peak at 0.25 eV 1 0 U-235 1 0 3
2 Cross 1 0 1 Section 1 0 0 (barns) 1 0
- 1 1 0
- 2 C1 ross 0 S ection (b)
- 3 1 0
- 4 1 0 U-238 - 5 1 0
- 6 1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 Energy (M eV) The majority isotope U-238 has a fission cross section which is 6 or 7 orders of magnitude smaller than the rare isotope U-235
This is largely because 238 is an even number and 235 is an odd number.
Fission reactor design & issues • Chain reactions • Need for uranium enrichment • Importance of delayed neutrons • Need for moderators • Breeding • Reactor designs • Neutron cycle • Reactor economics • Reactor control, Chernobyl accident • Fuel cycle • Waste issues Chain reaction
• Suppose a fission reaction produces k neutrons • Each of these will produce k more, so there will be k2 neutrons • Will get a series so the number of neutrons is
2 3 Nneutrons 1 k k k ... if k 1 kn 1 n 0 1k if k 1 Reactivity
2 3 Nneutrons 1 k k k ... if k 1 kn 1 n 0 1k if k 1 k 1 called sub-critical k 1 called critical k 1 called super-critical t k 1 the “reactivity”, is like the accelerator pedal k A serious problem
• While U-238 does not provide significant fission, it does absorb neutrons • This can reduce the effective multiplication constant k • Remember that U-238 is more abundant by factor 99.3/0.7=141 so the U-238 could kill multiplication if its neutron absorption cross section exceeded 1/141 the U-235 neutron multiplication Fission Cross-section v. U-238 capture 1 eV 1 Mev
5 1 0 U-235 fission At 0.25 eV peak, 4 Ratio is two orders of magnitude 1 0
1 0 3 Less than one order 2 of magnitude ratio 1 0
1 1 0
0 1 0
- 1 1 0 U-238 neutron capture
- 2 C1 ross 0 S ection (b)
- 3 1 0
- 4 1 0 U-238 fission
- 5 1 0
- 6 1 0 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 Energy (M eV) Solutions to the problem
1. Get rid of the U-238 (or at least reduce its proportion); this is called enrichment 2. Take advantage of the two order of magnitude ratio of U-235 reaction to U-238 absorption when the neutrons are slow (i.e., slow down the neutrons) 3. Some combination of the above, slow down and enrich Thermal v. fast reactors
• A thermal reactor uses neutrons that have been slowed down – If great care is taken to avoid neutron absorption by other isotopes in neighborhood, then can get criticality with natural uranium – Otherwise, need some enrichment (a few percent) • A fast reactor uses neutrons that have not been slowed down (i.e. MeV neutrons) and has highly enriched uranium Fission fragments and delayed neutrons • U-235 splits into many different types of fragments, most of which are radioactive and undergo further decay, releasing alpha particles, beta particles, gamma rays, and a few more neutrons • It takes time for all this to happen so these fragment neutrons are delayed in time by a few seconds on average • A thermal reactor is operated so that 0.65% of the criticality comes from the delayed neutrons • This makes it possible to make slow adjustments to the multiplication factor k • The reactor can easily be controlled by mechanical adjustments to reactivity Fission fragment distribution Moderators: slowing down the neutrons • Collision theory shows that most efficient energy transfer is when target mass is same as incident mass • Thus, slowing neutrons by protons should work well (get billiard ball type collisions)
• Could use water H2O, cheap, lots of protons • Unfortunately, protons tend to bind with neutrons (like completing a shell) so get neutron absorption
• Thus H20 moderates, but also absorbs neutrons • Works, but need slightly enriched uranium (few percent) • Called light water reactor (USA method) Head-On Collisions (elastic)
Suppose a particle of mass m1 and incident velocity v1i collides head-on with a stationary target particle of mass m2
Momentum and energy conservation imply m1v1i m1v 1f m2v2f 2 2 2 m1v1i m1v 1f m2v2f
Solve for v2f 2v 1i v2f 1 m2/m1
Energy transferred by collision to m2 is 2 2v 4m2 /m1 W 1 m v2 1 m 1i 1 m v2 2 2 2f 2 1 1i 2 2 2 1 m2/m1 2 1 m2/m1 2 Since x/ 1 x maximizes at the value 1/4 when x 1, it is seen that W2 peaks when m2 m1.
In this case all of m1’s energy is transferred to m2 (billiard ball collision).
1
0.8
0.6 W transfer 0.4
0.2
0 1 2 mass ratio 3 4 5
2 W2/ m1v1i /2 v. m2/m1 Heavy Water Moderators
• Another approach is to use deuterons as
moderator (D20) • Collisional energy transfer not quite as good, but OK • Virtue is little neutron absorption • Thus can use natural uranium • Canadian method (Heavy Water Reactor)
1
Fraction of 0.8
Energy transferred 0.6 Neutron W transfer In a collision 0.4 hitting 0.2 deuteron
0 1 2 mass ratio 3 4 5
2 W2/ m1v1i /2 v. m2/m1 Graphite moderators
• Carbon-12 is very stable (like 3 alpha particles combined) and so does not capture neutrons very much • It is like a filled atomic shell again and so does not have an affinity for an extra neutron • Energy transfer per collision is now just 28% per collision so need many collisions to slow down • Need to have large volume reactor so that neutrons are not lost during this long slowing down process • Use pure graphite (no contamination with neutron absorbers) Breeding
• Heavy nuclei with odd numbers of nucleons are more unstable than those with even numbers (e.g. U-235 v. U-238) • U-238 had a substantial cross-section for neutron absorption • The reaction is 238 239 239 239 92 U n 92 U 93 Np 94 Pu • Note that emission of a beta changes a neutron into a proton and so raises Z while keeping A the same • Plutonium-239 is also fissile (i.e., undergoes fission with thermal neutrons) • This way the 99.3% of natural uranium which is U-238 can be converted into fissile fuel • Increase fuel reserves by factor of 140 Breeder reactors
• Run a U-235 reactor in such a way that a large number of neutrons leak out of the reactor (need to use enriched uranium to allow for this loss) • Surround the reactor with a blanket of U-238 which absorbs these neutrons and gets transmutated into plutonium • Get more fissile material than what one starts with (breeding ratio) • Fast breeder reactors do this with unmoderated neutrons, highly enriched uranium in core • Breeders have been abandoned because there is ample supply of uranium and breeders have had serious technical problems Plutonium production in a thermal reactor • 235U fission requires one neutron • Conversion of 238U to 239Pu also requires one neutron • If fuel is natural or low enrichment uranium, then it is mainly 238U • neutrons captured by 238U result in 239Pu production • Neutrons captured by 239Pu give 240Pu Plutonium production in a thermal reactor, cont’d • Plutonium is fissile and so is a fuel • Produced plutonium gets burned to some extent and so partially makes up for depletion of 235U • 239Pu builds up as fuel burned • Fission products also build up as waste, many are large neutron absorbers (poisons) • Reprocessing chemically separates out 239Pu for use as fuel again • Benefits: – get more energy out of initial uranium – get rid of plutonium, reduces long term waste problem • Drawbacks: – Proliferation since 239Pu can be used to make weapons – If initial uranium is heavily used then substantial 240Pu produced which causes premature detonation in weapons (fizzle) – Weapons production involves putting natural uranium in reactor for only a short time so that 239Pu produced, but not much 240Pu • USA has decided not to do reprocessing because of proliferation issue Plutonium buildup in a reactor 240Pu is fissile but fizzles 239Pu is fissile Credit: Scientific American