Nuclear Forces Opposite charges attract, like charges repel - - PowerPoint PPT Presentation

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

F  e2 40r2 U  e2 40r

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

• rbit of a comet)
• Need some head-on collisions
• Rutherford shot alpha particles at aluminum and
• btained 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

U  Z Z Ale2 40r U 7.7 MeV Z  2 Z Al 13 r  Z Z Ale2 40U



2 13  1.6 1019 4 8.8 1012 7.7 106 4. 9 1015 m

Determining size of nucleus using neutron scattering

• Much easier, since no Coulomb interaction (no

electrostatic repulsion)

• Simply use
• Neutrons either hit or miss (no grazing)
• These give a nuclear radius scaling as

  R

R0 1 nd

r  r0A1/3 where r0  1.37 1015 m

Implications of r  r0A1/3

• Volume of a nucleus scales as A
• Densities of all nuclei are about the same
• Nuclear mass density is

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

  Am1u

4 3 r3 

1.66043 1027

4 3 1.37 10153  1. 5 1017 kg m 3

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

E  1

2 mv 2  m v2 2m



p 2 2m

Quantum mechanics: quantize momentum p k n where k n /nL Quantized energy for L  2r0

E n  p2 2m   22n2 8mr0

2

n 2 1.05 10342 2

8 1.6 1027 2 1.37 10152

1.1 1012n2 Joules 7n2 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

Neutron trapped In potential well L Potential seen by a neutron Well is due to attractive nuclear force from all other neutrons, protons (short range) Neutron outside well feels no force, can wander in

Coulomb Barrier, cont’d

Proton trapped In potential well L Nominal dimension Well is due to attractive nuclear force from all other neutrons, protons (short range) Ridge (lip) due to repulsive electrostatic force is called Coulomb barrier Proton outside well does not feel short range nuclear force but does feel repulsive force due to protons inside nucleus

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

• Any substantial change in configuration of

protons, neutrons in nucleus will generally involve releasing or absorbing energies of the order of MeV

U  e2 40r0  1.78 1013 J  1 MeV

Compare to energy of an electron

• rbiting 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

• Chemical processes involve rearranging

electron orbitals and so are an order of magnitude smaller, a few eV or less

Uatom  e2 40rBohr  30 eV

Cross-section can also be used to characterize rate of a process

• e. g. can define a cross-section for fission,
• r 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

 process  Rprocess R0 1 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 Er  Ze 40r2 The repulsive radial force on each of these charges is Fr  Ze2 40r2 Here e 1. 6 1019 is the charge on a proton and 0  8. 854 1012 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 Uper charge 



r

Frdr 



r

Ze2 40r2 dr  Ze2 40r



r

 Ze2 40r The work done on all the charges for doing this is U  Z2e2 40r This is the electrostatic potential energy of Z protons grouped together on surface

• f a sphere of radius r.

If charges are uniformly distributed in the volume of the sphere, then find (HW) that U  3 5 Z2e2 40r

Protons on surface of a sphere Protons filling up volume

• f a sphere

U 

Z2e 2 40r

U  3

5 Z2e 2 40r

U  3 5 Z2e2 40r The nuclear radius scaled as r r0A1/3 and since Z  A/2 the electrostatic potential energy of nuclei scales with A as U  3 5 Z 2e2 40r  3 5 A/22e2 40r 0A1/3  A5/3 and so the electrostatic energy per nucleon scales as U A  A2/3 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

Z=N P r

• t
• n

r i c h N e u t r

• n

r i c h Heavy elements are progressively more neutron rich

Nuclide Chart Trends

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

neutron Nucleus with odd A Nucleus with even A 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).

+ + + + + + + + + + + + + + + + + + + + + + + + + + +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ Vibrating nucleus Electrostatic repulsion

Uranium fission example

First neutron hits U-235 and excites it

92 235U 

n  92

236 U

where the asterisk means excited state (can vibrate) Suppose the U-236 splits into 56

144Ba, 36 89Kr and 3 neutrons

Using r r0A1/3 where r0 1. 37 1015 m the radii of the Barium and Krypton are rBa 1. 37 1015 1441/3 7. 18 1015m rKr 1. 37 1015 891/3 6. 12 1015m

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 UBa,Kr  ZBaZKre2 40r  56 36 1. 6 1019 4 8. 8 1012 7. 18 1015 

• 6. 12 1015

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

235U

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 Typical 235U fission

235U -> 147La + 87Br +n

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

 fission  Rfission R0 1 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
• dd-A nucleus is akin to completing an

atomic shell (like adding an electron to chlorine, for example)

• 9

1 0

• 8

1 0

• 7

1 0

• 6

1 0

• 5

1 0

• 4

1 0

• 3

1 0

• 2

1 0

• 1

1 0 1 0 1 1 0 E n e r g y ( M e V ) 1 0

• 6

1 0

• 5

1 0

• 4

1 0

• 3

1 0

• 2

1 0

• 1

1 0 0 1 0 1 1 0 2 1 0 3 1 0 4 1 0 5 C r o s s S e c t i o n ( b )

Comparison of U-235 and U-238 fission cross-sections Natural uranium is 99.3% U-238 and 0.7% U-235

Cross Section (barns) U-235 U-238 1 eV 1 MeV Peak at 0.25 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

Nneutrons  1  k  k2  k3  ... 

n 

kn   if k  1

1 1k if k  1

Reactivity

Nneutrons  1  k  k2  k3  ... 

n 

kn   if k  1

1 1k if k  1

k  1 called sub-critical k  1 called critical k  1 called super-critical t  k  1 k the “reactivity”, is like the accelerator pedal

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

• 9

1 0

• 8

1 0

• 7

1 0

• 6

1 0

• 5

1 0

• 4

1 0

• 3

1 0

• 2

1 0

• 1

1 0 1 0 1 1 0 E n e r g y ( M e V ) 1 0

• 6

1 0

• 5

1 0

• 4

1 0

• 3

1 0

• 2

1 0

• 1

1 0 0 1 0 1 1 0 2 1 0 3 1 0 4 1 0 5 C r o s s S e c t i o n ( b )

U-235 fission U-238 neutron capture U-238 fission 1 eV 1 Mev

Fission Cross-section v. U-238 capture

Less than one order

• f magnitude ratio

At 0.25 eV peak, Ratio is two orders of magnitude

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
• f 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 m1v1i

2 m1v 1f 2 

m2v2f

2

Solve for v2f v2f  2v 1i 1  m2/m1 Energy transferred by collision to m2 is W2  1 2 m2v2f

2  1

2 m2 2v1i 1  m2/m1

2

 1 2 m1v1i

2

4m2/m1 1  m2/m12 Since x/1  x2 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).

0.2 0.4 0.6 0.8 1 W transfer 1 2 3 4 5 mass ratio

W2/m1v1i

2 /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)

0.2 0.4 0.6 0.8 1 W transfer 1 2 3 4 5 mass ratio

W2/m1v1i

2 /2 v. m2/m1

Fraction of Energy transferred In a collision Neutron hitting deuteron

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
• 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

92 238U  n  92 239U 

 93

239Np 

 94

239Pu

Breeder reactors

• Run a U-235 reactor in such a way that a large number
• f 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
• ne 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