Nuclear PhysicsNuclear Physics

By: Cheung Kwok Tin Andy (2)

Lai Chin Kei Jacky (6)

Lam Chi Pak Kenneth (7)

Syllabus Notes

•The nucleus. •The Rutherford model of the atom.

•The mass energy relationship.

•The unified atomic mass unit (carbon scale). Binding energy.

•Energy release in fission and fusion.

•Interpretation of equations representing nuclear reaction.

•Nuclear power: advantages and disadvantages.

•The principle of the fission reactor.

•Qualitative treatment of fission and the chain reaction, and the role of fuel, moderator, coolant and controls rods are expected.

Syllabus

Thomson’s ModelIn 1897, after Thompson had discovered the electron, he put forth the ‘raisin bun’ model. He imagined that the negative electrons were embedded in an atom. Most of the mass is the positive charge.

Rutherford ScatteringA crucial experiment which tested the Thompson’s model was carried out by Geiger and Marsden under the direction of Rutherford. This experiment involved the scattering of particles from the atom, called Rutherford’s scattering.

Properties of particles

• The mass of an particle is about four times that of a hydrogen atom, so it is much heavier than an electron.   

• Each particle carries two units of positive charge.

• The particles are released with very high energy, about several MeV.

• A radioactive source emits relatively few particles. So we can consider one particle at a time.

Note: 1 eV = 1.6010-19C 1V = 1.60 10-19J

Scattering according to Thomson’s Model

• Suppose the particle passes outside the positive charge (Fig.1). The smallest distance to the center of positive charge is about the radius R of the atom, which is fairly large. So the particle is not subjected to a large force and it should not have a large deflection.

Fig. 1

R

Scattering according to Thomson’s Model (Continued)

• Suppose the particle penetrates near the center of the atom (Fig.2). In this case, the shaded part of the positive charge exerts a strong upward force on the particle. The unshaded part of positive charge exerts a strong downward force on the particle. However, these forces are in opposite directions, and the net force is small. So again the particle should not have a large deflection.

Fig. 2

Results of Rutherford Scattering• Most particles come out with very small

deflections.• Some particles come out with large

deflections.• Occasionally, the particles even come out

with deflections close to 180º.

• Thomson’s model cannot account for those scattering events with large deflection angles.

Rutherford’s Model

• Based on the experimental result, Rutherford put forth his model.

Rutherford’s model

electron

nucleus

Rutherford’s Model

particles propagated on the atomic nuclei of gold foil

Size of Nucleus

• Atomic radius 10-10m

• Nuclear radius 10-15m

Inside the Nucleus• Nucleus contains nucleons which are

protons and neutrons.Sub-atomic particles Proton Neutron Electron

Symbol p n e-

Mass 1.67310-27 kg 1.67510-27 kg 9.10910-28 kg

1.007276u 1.008665u 0.000549u

Relative mass 1 1 1/1837

Electric charge 1.6010-19C 0C -1.6010-19C

Electric charge relative to that on a proton

+1 0 -1

No. of particles per nuclide in

Z A-Z ZXA

Z

Packing of NucleonsNucleons inside a nucleus are packed very tightly, with almost no empty space. So total volume of a nucleus of radius R is proportional to the number of nucleons A:

3

1

0

3

1

3

or

3

4

ARR

AR

AR

where R0 1.210-15m

The equation implies that the densities of all nuclei are nearly the same.

Isotopes• Isotopes of the same element have the same

number of electrons and protons, but with different number of neutrons.

• Relative isotopic mass of a particular isotope of an element is the mass of one atom of that isotope on the 12C＝ 12.00000 scale.

• Each species of atoms characterized by its nucleus is called a nuclide. Two or more nuclides of the same element are called isotopes.

Atomic Mass Unit

• Atomic mass unit(unit: u) is exactly 1/12 of the mass of one atom of carbon-12.

• Mass of 12C = 12.000000u

1u = 1.6610-27 kg

• Avogadro number L is the number of atoms in 12g of 12C.

12327

11

1002.61066112

0.012

12

12

molkg.

molkg

u

molgL

-

-

Mass Defect

Consider neutral 12C atom:

6mp + 6mn + 6me = 12.098940u

But mass of 12C atom is exactly 12.000000u!!!

So 12C has a Mass Defect ∆m of 0.098940u!!!

In general,

X)m(ZmZ)m(AZmX)Δm( AZenp

AZ

X)m(Z)m(AH)Zm( AZn

11

Nuclear Binding1. Nuclear Force or Strong Force is

• force which holds the nucleons together.• very strong to overcome the repulsion

between protons.• very short range, about 2R 2.4×10-15m.• For distance r >> 2R, nuclear force is

negligible.• r > 2R: attraction• r < 2R: repulsion

Note: R is the radius of a nucleon.

Nuclear Binding

2. Binding energy Eb

• Energy required to break up the nucleus

• Eb small nuclide not stable

Reason: small amount of energy is needed to break the nuclide apart

• Eb large nuclide very stable

Reason: lots of energy is needed to break the nuclide apart

Nuclear Binding

From this graph,• Eb/A 8MeV for large nuclides• Eb/A is largest for A 55(Fe) • stand out as specially

stableOC, He, 16

8126

42

Explanation of Binding Energy

1. Bulk effectEach nucleon is bound to its neighbours. Attracting nucleons cause negative energy or positive binding energy let a = positive binding energy each nucleon has

3 effects affecting the trends of Eb/A versus A graph.

Explanation of Binding Energy

2. Surface effect

Nucleons on surface of nucleus have fewer neighbours less negative energy or a lower binding energy

Extra energy = reduction in Eb

no. of nucleons on the surface

surface area R2 A2/3

let reduction in Eb / A = bA-1/3

Explanation of Binding Energy

3. Electric effect

Protons repelling each other make the nucleus unstable and leads to reduction of Eb

reduction of Eb no. of proton pairs interactZC2

and electric PE of each pair R-1

reduction in Eb Z(Z-1)/R

For most nuclide, Z Z-1 ½A, R A1/3

reduction in Eb A2/A1/3 = A5/3

let reduction in Eb/A = cA2/3

Explanation of Binding Energy

Combining all effects,

3

2

3

1

cAbAaA

Eb

The data are well described if a 15.7 MeV, b 17.8 MeV, c 0.18 MeV.

Proton-Neutron Ratio• For small nuclei, N Z.

there are roughly equal no. of proton and neutron.

• For large nuclei, the ratio of neutron to proton increases to increase the stability of nuclide, for example,

1.692

92238

U,for 23892

ZN

Mass Energy Equivalence

• Einstein suggested that mass is just a form of energy.

• The relationship is E = mc2.

Example

1u = 1.6610-27 kg equivalent to 931.5 MeV energy

• In nuclear physics, the below relationship is used frequently.

1u c2 = 931.5 MeV

Mass Energy Equivalence

Applying E = mc2,

Eb= ∆mc2, where Eb is the binding energy.

Mass energy equivalence not just apply only to nuclear binding, but also to other binding, e.g. electrical binding.

Mass defects occurs in any forms of binding, but except for nuclear binding, the defects are usually negligible.

Fission• Fission occurs when an unstable nucleus split

up into a few large part.• Fission can be induced or spontaneous.• Uranium-235 and plutonium-239 are the main

examples.

Spontaneous fissionInduced fission

FissionConsider the fission of U-235:

• Neutron is used as a hammer to hit the U-235 gently to break it apart.

• The process can be divided into four steps:

Step 1: A neutron is captured by a U-235 nucleus and forms U-236 nucleus:

Step 2: The U-236 nucleus is not stable, so it spilt up usually into two parts, e.g.

UUn 23692

23592

10

BrLaU 8735

14957

23692

FissionStep 3: Neutron-proton ratio of two nuclide is too large for stability, so some neutron is ejected to decrease the no. of neutrons.

nBrBr

nLaLa10

8635

8735

10

14857

14957

Step 4: The no. of neutrons continues to decrease by a series of dacay.

NdPrCeLa 14860

β14859

β14858

β14857

KrBr 8636

β8635

Overall equation:

β4n2KrNdUn 01

10

8636

14860

23592

10

Fission

For reaction β4n2KrNdUn 01

10

8636

14860

23592

10

1.0087un)m(

85.9106uKr)m(

147.9169uNd)m(

235.0439uU)m(

10

8636

14860

23592

n)2m(Kr)m(Nd)m(n)m(U)m(Δm 10

8636

14860

10

23592

0.2077u

1.0087u285.9106u147.9169u235.0439u1.0087u

0.09%236u

0.2077u

m

Δm

i.e. 0.09% of mass is used up.

Using data

Fission

• Large amount of energy is released.• Although a neutron is needed to start the

reaction, there are more neutrons at the end, there is neutron multiplication.

• About 0.1% of mass is converted to energy.• About 1014J energy is released per kg of U-

235, while typical chemical reactions release only about 1010J per kg of reactant.

Fission – Chain ReactionChain reaction refers to a process in which neutrons released in a fission produce an additional fission in at least one further nucleus. This nucleus in turn produces neutrons, and the process repeats.

Uncontrolled chain reaction

Fission ReactorControlling a nuclear chain reaction, which depends on controlling the no. of neutrons available, is very important in a fission reactor.In fact, the balance of neutrons is affected by:

• There is a net gain of neutrons from the fission reaction.

• There is a loss due to escape from the reaction.• There is a loss due to absorption by other nuclei.• There is effectively a loss if a fraction of the neutrons

fails to be slowed down by proton in H2O. In a fission reactor, the overall neutron balance is achieved by adding or taking away neutron-absorbing material.

Fission Reactor

Most common fission reactors are• Boiling Water Reactor (BWR)• Pressurized Water Reactor (PWR)

which use U-235 as fuel and water as moderator.

Another type of reactor, Fast Breeder, is under experiment. Fast breeder use U-238 as fuel and liquid sodium as moderator.

Pressurized Water Reactor

• Fuel rods containing special treated uranium is used.

• Fuel rods are stored in a primary circuit consisting of a reactor vessel and pipes in which water is circulated. Water enters the vessel at 300oC.

• Water leaves the vessel at about 330oC. The pressure in the vessel is about 150 atm. Water will not boil at this condition.

Brief descriptions of pressurized water reactor (PWR).

Pressurized Water Reactor

• The hot water (330oC) from the primary circuit transfers heat to a secondary circuit through a heat exchanger.

• In the secondary circuit, the water coming out of the heat exchanger has a temperature of about 280oC.

• The pressure in the secondary circuit is lower, about 50 atm. As a result, the water at 280oC boils and become steam.

Pressurized Water Reactor

• The high pressure steam turns the turbine. Energy is transferred to mechanical motion.

• Since energy has been removed from the steam, it cools and condenses into a reservoir below.

• Water in reservoir is still too hot, so cooling water is piped in. All the energy carried away by the cooling circuit is wasted.

• The turbine is connected to the generator. Mechanical energy is converted to electrical energy.

The rest of the system is just like a conventional power plant for generating electricity.

Schematic of a Pressurized Water Reactor

Pressurized Water Reactor

Main Features in PWR• Fuel

– Natural uranium contain 0.3% U-235, which is not enough for chain reaction.

– Uranium is treated to increase the concentration of U-235 to 3%.

– Uranium in the form of UO2 is made into the form of small pallets.

– The pallets are assembled into rods and encased in the zirconium alloy tube. Fuel rods

Fuel pallet

Main Features in PWR

• Moderator – water– Water molecule contain two hydrogen

nucleus, i.e. protons, which have the mass about the same as neutron.

– KE transfer is most effective when two objects with the same mass collide.

– Act as moderator to slow down fast neutrons.

– Also act as absorber of heat from the fuel rods.

Main Features in PWR

• Control rods– Usually made of caesium or boron.– To absorb neutron and hence control the rat

e of chain reaction.– Pushed deeper into the reactor vessel to slo

w down the reaction rate.– Pulled up to speed up the reaction rate.– All rods are plunged in at once to stop the re

action if there is accident.

Main Features in PWR

• Coolant – water– Water condensed into the reservoir is still

too hot.– Cooling water is pumped in to cool down.– Sea water is usually used.– This is the reason why fission reactors are

often built beside the sea.

Main Features in PWRThree safety barriers are used in PWR to prevent the escape of radioactive material into the environment.

1. The zirconium tubes in which the fuel is encased can withstand very high temperatures and pressures. They seal the fission material released from the fuel and prevent it from escaping into the water in primary circuit.

2. The reactor vessel has an alloy steel wall of about 20cm thickness. The steel piping of the primary circuit is about 7cm thick.

Main Features in PWR

3. The whole primary circuit is placed inside a containment building founded on a layer of bedrock. The building has a 5.5cm thick base of reinforced concrete and a 90cm thick reinforced concrete wall with 6mm thick steel lining on the inside.

Fusion

• Fusion occurs when two small nuclei collide and combine into one bigger nuclei.

• Some examples:

H2HeHeHe

γHeHH

νeHHH

11

42

32

32

32

11

21

01

21

11

11

where e+ and are positron and neutrino respectively

Overall equation: 2224 01

42

11 eHeH

Fusion

• DD fusion reaction

• DT fusion reaction

• is deuterium or D. is tritium or T. They are the isotopes of hydrogen.

HHHH 11

31

21

21

nHeHH 10

42

31

21

H21 H3

1

FusionFor reaction 22e2HeH4 0

142

11

0.0005u)m(e

4.0026uHe)m(

1.0078uH)m(42

11

]2mnucleus) He[m(-nucleus) H4m(Δm e42

11

= 4×1.0078u - 4m(e-) – [4.0026u - 2m(e-)] - 2m(e+)

= 0.0266u

0.7%4u

0.0266u

m

Δm

Percentage of mass used up is much larger than in fission(~0.1%)!!!

Using data

Fusion – Coulomb Barrier

1. The gas must be so hot that the atoms are ionized. Then the bare nuclei can collide with each other without the ‘protective coats’ – the electrons. A temperature of about 105 K is necessary for ionization.

2. Actually, the gas must be even hotter for fusion to occur. The reason is that the nuclei must have enough KE to overcome the mutual Coulomb repulsion. This effect is very important and a temperature of 107 – 108 K is necessary. For example, the DD reaction requires about 5×108 K, while the DT reaction requires 5×107 K.

The following two conditions needed.

Fusion – Future Source of Energy

• Hydrogen is almost limitless in the form of H2O.

• End products are light elements, which are not strongly radioactive.

• Technical problems:

Temperature needed is 107-108K!!!

No physical container can withstand this temperature.

Tremendous gas pressure and very difficult to contain the reacting gas.

Experimental fusion reactor

Fission and Fusion:Advantages and Disadvantages

Energy Sources

Advantages Disadvantages

Fission -Clean, no CO2

-Does not produce immediate pollution

-Waste disposal is difficult

-Safety concerns

Fusion -Inexhaustible supply of water, the fuel of fusion

-Fuel is accessible worldwide

-Clean

-Fusion reactors are inherently safe, cannot explode or overheat

-Huge research and development costs

-React or vessel core becomes radioactive

THE ENDTHE END