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Particle Physics

Conservation

In nuclear reactions Mass – energy is conserved ( E = mc2) ( K.E. = ½mv2)Momentumis conserved ( momentum = mass x velocity )Electric charge is conserved

In some nuclear reactions in particular β decay,energy and momentum appeared not to be conservedIn 1930 Pauli suggested that another particle a neutrino () was emitted in the decayThe neutrino was discovered in 1956 by Cowan and Reines .

Worked example

U 238 ->Th234 + He 4Kinetic energy of the products = 6.8x10-13Joules 92 90 2 Calculate the energy of the particle

Solution : Total momentum before the interaction = total momentum after the interaction Momentum of the U is zero = Momentum of the (Th + He ) (He is an particle) => - Momentum of Th = Momentum of He () - Mass Th x velocityTh = Mass He x Velocity He

Ratio of the masses 234 : 4 = 58.5 : 1 ( Th : He )Ratio of the velocities 1Th : 58.5 HeMomentum = mass x velocity Kinetic energy = ½ mass (velocity)2

Therefore the ratio of the kinetic energies depends on the ratio of the velocities KE of Th + He () -> 6.8 x 10 -13 Joules 1 : 58.5 -> 6.8X10 -13Joules( 59.5) -> 6.8X10-13 Joules KE of Th = 6.8x10 -13 = 1.14 x 10 -14 Joules 59.5KE of the He ( ) = ( KE of Th x 58.5 ) = ( 1.14 x 10 -14 ) x ( 58.5 ) = 6.69x10 -13 Joules

Problem :Americium 241 emits an ( He42 )particle with a velocity of 3 x 107 ms-1 .

Calculate the recoil velocity of the resulting nucleus .

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1 7 4 4

H1 + Li 3 = He 2 + He 2 + Energy

This experiment is important because 1. It is the first particle accelerator 2. It was the first experimental proof for Einstein`s equation (E = mc2 ) 3. It was the first artificial splitting of a nucleus4. It was the first transmutation of an element using artificially accelerated particles .

Transmutation=to change one element into another

1919 Rutherford performed the first artificial transmutation when he changed Nitrogen into Oxygen4 14 17 1

He () + N = O + H 2 7 8 1

Problem( units given in Kg) When Lithium is bombarded with a proton two alpha particles are produced. How much energy is released in this reaction? Give your answer in both Joules and electron volts(eV).Lithium nucleus = 1.165007 x 10-26Kg, the mass of the proton = 1.673493x10-27Kg and the mass of the alpha particle is 6.646322x10-27Kg. ( speed of light = 3 x 108 )2

The reaction is 7 Li + 1H 4He + 4He + Energy 3 1 2 2

Total mass of reactants = (1.165007 x 10-26) + (1.673493x10-27) = 1.332356x10-26 KgTotal mass of the products = 2 x (6.646322x10-27) = 1.329264x10-26KgDecrease in mass = (1.332356x10-26) - (1.329264x10-26) = 3.092x10 -29 Kg

E = m c2 = ( 3.092 x 10-29 ) ( 3 x 108 )2 = 2.7828 x 10 -12 Joules.

Energy released = 2.7828 x 10 -12 J

Convert Joules to electron volts divide by the charge on the electron ( 1.6 x 10 -19 Coulomb )

Energy released 2.7828 x 10 – 12 = 17.39 x 10 6 e V 1.6x10-19

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In particle physics the mass is often expressed in unified atomic mass units u

1u =1.66 x 10 -27 Kg

Repeat of previous problem with the units given in unified atomic mass unitsu When Lithium is bombarded with a proton two alpha particles are produced. How much energy is released?Lithium = 7.01600u, a proton = 1.00783u and an alpha particle = 4.00260u ( 1 u = 1.66 x 10 -27 Kg ) and ( speed of light = 3 x 10 8 ms-1)

The reaction is 7 Li + 1H 4He + 4He + Energy312 2

Total mass of the reactants : 7.01600 + 1.00783 = 8.02383 uTotal mass of the products : 2 x ( 4.00260 ) = 8.00520 uMass Defect = (8.02383) - (8.00520) = 0.01863 u

Now convert (unified atomic mass units) u to Kg (0.01863) X (1.66 X 10 -27) = 3.0925 X 10 -29 Kg

Energy released E = mc2Energy = ( 3.0925 x 10 -29 ) ( 3 x 10 8)2 = 2.78 X 10 -12 Joules

If asked convert Joules to electron volts ( see previous example )

Problem : Show that 1 u is equivalent to an energy of 931 MeV ( 1u = 1.66 x 10 -27 kg ) ( c = 3 x 10 8 ms-1 ) ( e- = 1.602 x 10-19 C )

E = m c2 = ( 1.66 x 10 -27 ) x ( 3 x 10 8 ) 2 Joules of energy

To convert Joules to e V ( electron volts ) :

Energy injoules = ( 1.66 x 10 -27 ) ( 3 x 10 8 ) 2 = 931 x 10 6eV = 931 MeVCharge on the electron ( 1.602 x 10 -19 )

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2007 ( higher ) Q. 10 a( calculations only )

What is the velocity of the proton when it is accelerated through a potential difference of 700kV?( mass of proton = 1.6726 x 10-27 kg ) ( charge on the electron = 1.6022 x 10 -19 C )

Work ( energy ) = charge on the electron x voltage Work ( kinetic energy ) = ½ mass x (velocity) 2

W = q x V W = ½mv 2

q x V = ½mv2 ( V = VOLTAGE v = velocity ) 2 ( q x V ) = v2

m2 x ( 1.6022 x 10 -19 ) x (700 x 10 3 ) = v 2

1.6726 x 10-27

1.6 x 10 -7 m s -1 = v

Write a nuclear equation to represent the disintegration of a lithium nucleus when bombarded with a proton , and calculate the energy released in the disintegration .( mass of Lithium nucleus = 1.1646 x 10 -26 kg) ( mass of Helium nucleus = 6.6443x 10-27kg )( speed of light = 2.9979 x 10 8 ms-1)

Li73 + H1

1 ->2 He42

Mass of the reactants = ( 1.646 x 10 -26 ) + (1.6726x 10-27) = 1.33186 x 10 -26 Kg Mass of the products = 2 x ( 6.6443 x 10-27 )= 1.32886 x 10 -26 kgLoss of mass = ( 1.33186 x 10-26 ) – (1.32886 x 10-26) = 0.003 x 10 -26 = 3 x 10 -29 kg

Energy released = E = mc2 ( 3 x 10 –2 9 ) x ( 2.9979 x 108)2 = 2.7 x 10 -12 Joules

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Converting energy into mass

Antimatter :In 1928 Paul Dirac predicted the existence of antimatter In 1932 Carl Anderson discovered the positron ( e+) , the antiparticle of the electron .We now know that every particle has an antiparticle

Pair production : Is the production of a particle and its antiparticle from a photon of energy

Pair production occurs when a high energy gamma ray () photon loses its energy (E =hf ) when it collides with a nucleus. Some of its energy is converted into the mass of the electron and positron , the rest of its energy appears as the kinetic energy of the particles.

Photon electron + positron e- + e+

Mass – Energy is conserved : energy of the photon ( E = h f ) is converted into the mass and kinetic energy of the electron and positron.Momentum is conserved :Theincident gamma ray had zero momentum The electron - positron pair move with equal velocity inopposite directions’Electric charge is conserved : The gamma ray photon had no charge Sum of the charges on the positron(e+) + electron(e-) = zero

Problem: An electron and a positron are produced from a photon of energy in the electric field of the Nucleus. What is the minimum energy photon necessary for this? (melectron = 0.0005486 u ) ( 1 u = 1.66x10-27Kg) ( e- = 1.602x 10-19C ) ( c = 3 x 10 8 ms-1)

Note : The minimum energy assumes all the energy is turned into matter , and the particles produced have zero kinetic energy .

The electron and positron have equal mass ( melectron = 0.0005486 u )Mass of electron + positron = 2 ( 0.0005486 ) = 0.0010972 u

Convert u to kg ( 0.0010972 ) x ( 1.66 x 10-27) = 1.8213 x 10-30 kg

E = mc2 = ( 1.18213 x 10 -30 ) x ( 3 x 10 8 )2 = 1.63917 x 10 -13 Joules of energy

Convert joules to eV = energy in joules = 1.63917 x 10 -13 = 1023202.247 eV = 1.02 x 10 6 eV Charge on electron 1.602 x 10 -19

When the photon has greater energy than this the excess shows as kinetic energy of the particles

If the photon has less energy than this pair production will not happen

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Converting matter into energy

Pair annihialation : When an electron and a positron meet they annihilate each other and are replaced by two photons of energy moving in opposite directions e- + e+ = + e- + e+ = 2hf ( electron + positron 2 photons of energy )

Conservation of mass - energy : All the mass of the electron and positron ischanged into two photons of energy Conservation of momentum : Initially the particles are at rest ( momentum = 0 )two photons of energy moving in opposite directions are produced Conservation if electric charge : The positron and the electron have equal but opposite chargeAfter the interaction there is only two photons of energy ( zero charge )

Note : In pair production we calculated the minimum energy of the photon required to produce an electron positron pair to be 1.02 x 10 6 eV

As annihilation produces two ray photons , the energy of each photon must be 0.51 x 10 6 eV

The PET scanner ( uses accelerated positrons ) to examine inside the brain

Antiprotons : The antiproton was discovered in 1955 by Segre and Chamberlain Because the mass of the proton is so much bigger than the electron ittakes considerably more energy to produce a proton – antiproton pair

When a proton and an antiproton collide they do not annihilate each other but produce particles that decay further . Example proton – antiproton pion – antipion ( these in turn decay )p+ + p-+ + -

Problem: A photon ( ) strikes a hydrogen nucleus ( p+) and produces a neutral lambda hyperon (0) and a K meson ( K+). Calculate the minimum energy the photon requires for this reaction .Mass of p+ = 1.007825 u mass of photon () = Zero ( energy only )Mass of K+ = 0.530344 u mass of 0 = 1.198646 u( 1u = 1.66 x 10 -27 kg ) ( c = 3 x 10 8 )

+ p+ = K+ + 0

Mass of reactants ( + p+ ) = 0 + 1.007825 u = 1.007825 uMass of products ( K+ + 0 ) = ( 0.530344 ) + ( 1.198646 ) = 1.72899 u

Mass created( mass of products - mass of reactants ) 1.72899 - 1.007825 = 0.721165 u

Convert u to kg( 0.721165 ) x ( 1.66x10-27 ) = 1.1971339 x 10 -27 Kg

E = mc2 E = ( 1.1971339 x 10 -27 ) x ( 3 x 10 8 ) 2 = 10.774205 x 10 -11Joules

Convert Joules to eV10.774205 x10 -11 = 6.725 x 10 8 eV = 672 MeV1.602 x 10 – 19

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Families of particles

All particles are now classifed as either Leptons or Hadrons

Leptons : Are not effectedby the strong nuclear force( feel only 3 of the 4 forces ) They are elementary ( fundamental ) particles , they cannot be subdivided Examples : electron ( e-), tau ( ), muon ( ) their neutrinos and their antiparticles .

Hadrons : Are effected by the strong nuclear force ( feel all 4 forces ) Hadrons are made of elementary particles called quarks

Hadrons are subdivided intoMesonsand Baryons

Mesons masses between the electron and the proton Consist of a quark and an antiquark Examples the pion ( +) and the kaon ( k+)

Baryons masses greater than or equal to the mass of the proton Consist of 3 quarks or 3 antiquarksExamples proton, antiproton, neutron, antineutron and sigma ( +)

Quarks Anti Quarks

Name Symbol Charge Name Symbol ChargeUpDown

u d

+2/3 e -1/3 e

Anti -- UpAnti-- Down

u d

-2/3e+1/3e

StrangeCharmed

s c

-1/3 e +2/3 e

Anti -- StrangeAnti-- Charmed

sc

+1/3e-2/3e

Bottom Top

bt

-1/3 e +2/3 e

Anti --BottomAnti --Top

bt

+1/3e-2/3e

Matter

Leptons Hadronse.g Electrons and neutrinos

Mesons Baryons2 quarks 3 quarks

( quark and ananti-quark ) e.g. Protons and neutrons

e.g. pions, kaons

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Search for the fundamental particles of matter

1. The ancient Greeks proposed the concept of the atom2. Dalton( 1800s) said the atom was the fundamental particle of matter 3. Thompson( 1897 ) discovered the electron , it was named by G. Stoney4. In 1905 Einstein predicted mass – energy would be conserved in nuclear reactions (E = mc2 )5. Rutherford( 1917 ) reported the discovery of the proton.6. Chadwick( 1932 ) discovered the neutron 7. Cockroft and Walton first experimental proof of Einstein`s E = mc2

8. However in some nuclear reaction in particular β decayenergy and momentum appeared not to be conserved 9. Pauli (1931) explained the apparent breach by proposing that another particle the neutrino ()carried off the missing energy and momentum . 10. Cowan and Reines (1956 ) discovered the neutrino and antineutrino .11. Paul Dirac (1928) predicted the existence of antimatter 12. In 1932 Anderson discovered the thepositron ( first antimatter )13. Improved particle accelerators allowed the detection of a vast array of particles referred to as the particle Zoo14. Murray Gell – Mann( 1964 ) introduced the idea of quarks (named after a word in a James Joyce book )

Formula

Change in energy= change in mass x (speed of light )2E = mc2

Note : when using this formula energy must be in joules and the mass in Kg

( Energy in eV) x ( charge on the electron ) = Energy in joules

Energy in joules = energy in eVCharge on the electron

1 u = 1.66 x 10 -27 kg ( u = unified atomic mass unit )

Momentum = mass x velocity m = m x v

Kinetic energy = ½ ( mass ) x ( velocity )2JoulesK.E. = ½ mv2 Joules

Velocity of a wave = frequency x wavelength c = f x meter second -1 = Hertz meter

Energy = Planks constant x frequency E = hf