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1 Nuclear and Particle Physics 1 Nuclear and Particle Physics

1 Nuclear and Particle Physics - PowerPoint Presentation

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1 Nuclear and Particle Physics - PPT Presentation

2 Nuclear Physics Back to Rutherford and his discovery of the nucleus Also coined the term proton in 1920 and described a neutron in 1921 Neutron discovered by Chadwick in 1932 Ernest Rutherford ID: 643079

mev mass protons number mass mev number protons neutrons positron electron nucleus atomic decay emission energy isotopes kev 931

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Slide1

1

Nuclear and Particle PhysicsSlide2

2

Nuclear PhysicsBack to Rutherford and his discovery of the nucleusAlso coined the term “proton” in 1920, and described a “neutron” in 1921Neutron discovered by Chadwick in 1932

Ernest Rutherford

1871-1937

m

e

=

9.1 x 10

-31 kg mN = 1.6749 x 10-27 kgmP = 1.6726 x 10-27 kg

James Chadwick 1891-1974

nucleonsSlide3

3

Nuclides and IsotopesTo specify a nuclide:

Z is the atomic number = number of electrons

or

protons

A is the mass number = number of neutrons + protons

So number of neutrons = A-Z

Number of protons = Z

Isotopes – same atomic number, different mass numbere.g. carbon:

Many isotopes do not occur naturally, also elements > USlide4

4

SizesWe saw with the Bohr model that radius of the atom depended on atomic numberNucleus = protons + neutrons = mass numberThe volume of a nucleus is proportional to the mass numberSlide5

5

MassesMass spectrometer

1 atomic mass unit (u.) = 1.6606 x 10

-27

kg = 931.5 MeV

Fixed so that carbon = 12.00000 u

m

N = 1.6749 x 10-27 kg = 1.0087 u

mP = 1.6726 x 10-27 kg = 1.0078 uSlide6

6

Binding EnergyTotal mass of a nucleus < sum of massesExample:Mass of helium nucleus = 6.6447 x 10-27 kg

Contains 2 protons and 2 neutrons

Mass = 2 x (1.6749 x 10

-27

+ 1.6726 x 10

-27

) kg

= 6.6950 x 10-27 kg

Difference = (6.6950 – 6.6447) x 10-27 = 0.0503 x 10-27 kgEnergy = mc2 = 0.0503 x 10

-27 x c2 = 4.53 x 10-12 J = (4.53 x 10

-12) / (1.6 x 10-19) = 2.83 x 107 eV = 28.3 MeVSlide7

7

Atomic Mass Units1 u = 931.5 MeVmN = 1.6749 x 10-27 kg = 1.0087 um

P

= 1.6726 x 10

-27

kg = 1.0078 uMass of helium nucleus = 4.0026 uSlide8

8

Atomic Mass UnitsSame calculationMass of 2p + 2n = 2 x (1.0078 + 1.0087) = 4.0330 u

Difference = 0.0304 u

Binding energy = 0.0305 x 931.5 = 28.3 MeV

4.0330 uSlide9

9

Average Binding EnergyGraph

He – 4 nucleons, 28.3 MeV total: average = 7.075MeVSlide10

10

Attractive?How does nucleus stay together? Like charges repel!Force stronger than electric force Strong nuclear forceShort range (~10-15

m)

Stable nuclides N = Z

A > 30-40 – more neutrons

Z > 82 – no stable nuclides

Strong force can’t overcome repulsionSlide11

11

RadioactivityBecquerel, 1896Emission of radiation without external stimulusCuries – polonium (Po) and radium (Ra)

Henri Becquerel

1852-1908

Marie Curie

1867 - 1934

Pierre Curie

1859 - 1906

1903 (Physics)

1911 (Chem)

Radioactivity unaffected by heating, cooling, etc.Slide12

12

ClassificationRutherford classified 3 types of radioactivity according to penetration powerAlso different charge

Important factor: Conservation of

nucleon number

(neutrons + protons) = (neutrons + protons)

Video:

People Pretending to be Alpha Particles

”Slide13

13

Alpha DecayLeast penetrating – nucleus of

Radium 226 is an alpha emitter:

Parent

Daughter

transmutation

Mass of parent > mass of daughter + mass of alpha

Difference = kinetic energySlide14

14

Example232.03714 u  228.02873 u + 4.002603 u

total = 232.03133 u

Lost mass = 232.03714 – 232.03133 = 0.00581 u

0.00581u

x

931.5 MeV/u

= 5.4 MeV (some recoil)Slide15

15

Beta decay One electron

What is lost is NOT an orbital electron

Instead a neutron changes to a proton + electron

So (6p + 8n) => (7p + 7n) + e

-

- decaySlide16

16

ExampleKeep track of electrons!Carbon 14 has m = 14.003242 u 6 electronsNitrogen 14 has m = 14.003074 u normally 7 electronsBut in the decay, the nitrogen would have 6 electronsHowever the total on the r.h.s. of the equation has 7

So difference = 0.000168 u = 0.156 MeV = 156 keVSlide17

17

Conservation of energyEnergy of decay = 156 keV = problem!

?Slide18

18

A new particleProposed by Pauli (1930) - neutrinoTheory by FermiDiscovered 1956Zero charge, ~0 rest mass

Wolfgang Pauli

1900-1958

Enrico Fermi

1901-1954

antineutrino

“Zero rest mass” – speed of light

1998 – Super Kamiokande – some mass

Cosmic neutrino detectionSlide19

19

More on positronsMany isotopes have more neutrons than protonsDecay by emission of electronOther isotopes have more protons than neutronsDecay by emission of positron

Proton changes to a neutron + positron

+

decaySlide20

20

AnnihilationProton changes to a neutron + positron

+

decay

Positron annihilation

Application – positron emission tomographySlide21

21

Positron Emission TomographyPET – basis – use radio-labelled compounds, i.e. those containing a radionuclide.Positron emitters:

As an example, oxygen-15 can be used to look at oxygen metabolism and blood flow. Fluorine-18 is commonly used to examine cancerous tumours.Slide22

22

PET - methodAnnihilation produces two back-to-back 511 keV photons

Simultaneous detectionSlide23

23

Electron captureNucleus absorbs orbiting electron

Proton changes to neutron

Usually K electron

X-ray emission as outer electron jumps down to KSlide24

24

Gamma decay

Most penetrating

 = photon. High energy

*Excited nucleus  lower energy state

Energy levels far apart = keV or MeV

-

(13.4 MeV)

-

(9.0 MeV)

(4.4 MeV)Slide25

25

Homework . . . p.902,#6; p.908, Practice 25B; p.912,Section Review

p.928, 30-37;

p. 930, 56,60;

Read through lab for next time; answer

pre-lab questions