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Nuclear and Particle Physics

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Nuclear Physics

Back 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 Rutherford1871-1937

me = 9.1 x 10-31 kg

mN = 1.6749 x 10-27

kg

mP = 1.6726 x 10-27

kgJames Chadwick 1891-1974

nucleons

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Nuclides and IsotopesTo specify a nuclide: XA

Z

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 = ZIsotopes – same atomic number, different mass number

e.g. carbon: C126 C13

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Many isotopes do not occur naturally, also elements > U

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Sizes

We saw with the Bohr model that radius of the atom depended on atomic number

Nucleus = protons + neutrons = mass number

The volume of a nucleus is proportional to the mass number

r (1.2 x10 15m) A3

)(mA)(1.2x103

4r

3

4V 33153 π π

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MassesMass spectrometer

1 atomic mass unit (u.) = 1.6606 x 10-27 kg = 931.5 MeVFixed so that carbon = 12.00000 u

mN = 1.6749 x 10-27 kg = 1.0087 u

mP = 1.6726 x 10-27 kg = 1.0078 u

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Binding Energy

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

He42

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 kg

Energy = 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 MeV

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Atomic Mass Units1 u = 931.5 MeV

mN = 1.6749 x 10-27 kg = 1.0087 u

mP = 1.6726 x 10-27 kg = 1.0078 u

Mass of helium nucleus = 4.0026 u

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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 u

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Average Binding Energy

• Graph

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

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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 repulsion

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Radioactivity

Becquerel, 1896Emission of radiation without external stimulus

Curies – polonium (Po) and radium (Ra) Henri Becquerel1852-1908

Marie Curie1867 - 1934

Pierre Curie1859 - 1906

1903 (Physics)1911

(Chem)

Radioactivity unaffected by heating, cooling, etc.

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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

”

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Alpha Decay

Least penetrating – nucleus of He42

Radium 226 is an alpha emitter:

HeRnRa 42

22286

22688

Parent Daughter

transmutation

Mass of parent > mass of daughter + mass of alpha

Difference = kinetic energy

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Example

232.03714 u 228.02873 u + 4.002603 utotal = 232.03133 u

Lost mass = 232.03714 – 232.03133 = 0.00581 u0.00581u x 931.5 MeV/u = 5.4 MeV

(some recoil)

HeThU 42

22890

23292

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Beta decay One electron

eNC 01

147 14

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What is lost is NOT an orbital electron

Instead a neutron changes to a proton + electron

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

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Example

Keep 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 keV

eNC 01

147 14

6

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Conservation of energy

• Energy of decay = 156 keV = problem!

?

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A new particle

• Proposed by Pauli (1930) - neutrino• Theory by Fermi• Discovered 1956• Zero charge, ~0 rest mass

Wolfgang Pauli

1900-1958

Enrico Fermi

1901-1954

eNC 01

147

146

antineutrino

“Zero rest mass” – speed of light

1998 – Super Kamiokande – some mass

Cosmic neutrino detection

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More on positrons

Many isotopes have more neutrons than protons® Decay by emission of electron

Other isotopes have more protons than neutrons® Decay by emission of positron Proton changes to a neutron + positron

eFNe 01

199

1910

+ decay

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Annihilation

Proton changes to a neutron + positron

eFNe 01

199

1910

+ decay

Positron annihilation

Application – positron emission tomography

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Positron Emission Tomography

PET – basis – use radio-labelled compounds, i.e. those containing a radionuclide.

Positron emitters:

I,F,C,N,O 12453

189

116

137

158

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.

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PET - method

Annihilation produces two back-to-back 511 keV photons

Simultaneous detection

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Electron capture

• Nucleus absorbs orbiting electron

LieBe 73

01

74

Proton changes to neutron

Usually K electron

X-ray emission as outer electron jumps down to K

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Gamma decay

Most penetrating = photon. High energy*Excited nucleus lower energy state

eCB 01

126 12

5

e*CB 01

126 12

5

Energy levels far apart = keV or MeV

- (13.4 MeV)

C126

- (9.0 MeV)

(4.4 MeV)

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Homework . . .

1.p.902,#6; 2.p.908, Practice 25B; 3.p.912,Section Review4.p.928, 30-37;5.p. 930, 56,60; 6.Read through lab for next

time; answer pre-lab questions