5.3 Nuclear physicsG485 Fields, Particles, Frontiers of

Physics

KS5 OCR PHYSICS H158/H558Mr Powell 2012

Index

5.3.1 The Nuclear Atom

5.3.2 Fundamental

Particles

5.3.3 Radioactivity

5.3.4 Fission and Fusion

Mr Powell 2014

5.3 Nuclear Physics

Assessable learning outcomes..(a) describe qualitatively the alpha-particle scattering experiment and the evidence this provides for the existence, charge and small size of the nucleus.(b) describe the basic atomic structure of the atom and the relative sizes of the atom and the nucleus; (c) select and use Coulomb’s law to determine the force of repulsion, and Newton’s law of gravitation to determine the force of attraction, between two protons at nuclear separations and hence the need for a short-range, attractive force between nucleons

(d) describe how the strong nuclear force between nucleons is attractive and very short-ranged; (e) estimate the density of nuclear matter; (f) define proton and nucleon number; (g) state and use the notation.... for the representation of nuclides;

(h) define and use the term isotopes; (i) use nuclear decay equations to represent simple nuclear reactions; (j) state the quantities conserved in a nuclear decay.

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Alpha Decay....

Alpha particles tracks

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(a) describe qualitatively the alpha-particle scattering experiment and the evidence this provides for the existence, charge and small size of the nucleus.

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Probing the Nucleus – Rutherford Scattering

Rutherford alpha particle scattering experiment

scattered alphaparticles

microscope to viewzinc sulphide screen,and count alphaparticles

vary angle ofscatteringobserved

radiumsource ofalphaparticles

thin goldfoil

alpha particlebeam

zinc sulphidescreen, tiny dots oflight where struckby alpha particle

lead block toselect narrowbeam of alphaparticles

nucleus

1

2

34

5

1

2

34

5

135

90

5030

20

By firing alpha particles at a heavy gold nucleus Rutherford could easily see that atoms were mostly space with a large positive nucleus in the centre.

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Scattering Formulae (extension)

Rutherford’s picture of alpha scattering

Assumptions: alpha particle is the He nucleus, charge +2e gold nucleus has charge + Ze, and is much more massive than alpha particles scattering force is inverse square electrical repulsion

scattering angle

‘aiming error’ bgold nucleuscharge + Ze

alpha particlescattered

equal force F butnucleus is massive,so little recoil

charge +2e

For calculations

d

force F = 2Ze2

40d2

There are complex formulae that we can use to work out the distance of closest approach. You will not have to learn these but should appreciate some of the maths involved. At the very least the idea of the forces involved

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Distance of Closest Approach (extension)Distance of closest approach

5 MeV

variation of potential1r

d

alpha particle stopswhere potential hill is5 MeV high

alpha particle with5 MeV initial kineticenergy

alpha particlescattered through 180

charge+ Ze(Z = 79 for gold)

Initial kinetic energy

= 5 MeV

= 5 106 eV 1.6 10–19 J eV–1

= 8.0 10–13 J

Alpha particle stops where

initial kinetic energy = electrical potential energy

8.0 10–13 J =+ 2 Ze2

40d

substitute values of Z, e, 0:

d = 4.5 10–14 m

Electrical potential energy

V = + 2 Ze2

40d

Z = 79, e = 1.6 10–19 C,

0 = 8.9 10–12 C2 N–1 m–2

Radius of gold nucleus must be less than of the order of 10–14 m

Atoms are 10000 times larger than their nuclei

Where does the alpha particle stop?

initial kinetic energy= electrical potential energy

distance r

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Distance of Closest Approach (extension)

Distance of closest approach

5 MeV

variation of potential1r

d

alpha particle stopswhere potential hill is5 MeV high

alpha particle with5 MeV initial kineticenergy

alpha particlescattered through 180

charge+ Ze(Z = 79 for gold)

Initial kinetic energy

= 5 MeV

= 5 106 eV 1.6 10–19 J eV–1

= 8.0 10–13 J

Alpha particle stops where

initial kinetic energy = electrical potential energy

8.0 10–13 J =+ 2 Ze2

40d

substitute values of Z, e, 0:

d = 4.5 10–14 m

Electrical potential energy

V = + 2 Ze2

40d

Z = 79, e = 1.6 10–19 C,

0 = 8.9 10–12 C2 N–1 m–2

Radius of gold nucleus must be less than of the order of 10–14 m

Atoms are 10000 times larger than their nuclei

Where does the alpha particle stop?

initial kinetic energy= electrical potential energy

distance r

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Summary Question...

For an alpha particle with an initial KE of 6MeV fired at a gold nucleus find the distance of closest approach of the alpha and the nucleus...

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Small Nucleus....

Nucleons....

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Estimate Nuclear Diameter with “Electron Diffraction”

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Quick Question

d

E

hc

22.1sin

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(b) describe the basic atomic structure of the atom and the relative sizes of the atom and the nucleus;

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(c) select and use Coulomb’s law to determine the force of repulsion, and Newton’s law of gravitation to determine the force of attraction, between two protons at nuclear separations and hence the need for a short-range, attractive force between nucleons

Mr Powell 2014

(d) describe how the strong nuclear force between nucleons is attractive and very short-ranged;

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Where does it come from?

The strong force actually acts between quarks which are found inside nucleons.

It's the strong force that causes nucleons to attract.

The carrier of this force is the gluon.

The force ensures that the protons and neutrons in the nucleus of the atom stay together without flying apart. The nucleus of the atom is formed in this way.

This force is so strong that it almost causes the protons and neutrons within the nucleus to bind to each other.

This is why the minute particles that possess this force are called "gluon" meaning "glue" in Latin.

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1 2 3 4 5

Equilibrium separation

Separation/ fm

attraction

repulsive

Forc

e (n

o un

its)

Interactions combined..

The overall graph is a combination of electrostatic repulsion (charged quarks) and the strong force (quarks) +

+

1 2 3 4 5

Typical equilibrium separation

Separation/ fm

Electric force dominates at larger separations

strong nuclear force dominates at smaller separations

attraction

repulsive

Forc

e (n

o un

its)

uu d

du d

200 400

Separation/ fm

Repulsive Force (N)

600

400

200

104

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Main properties of the strong nuclear force are:

1. At typical nucleon separation (1.3 x 10-15m) it is a very strong attractive force (104 N).

2. At much smaller separations between nucleons the force is very powerfully repulsive.

3. Beyond about 1.3 x 10-15m separation, the force quickly dies off to zero.

4. Thus, the strong nuclear force is a very short-range force.

5. The much smaller Coulomb force between protons has a much larger range and becomes the only significant force between protons when their separation exceeds about 2.5 x 10-15m.

6. The strong nuclear force is not connected with charge. Proton-proton, proton-neutron and neutron-neutron forces are the same. (The force between protons, however, must always be modified by the Coulomb repulsion between them.)

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Fundamental Forces

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Element A A1/3 R in fm

Carbon 12 2.29 2.75

Oxygen 16 2.52 3.02

Silicon 28 3.04 3.64

Calcium 40 3.42 4.10

Vanadium 50 3.68 4.42

Strontium 88 4.45 5.34

Indium 115 4.86 5.84

Atomic Radius.... MVarious types of scattering experiments suggest that nuclei are roughly spherical and appear to have essentially the same density.

The data are summarized in the expression called the Fermi model;

r0 = 1.2 x 10-15m = 1.2fm r = atomic radiusA = mass number (nucleons)

Can you plot a graph and show that this formulae is correct and the constant is 1.2fm

3

1

Arr o FACT

The diameter of the nucleus is in the range of 1.6fm (1.6 × 10−15 m) (for a proton in light hydrogen) to about 15 fm (for the heaviest atoms, such as uranium).

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Finding R0 P

2.00 2.50 3.00 3.50 4.00 4.50 5.00 5.500.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

f(x) = 1.19999999999997 x + 1.01252339845814E-13R² = 1

Nuclear Radius

cuberoot of A ( Nucleon Number)

R in

fm

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(e) estimate the density of nuclear matter;

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Summary & Question

1. Calculate the radius of an oxygen nucleus which has 16 nucleons..

r=r0A 1/3= 1.4 x 10-15 x (16)1/3

r=3.5 x 10-15m (3.5fm)

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Nuclear Density – is massive!

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(f) define proton and nucleon number & (g) state and use the notation.... for the representation of nuclides;

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(h) define and use the term isotopes;

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(i) use nuclear decay equations to represent simple nuclear reactions;

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(i) use nuclear decay equations to represent simple nuclear reactions;

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(j) state the quantities conserved in a nuclear decay.

5.3.1 The Nuclear Atom Covered in my lesson

Revised/ Made my own notes or reviewed at

home

Attempted Exam or Revision Questions

(a) describe qualitatively the alpha-particle scattering experiment and the evidence this provides for the existence, charge and small size of the nucleus

(b) describe the basic atomic structure of the atom and the relative sizes of the atom and the nucleus; (c) select and use Coulomb’s law to determine the force of repulsion, and Newton’s law of gravitation to determine the force of attraction, between two protons at nuclear separations and hence the need for a short-range, attractive force between nucleons

(d) describe how the strong nuclear force between nucleons is attractive and very short-ranged; (e) estimate the density of nuclear matter; (f) define proton and nucleon number;

(g) state and use the notation A,Z,X for the representation of nuclides;

(h) define and use the term isotopes; (i) use nuclear decay equations to represent simple nuclear reactions; (j) state the quantities conserved in a nuclear decay.

Next Steps for me? / (what do I need to ask for help on)

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Q2 June 07.... EThe table shows data for some nuclei. 1 eV = 1.6 ×10-19 J & speed of electromagnetic radiation = 3.0 × 108 ms-1

(a) (i) Show that these data support the rule that where R0 is a constant;

R = R0A(1/3)

(ii) The mass of a nucleon is about 1.7 × 10-27 kg. Calculate the density of nuclear matter. (6 marks)

(b) (i) Explain what is meant by the binding energy of a nucleus.

(ii) Show that the total binding energy of a sodium-23 nucleus is about 3 × 10-11 J.

(iii) Calculate the mass-equivalent of this binding energy. (5 marks)

Mr Powell 2014