Name……………….
Class……….
Plymstock School Physics Department
Module G485.3 Nuclear Physics student booklet
Lesson 33 notes - The Nuclear AtomObjectives(a) describe qualitatively the alpha-particle scattering experiment and the evidence this provides for the existence, charge and small size of the nucleus (HSW 1, 4c);(b) describe the basic atomic structure of the atom and the relative sizes of the atom and the nucleus;
OutcomesBe able to describe what makes up an atom and recall the relative sizes of the atom and the nucleus.Be able to describe qualitatively the alpha-particle scattering experiment.Be able to describe the observations made from this experiment.Be able to explain how these observations are evidence for the existence, charge and small.
Rutherford ScatteringRutherford alpha particle scattering experiment
scattered alphaparticles
microscope to view zincsulphide screen, and countalpha particles
vary angle ofscatteringobserved
radium source ofalpha particles
thin goldfoil
alpha particlebeam
zinc sulphide screen,tiny dots of light wherestruck by alpha particle
lead block to selectnarrow beam of alphaparticles
The experiment takes place in a vacuum to avoid problems of absorption by air.Alpha particles are shot at the Gold Leaf. Some were deflected, some rebounded (1 in 10000). This led to the explanation that the gold leaf was
made up of very small particles that had the majority of there charge and mass concentrated in a very small space at the centre of the particle: The nucleus.
The AtomWe now think of atoms in terms of a positively charged nucleus with negative electrons orbiting it. In the Bohr atom these electron shells (energy levels) are just the most likely place that an electron will be found and explained using Quantum Theory.
Atomic Sizes•Electron size ~ 10-18 m•Nucleus sizes ~ 1 fm•Atomic sizes ~ 1Å•Hydrogen molecule length ~ 1.28Å•Water Molecule size ~ 2.8Å
Lesson 33 questions – The Atom
1. Describe briefly the two conflicting theories of the structure of the atom.
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2. Why was the nuclear model of Rutherford accepted as correct?
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3. What would have happened if neutrons had been used in Rutherford’s experiment? Explain your answer.
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4. What would have happened if aluminium had been used instead of gold in the alpha scattering experiment? Explain your answer.
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5. What three properties of the nucleus can be deduced from the Rutherford scattering experiment? Explain your answer.
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6 Describe briefly one scattering experiment to investigate the size of the nucleus of the atom.Include a description of the properties of the incident radiation which makes it suitable for this experiment.
In your answer, you should make clear how evidence for the size of the nucleus follows from your description.
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Lesson 34 – The Strong Nuclear ForceObjectives(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 (HSW 1, 2, 4);(d) describe how the strong nuclear force between nucleons is attractive and very short-ranged;
OutcomesBe able to describe why there is a need for the short range attractive force between nucleons.Be able to use Coulomb’s Law and Newton’s Law of gravitation to be able to calculate forces within nuclei to explain why there is a need for the short range attractive force between nucleons.Be able to describe the properties of th strong nuclear force.Be able to interpret graphical representations of the strong nuclear force.
Gravitation and Electrostatic forcesImagine 2 particles with a charge +e (e.g. 2 protons).Gravity will pull them together (gravity is an attractive force).Since they have the same charge Coulomb’s electrostatic force will push them apart (a repulsive force).
Calculate the Gravitational force and Electrostatic force between them:•Data•p = 1.602x10-19C•ε0 = 8.85x10-12Fm-1 •mp = 1.673x10-27kg•G = 6.673x10-11Nm2•r = 1fm
You should find that the ratio between the forces Fe:Fg is 1.2x1036
In other words it’s not gravity that’s holding nuclei together; its contribution is negligible – there must be another force
– A NUCLEAR FORCE.
The Strong Nuclear ForceCharacteristics:I. It is an attractive force between nucleons - We have discovered this from our calculations of the Coulomb force.II. It is repulsive at very short range - If it were not then all nucleons without a charge would be able to pull each other into an infinitesimally small point. – This does not happen.III. It does not extend beyond distances of a few femtometres - Evidence from the Rutherford Scattering experiment shows that alpha particles describe paths explained by Coulomb repulsion until thy get very close to the nucleus whereby explanations need an extra force – the Strong Force – to describe their paths.IV. It does not depend on the charge of the nucleons - both neutrons and protons feel the strong force – the evidence for this again comes from scattering experiments.V. It is readily saturated by surrounding nucleons - The strong force is stronger than the electrostatic force, but acts over a shorter distance. Adding more nucleons is favoured with small nuclei but not with large. e.g. adding a proton to a small vs. large nucleus.
Force vs SeperarationThe graph shows how the Force between 2 nucleons varies with the separation between the nuclei.
Lesson 34 questions – Nuclear Forces1. Fig. 1 shows two protons A and B in contact and at equilibrium inside a
nucleus.
A B
Fig. 1
Proton A exerts three forces on proton B. These are an electrostatic force FE, a gravitational force FG and a strong force FS.
(a) On Fig. 1, mark and label the three forces acting on proton B. Assume that every force acts at the centre of the proton.
[2]
(b) Write an equation relating FE, FG and FS.
[1]
(c) The radius of a proton is 1.40 × 10–15 m.
Calculate the values of
(i) FE
FE = ..................................... N
[2]
(ii) FG
FG = ..................................... N
[2]
(iii) FS.
FS = ..................................... N
[1]
(d) Comment on the relative magnitudes of FE and FG.
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[1]
(e) Fig. 2 shows two neutrons in contact and at equilibrium inside a nucleus.
Fig. 2
Without further calculation, state the values of FE, FG and FS for these neutrons.
(i) FE = .................................................................................. N
[1]
(ii) FG = .................................................................................. N
[1]
(iii) FS = ................................................................................. N
[1]
[Total 12 marks]
2. This question is about the strong and electrostatic forces inside a nucleus.
The figure below shows how the strong force (strong interaction) and the electrostatic force between two protons vary with distance between the centres of the protons.
force
00
strongforce
electrostaticforce
distancebetween centres
(a) Label on the figure the regions of the force axis which represent attraction and repulsion respectively.
[1]
(b) (i) On the figure above, mark a point which represents the distance between the centres of two adjacent neutrons in a nucleus. Label this point N.
Explain why you chose point N.
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(ii) On the figure, mark a point P which represents the distance between two adjacent protons in a nucleus.
Explain why you chose point P.
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(c) On the figure, sketch a line to show how the resultant force between two protons varies with the distance between their centres. Pay particular attention to the points at which this line crosses any other line.
[3]
(d) (i) Write an expression for the electrostatic force between two point charges Q which are situated at a distance x apart.
[1]
(ii) The electrostatic force between two protons in contact in a nucleus is 25 N.Calculate the distance between the centres of the two protons.
distance = ...................................... m[2]
[Total 11 marks]
Lesson 35 notes – Nuclear PropertiesObjectives(e) estimate the density of nuclear matter;(f) define proton and nucleon number;(g) state and use the notation A
ZX for the representation of nuclides;(h) define and use the term isotopes;
OutcomesBe able to define proton and nucleon number.Be able to define and use the term isotope.Be able to state and use the notation A
ZX for the representation of nuclides.Be able to estimate the density of nuclear matter.
Nuclide notation
A is also the total number of nucleons – The Nucleon Number.
IsotopesWe can use the isotopes of carbon as an example of this notation. All elements with the same number of protons have near identical chemical properties. An Isotope has the same number of protons but different numbers of neutrons. i.e. The same proton number (Z) but different nucleon number (A). Isotopes generally have different nuclear stability, (the ability to stick around)
In the example of Carbon, the element is determined by the atomic number 6. Carbon-12 is the common isotope, with carbon-13 as another stable isotope which makes up about 1%. Carbon 14 is radioactive and the basis for carbon dating.
The Nuclear RadiusYou may expect that (and it can be proven that) the volume V of the nucleus is proportional to of the number of nucleons A. V is also proportional to the cube root of the radius r, so it follows that A is directly proportional to r3. Or
r = r0A1/3
where r0 is the radius of one nucleon.
Nuclear DensityIt can be show that ρ=3m/(4r0
3) where ρ is the nuclear density (the density of any nuclide) and r0 is the radius of a nucleon.
ρ=1017kgm-3
This is the same for more or less all atoms with a few exceptions. Which is very dense. These densities are similar to those of neutron stars.
Lesson 35 questions – Nuclear Properties
A periodic table may be needed
1 Write down the nuclear notation ( ) for:
(a) an alpha particle
(1)
(b) a proton
(1)
(c) a hydrogen nucleus
(1)
(d) a neutron
(1)
(e) a beta particle
(1)
(f) a positron.
(1)
2 Write down the nuclear notation ( ) for
(a) carbon 13
(1)
(b) nitrogen 14
(1)
(c) neon 22
(1)
(d) tin 118
(1)
(e) iron 54
(1)
3 Explain what is meant by the statement that the strong interaction is a short-range force and explain what this implies about the densities of nuclei of various sizes.
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Lesson 36 – Nuclear ReactionsObjectives(i) use nuclear decay equations to represent simple nuclear reactions;(j) state the quantities conserved in a nuclear decay.
OutcomesBe able to describe that heaver elements are more radioactive.Be able to use nuclear decay equations to represent simple nuclear reactions.Be able to state the quantities conserved in a nuclear decay.Be able to explain why heaver elements are more radioactive.
N-Z plot (Segrè plot).
The graph shows Neutron Number against Proton number and you can see that in heavier elements the number of neutrons is bigger than the number of protons. This can be explained with an understanding of the forces inside a nucleus. But as these numbers become more and more unbalanced the nucleus becomes more and more unstable until they break down radioactively.
Proton number
Neutron number
+
-A
Bp n + + +
n p + - +
/N
/Z
The grids below show some of the emissions given by unstable nuclei:
Grid showing change in A and Z with different emissionsWorked examples:
Nucleon number (A)
Proton number (Z)
A, Z+1A, Z-1
A-4, Z-2
+ -
A, Z
Proton number (Z)
Neutron number (N)
N-1, Z+1
+
-
N+1, Z-1
N, Z
N-2, Z-2
Equations for alpha, beta and gamma decayNuclear decay processes can be represented by nuclear equations. The word equation implies that the two sides of the equation must ‘balance’ in some way.
Examples of equations for the sources used in school and college labs.
sources are americium-241,
- sources are strontium-90,
(Extension
The underlying process is:
n –> p + e- +
Here, is an antineutrino.
You can translate n –> p + e- into the AZ notation:
)
sources are cobalt-60 . The radiation comes from the radioactive daughter of the decay of the . The is formed in an ‘excited state’ and so almost immediately loses the energy by emitting a ray. They are only emitted after an or decay, and all such rays have a well-defined energy. (So a cobalt-60 source, which is a pure gamma emitter, must be designed so that betas are not emitted. How? – (by encasing in metal which is thick enough to absorb the betas but which still allows gammas to escape.)
Decay processes
Radioactive decay processes decay
2 fewer protons2 fewer neutronsproton number Z
ZN
Z–2N–2
– decay
1 more proton1 less neutronproton number Z
+ decay
1 less proton1 more neutronproton number Z
decay
same protonsand neutronsproton number Z
ZN
–
+
Z+1N–1
ZN
Z–1N+1
ZN
Summary of Nuclear DecayFor nuclear decay to occur the mass of the daughter nuclei must be less than the parent. In radioactive decay, A, Z and N are all conserved as is the number of particles. We get neutrinos and antineutrinos in some decay because of this. The mass that is lost is given as energy by the equation E=mc2. (more on this later!)
Lesson 36 background notes – history of radioactivityWilhelm Conrad Roentgen (1845-1923)On November 8th 1895 Roentgen discovered X-Rays. A selfless man and a very great investigative scientist, he had noticed the X-rays penetrating an opaque paper around a cathode ray tube to make a fluorescent screen glow.
Antoine Henri Becquerel (1852-1908)Henri Becquerel came from a scientific background and once he heard of Roentgen’s discovery began his own on fluorescent materials. At first he thought that the Uranium crystals absorbed the Sun’s light and reradiated X-rays out, but he soon found that this was not the case. Uranium was spontaneously emitting another type of radiation.
Becquerel experimented with the radiation and found that although similar to X-rays, the radiation could be deflected by magnets and so must contain negatively charged particles.
Pierre Curie (1859-1906)Marie Curie (1867-1934)After they married in 1895, Pierre and Marie Curie researched the phenomenon of radioactivity together (although Henri Becquerel discovered it, it was Marie who coined the phrase “Radioactivity”).
They researched a uranium ore called pitchblende and found that after extraction of pure uranium it was less
active then when in the ore. This led her to the discoveries of polonium and radium.
Pierre died in a road accident in 1906 and Marie was awarded his teaching post. She died in 1934 from pernicious anaemia; probably caused by overexposure to nuclear radiation.
Ernest Rutherford (1871-1937)The Grandfather of Nuclear physics made some amazing discoveries about the changes in radioactive particles that occur.
He also is the man that pretty much described the atom as we think of it today. He was experimenting by firing alpha particles at a gold leaf when he noticed that 1 in 8000 of the particles bounced off it instead of going through. He surmised that this was due to most of the mass of an atom being concentrated in one place with nothing much around it apart from electrons at a great (atomically speaking) distance.
The amazed Rutherford commented that it was "as if you fired a 15-inch naval shell at a piece of tissue paper and the shell came right back and hit you."
Lesson 37 notes – QuarksObjectives(a) explain that since protons and neutrons contain charged constituents called quarks they are, therefore, not fundamental particles;(b) describe a simple quark model of hadrons in terms of up, down and strange quarks and their respective antiquarks, taking into account their charge, baryon number and strangeness;(c) describe how the quark model may be extended to include the properties of charm, topness and bottomness;(d) describe the properties of neutrons and protons in terms of a simple quark model;
OutcomesBe able to explain that since protons and neutrons contain charged constituents called quarks they are, therefore, not fundamental particles.Be able to describe a simple quark model of hadrons in terms of up, down and strange quarks and their respective antiquarks, taking into account their charge, baryon number and strangeness.Be able to describe the properties of neutrons and protons in terms of a simple quark model.Be able to describe how the quark model may be extended to include the properties of charm, topness and bottomness.
What’s stuff made of?Just over 100 years ago we’d find out about radioactive particles and had
started discovering what stuff was actually made of. The Greeks had come up with the idea of an indivisibly small atom that makes up everything. This was not quite right when we found out that inside atoms were nuclei, and inside these were nucleons (protons and neutrons) and around these were electrons. Theoretical physicists came up with the idea of Quarks in order to explain what electrons and other cosmic particles were made from. With the use of particle accelerators these Quarks were proven to exist and many exotic fundamental particles left their trails.
Fundamental ParticlesThe diagram shows the different classes of fundamental particles. There are 3 types of fundamental particle: Bosons, Leptons and Quarks. On the diagram; Mesons and below are made up of Quarks.
FlavoursThere are 6 types or “flavours” of Quarks each with different properties as shown in the table:
Quark (Flavour) Symbol Spin Charge BaryonNumber S C B T
Up U 1/2 +2/3 1/3 0 0 0 0
Down D 1/2 -1/3 1/3 0 0 0 0Charm C 1/2 +2/3 1/3 0 +1 0 0
Strange S 1/2 -1/3 1/3 -1 0 0 0Top T 1/2 +2/3 1/3 0 0 0 +1
Bottom B 1/2 -1/3 1/3 0 0 -1 0S – Strangeness, C – Charm, B – Bottomness, T – TopnessOn top of this each individual Quark of the same flavour can have a different colour that represent forces.There are also corresponding Anti-Particle pairs for each of these Quarks. Ant-Particles will be discussed in the next lesson.
Baryons, Mesons and LeptonsThe Quark Model says that different combinations of different flavoured quarks make up all other particles.
So a Proton is made up of 2 ups and a down since that makes a charge of +1.
And a neutrons is made of 2 downs and an up since that makes a charge of zero.
Bosons
Neutrons and Protons are both heavy Baryons since they are made of 3 quarks.
Mesons are made up of a quarks and an antiquarks and are the medium sized particles.
Both mesons and baryons interact via the strong nuclear force.
Leptons are a class of their own and another fundamental particle. They are some of the smallest particles like electrons. They have a charge of +/-1 and vary in mass and “size”.
Extension: The Standard Model
Leptons and Quarks are fundamental particles associated with matter and are made up of Quarks and Leptons. Bosons are also fundamental particles that are mostly associated with forces.
Timeline of fundamental particles discoveries: http://en.wikipedia.org/wiki/Timeline_of_particle_discoveries
Lesson 37 questions – QuarksName………………… ( /23)…….%…….
MOST
1. (i) State the quark composition of the neutron.
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[1]
(ii) Complete the table to show the charge Q, baryon number B and strangeness S for the quarks in the neutron.
quark Q B S
[2]
(iii) Hence deduce the values of Q, B and S for the neutron.
Q …………… B …………… S ……………[1]
[Total 4 marks]
2. (i) Name the group of particles of which the electron and the positron are members.
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[1]
(ii) Name another member of this group.
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[1]
[Total 2 marks]
3. Describe briefly the quark model of hadrons.
• Illustrate your answer by referring to the composition of one hadron.
• Include in your answer the names of all the known quarks.
• Give as much information as you can about one particular quark.
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[Total 5 marks]
4. This question is about the properties of baryons.
Choose two examples of baryons
For each example discuss
• their composition
• their stability.
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[Total 6 marks]
5. This question is about the properties of leptons.
Choose two examples of leptons
For each example discuss
• their composition
• the forces which affect them
• where they may be found.
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[Total 6 marks]
Lesson 38 notes – Beta DecayObjectives(e) describe how there is a weak interaction between quarks and that this is responsible for β decay; (f) state that there are two types of β decay;(g) describe the two types of β decay in terms of a simple quark model;(h) state that (electron) neutrinos and (electron) antineutrinos are produced during β+ and β- decays, respectively;(i) state that a β- particle is an electron and a β+ particle is a positron;(j) state that electrons and neutrinos are members of a group of particles known as leptons.
OutcomesBe able to state that there are two types of β decay.Be able to state that a β- particle is an electron and a β+ particle is a positron.Be able to describe the two types of β decay in terms of a simple quark model.Be able to state that (electron) neutrinos and (electron) antineutrinos are produced during β+ and β- decays, respectively.Be able to state that electrons and neutrinos are members of a group of particles known as leptons.Be able to describe how there is a weak interaction between quarks and that this is responsible for β decay.Be able to link ideas about Momentum into forces between particles.
RadioactivityAlpha, Beta and Gamma are the 3 most common forms of Nuclear Radiation. We will study this in more detail later on but we are concentrating on Beta decay in this lesson.
Beta DecayBeta particles are electrons (or positrons) emitted from a nucleus. Beta Radiation comes with an associated anti-neutrino that shares the momentum and energy of the decay. This is shown in the before and after diagrams:
Beta decay is the decay of one of the neutrons to a proton via the weak interaction:Example:
Decay equation:
Quark Equation udd → uud + e- + ῡ
This can be simplified tod → u + e- + ῡ
Explanation:
The flow diagram above shows how a neutron changes into a proton and in doing so emits an electron and an anti-neutrino. Beta decay occurs where there is an unbalanced amount of protons and neutrons. (a free neutron is unstable and will decay in about 15 minutes but in a nucleus with other protons the binding energy (discussed in a later lesson) keeps the nuclei stable).
1 A neutron (2 down quarks and an up quark).2 One of the down quarks changes flavour to an up quark. The down
quark has a charge of -1/3 and the up quark has a charge of 2/3 so
a virtual W- boson (a particle associated with The Weak Force) is needed, which carries away a (-1) charge (conserving charge).
3 The new up quark bounces back because of conservation of momentum and we now have a proton (2 ups and a down).
4 The W- splits into an electron and an antineutrino (both leptons).5 All three new particles: The proton, the electron and the antineutrino
move away from each other.
Positron Emission β+Positron emmision happens when an up quark changes into a down quark. Isotopes which increase in mass under the conversion of a proton to a neutron, or which decrease by less than me, do not spontaneously decay by positron emission. Example:
The energy emitted depends on the isotope that is decaying; the figure of 0.96 MeV applies only to the decay of carbon-11. Isotopes that spontaneously decay are used in PET (positive emission topography) scans – used in medical diagnosis.
The Weak ForceAlthough there are 6 quarks, all matter in the universe seems to be made from just the Up and Down Quarks, the least massive charged lepton (the electron) and neutrinos. The weak force (or weak interaction) acts between quarks over a very small distance and is responsible for the decay of massive quarks and leptons into lighter quarks and leptons. When fundamental particles decay, they disappear, being replaced by two or more different particles. The total of mass and energy is conserved, although some of the original particle's mass is converted into kinetic energy, and the resulting particles always have less mass than the original particle that decayed.
When a quark changes flavour because of the weak interaction or a lepton changes type (a muon changing to an electron, for instance), the carrier particles of the weak interactions are the W+, W-, and the Z particles. The W's are electrically charged and the Z is neutral.
Lesson 38 questions – Beta DecayMOST
1. (i) Name the group of particles of which the electron and the positron are members.
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[1]
(ii) Name another member of this group.
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[1]
[Total 2 marks]
2. Tritium-3 ( H31 ) decays to helium-3 ( He3
2 ) with the emission of a – particle.
(i) Name the force responsible for this decay process.
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[1]
(ii) Write a nuclear equation to represent this process.
[1]
(iii) Write a quark equation, in its simplest form, to represent this process.
[2]
[Total 4 marks]
3. (a) The table of Fig. 1 shows four particles and three classes of particle.
hadron baryon lepton
neutron
proton
electron
neutrino
Fig. 1
Indicate using ticks, the class or classes to which each particle belongs.
[2]
(b) The neutron can decay, producing particles which include a proton and an electron.
(i) State the approximate half-life of this process.
..................................................................................................[1]
(ii) Name the force which is responsible for it.
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(iii) Write a quark equation for this reaction.
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(iv) Write number equations which show that charge and baryon number are conserved in this quark reaction.
charge .........................................................................................
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baryon number .....................................................................................
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(c) Fig. 2 illustrates the paths of the neutron, proton and electron only in a decay process of the kind described in (b).
neutronproton
electron
Fig. 2
Fig. 3 represents the momenta of the neutron, pn, the proton, pp and the electron, pe on a vector diagram.
p
p
p
p
e
n
Fig. 3
ALL
(i) Draw and label a line on Fig. 3 which represents the resultant pr of vectors pp and pe.
[1]
SOME
(ii) According to the law of momentum, the total momentum of an isolated system remains constant.Explain in as much detail as you can, why the momentum pr is not the same as pn.
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[Total 12 marks]
Lesson 39 notes – Radioactive propertiesObjectives(a) describe the spontaneous and random nature of radioactive decay of unstable nuclei;(b) describe the nature, penetration and range of α-particles, β-particles and γ-rays;
OutcomesBe able to describe the spontaneous and random nature of radioactive decay of unstable nuclei.Be able to describe the nature, penetration and range of α-particles, β-particles and γ-rays.
Nuclear RadiationUnstable nuclei break down and release different types of Nuclear Radiation. The 3 most common are:
Alpha particles (α)
The emission of an alpha particle (a helium nucleus), only occurs in very heavy elements such as uranium, thorium and radium. The reason alpha decay occurs is because the nucleus of the atom is unstable. With too many protons that repel each other. In an attempt to reduce the instability, an alpha
particle is emitted. The alpha particles are in constant collision with an energy barrier in the nucleus and because of their energy and mass, there exists a nonzero probability of transmission. That is, an alpha particle (helium nucleus) will tunnel out of the nucleus of the heavy element.
Beta Particles (β)Beta decay occurs when the neutron to proton ratio is too great in the nucleus and causes instability. In basic beta decay, a neutron is turned into a proton and an electron. The electron is then emitted.
There is also positron emission when the neutron to proton ratio is too small. A proton turns into a neutron and a positron and the positron is emitted. A positron is basically a positively charged electron.
Gamma Rays (γ)After a decay reaction, the nucleus is often in an “excited” state. This means that the decay has produced a nucleus that still has excess energy to get rid of. Rather than emitting another beta or alpha particle, this energy is lost by emitting a pulse of electromagnetic radiation called a gamma ray. The gamma ray is identical in nature to light or microwaves, but of very high energy. Like all forms of electromagnetic radiation, the gamma ray has no mass and no charge.
Gamma rays interact with material by colliding with the electrons in the shells of atoms. They lose their energy slowly in material, being able to travel significant distances before stopping. Depending on their initial energy, gamma rays can travel from 1 to hundreds of meters in air and can easily go right through people.
It is important to note that most alpha and beta emitters also emit gamma rays as part of their decay process. There is no such thing as a “pure” gamma emitter.
leadaluminiumpaperStopped by:
Several kms3m50mmRange (in air)
highmediumlowPenetrating power
0mediumhighIonising ability
very fast (speed of light)Fast**Slow*Speed
0-1+2Charge
01/20004Mass (atomic mass units)
Symbol
Gamma rayBeta particleAlpha particleType of Radiation
Nuclear Properties
*An alpha particle emitted by a uranium nucleus has an initial speed of about 15 million meters/second (about 0.05 light speed.**Beta particles travel with an initial speed of about 180 million m/s, or about 0.6 light-speed
Deflection in a magnetic/electric field.When Alpha, Beta or Gamma travel through either field they will feel a force as described in the diagrams:
Lesson 39 questions – Radioactive PropertiesName………………………………. ( /17)…..%………Class……………ALL
1. (a) Complete the table below for the three types of ionising radiation.
radiation nature range in air penetration ability
α 0.2 mm of paper
β electron
γ several km
[3]
(b) Describe briefly, with the aid of a sketch, an absorption experiment to distinguish between the three radiations listed above.
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[3]
[Total 6 marks]
2. State three ways in which decay by emission of an -particle differs from decay by emission of a -particle.
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.....................................................................................................................[Total 3 marks]
3. In this question, two marks are available for the quality of written
communication.
State and compare the nature and properties of the three types of ionising radiations emitted by naturally occurring radioactive substances.
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Quality of Written Communication [2]
[Total 8 marks]
Lesson 40 notes – Radioactive Decay and Half LifeObjectives(c) define and use the quantities activity and decay constant;(d) select and apply the equation for activity A = λN;(e) select and apply the equations A = A0e-λt and N = N0e-λt where A is the activity and N is the number of undecayed nuclei;(f) define and apply the term half-life;(g) select and use the equation λt1/2 = 0.693;(h) compare and contrast decay of radioactive nuclei and decay of charge on a capacitor in a C–R circuit (HSW 5b); See also lesson 25(i) describe the use of radioactive isotopes in smoke alarms (HSW 6a);(j) describe the technique of radioactive dating (ie carbon-dating).
OutcomesBe able to describe the use of radioactive isotopes in smoke alarms (HSW 6a).Be able to describe the technique of radioactive dating (ie carbon-dating).Be able to define and apply the term half-life.Be able to define and use the quantities activity and decay constant.Be able to select and apply the equation for activity A = λN.Be able to select and apply the equations A = A0e-λt and N = N0e-λt where A is the activity and N is the number of undecayed nuclei.Be able to select and use the equation λt1/2 = 0.693.Be able to compare and contrast decay of radioactive nuclei and decay of charge on a capacitor in a C–R circuit (HSW 5b). See also lesson 25Be able to derive and apply the equations N = N0e-λt where N is the number of undecayed nuclei.
Definitionsλ is the decay constant (s-1 ) is the fraction of substance that decays per second, so it always is a number less than 1.
Activity, A is the number of radioactive emissions from a mass of substance / unit time.
The unit is the Becquerel with 1 Bq = 1 emission /second
The Activity depends on two things: The mass of the substance and how active the isotope is.
A= λN
A is the Activity (Bq)N is the number of undecayed atoms of an isotope.λ is the decay constant (s-1 ) Different isotopes have different decay constants.
The received count rate, C is the number of emissions detected per unit time with units of s-1 or Hz.
So, C A
C = kA (with k being related the sensitivity of the detector)
The rate of reduction of N, -dN/dt = λN
The solution to this is
N=N0e-λt
Where N0 is the original is the original number of undecayed atoms at t=0
since A= λN and C = kA
A=A0e-λt
C=C0e-λt
We cannot say which atom will decay but we can describe very accurately how much of a substance will decay knowing the decay constant. The half-life of a substance can then be found. This the time it takes for the number of undecayed atoms in a substance to halve.
So we can write
From: N=N0e-λt1/2
N0 /2 =N0 e-λt1/2
½ = e-λt1/2
2 = eλt1/2
ln 2 = λt1/2
The half life for an isotope with decay constant λ is:
0.693 / λ = t1/2
Finding the decay constant graphicallyN = N0e-λt
lnN=ln(N0e-λt)lnN=lnN0 + ln(e-λt)
lnN=lnN0 -λtlnN = -λt + lnN0
In the form of y = mx + Cλ = m, the gradientlnN0 = C, the intercept and N0 can be found by eC
Extension Derivation of A = A0e-λt
Where A is the radioactivity in nuclei per units of time (SI unit: Bq) after time t has elapsed and A0 is the initial radioactivity.
Starting with N = N0e-λt
A = dN/dt = -λN0e-λt
If t = 0 A0 = -λN0
Substitute: A = -λN0e-λt
A = A0e-λt
Lesson 40 questions – Radioactive Decay and Half Life
Name…………………… ( /45)……..%………Class…………….
ALL
1. The activity A of a sample of a radioactive nuclide is given by the equation
A = N
Define each of the terms in the equation.
A ............................................................................................................
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N ............................................................................................................
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2. The radioactive nickel nuclide Ni6328 decays by beta-particle emission with
a half-life of 120 years.
(a) A copper nucleus is produced as the result of this decay. State the number of nucleons in the copper nucleus which are
protons ..................................................................................................
neutrons ..................................................................................................
[2]
(b) Show that the decay constant of the nickel nuclide is 1.8 × 10–10s–1.
1 year = 3.2 × 107 s
[1]
(c) A student designs an electronic clock, powered by the decay of nuclei of Ni63
28 . One plate of a capacitor of capacitance 1.2 × 10–12 F is to be coated with this isotope. As a result of this decay, the capacitor becomes charged. The capacitor is connected across the terminals of a small neon lamp. See Fig. 1. When the capacitor is charged to 90 V, the neon gas inside the lamp becomes conducting, causing it to emit a brief flash of light and discharging the capacitor. The charging starts again. Fig. 2 is a graph showing how the voltage V across the capacitor varies with time.
100
50
00 1.0 2.0
tim e / s
V / V
3.0
V1.2 × 10 F–12
neon lam p
Fig. 1 Fig. 2
(i) Show that the maximum charge stored on the capacitor is 1.1 × 10–10 C.
[2]
(ii) When a nickel atom emits a beta-particle, a positive charge of 1.6 × 10–19 C is added to the capacitor plate. Show that the number of nickel nuclei that must decay to produce 1.1 × 10–
10 C is about 7 × 108.
[2]
(iii) The neon lamp is to flash once every 1.0 s. Using your answer to (b), calculate the number of nickel atoms needed in the coating on the plate.
number = .......................[3]
(iv) State, giving a reason, whether or not you would expect the clock to be accurate to within 1% one year after manufacture.
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[Total 11 marks]
3. Uranium-238 U23892 decays to lead-206 Pb206
82 by means of a series of decays.
One nucleus of U23892 decays eventually to one nucleus of Pb206
82 .
This means that, over time, the ratio of lead-206 atoms to uranium-238 atoms increases. This ratio may be used to determine the age of a sample of rock.
In a particular sample of rock, the ratio
21
atoms 238-uranium of numberatoms 206-lead of number
.
(a) Show that the ratio
32
initially atoms 238-uranium of numberleft atoms 238-uranium of number
.
Assume that the sample initially contained only uranium-238 atoms and subsequently it contained only uranium-238 atoms and lead-206 atoms.
[2]
(b) Calculate the age of the rock sample.
The half-life of U23892 is 4.47 × 109 years.
age = ................................................ years[3]
(c) The rock sample initially contained 5.00 g of uranium-238. Calculate the initial number N0 of atoms of uranium-238 in this sample.
number = .........................................................[2]
(d) On the figure below, sketch graphs to show how the number of atoms of uranium-238 and the number of atoms of lead-206 vary with time over a period of several half-lives.
Label your graphs ‘U’ and ‘Pb’ respectively.
tim e
num berof atom s
00
N 0
[3]
[Total 10 marks]
4. The activity of the potassium source is proportional to the count rate minus the background count rate, that is
activity = constant × (count rate – background count rate).
(i) The radioactive decay law in terms of the count rate C corrected for background can be written in the form
C = Coe–t
where is the decay constant.Show how the law can be written in the linear form
ln C = –t + lnCo
[2]
(ii) Fig. 2 shows the graph of ln C against time t for the beta-decay of potassium.
4.0
4.2
4.4
4.6
ln C
0 2 4 6 8 1 0t / h
Fig. 2
Use data from the graph to estimate the half-life of the potassium nuclide.
half-life = ………………….h[3]
[Total 5 marks]
5. A radioactive material is known to contain a mixture of two nuclides X and Y of different half-lives. Readings of activity, taken as the material decays, are given in the table, together with the activity of nuclide X over the first 12 hours.
time / hour activity ofmaterial / Bq
activity ofnuclide X /Bq
activity ofnuclide Y /Bq
0
6
12
18
24
30
36
4600
3713
3002
2436
1984
1619
1333
4200
3334
2646
1323
400
296
(a) State the meaning of the terms
(i) radioactive
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(ii) nuclide
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(iii) half-life.
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(b) (i) The half-life of nuclide X is 18 hours. Complete the activity of nuclide X column.
[3]
(ii) Using your answer to (i) complete the activity of nuclide Y column.
[2]
(c) Calculate, or use a graph to determine, the half-life of nuclide Y.
0 20 40 60 80tim e/hour
half-life of Y = ......................... hours[3]
(d) Indicate briefly how it would be possible experimentally to obtain the initial activity (4200 Bq in this case) of nuclide X by itself.
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(e) Explain why it is not possible to give a half-life for a mixture of two nuclides.
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[Total 16 marks]
Energy = mass x speed of light squared or E = mc2
Lesson 41 notes – Binding energy and E=mc2
Objectives(a) select and use Einstein’s mass–energy equation ΔE = Δmc2 ;(b) define binding energy and binding energy per nucleon;(c) use and interpret the binding energy per nucleon against nucleon number graph;(d) determine the binding energy of nuclei using ΔE = Δmc2 and masses of nuclei;
OutcomesBe able to select and use Einstein’s mass–energy equation ΔE = Δmc2 .Be able to define binding energy and binding energy per nucleon.Be able to use and interpret the binding energy per nucleon against nucleon number graph.Be able to determine the binding energy of nuclei using ΔE = Δmc2 and masses of nuclei.Be able to derive E=mc2 from Einstein’s theory of Special Relativity!
Einstein Einstein’s Special theory of relativity says that matter and energy are equivalent
E is the amount of energy produced when a mass m is completely converted to energy and c is the speed of light (3x108 ms-1).
Binding energy and mass defectIf you take an atom and split it into its bits (protons and neutrons), the mass
before does not equal the mass of all the individual bits!For a nucleus this difference in mass is called the mass defect of the nucleus.Below is a table of some mass defects:
The difference in mass is the difference in energy it takes to split the atom up given by E = mc2. This is called the binding energy of a nucleus.
Mass defect = (unbound system calculated mass) - (measured mass of nucleus)
The energy given off during either nuclear fusion or nuclear fission is the difference between the binding energies of the fuel and the fusion or fission products.
Binding Energy per Nucleon
The Graph of Binding Energy per Nucleon against Nucleons in the Nucleus shows that, generally, from Hydrogen up until Sodium, the binding energy per nucleon increases as the number of nucleons in the nucleus are added. This is due to the Strong Force exerting itself on each nucleon and binding them all together tightly.
From magnesium to xenon, the nuclei have become large enough that nuclear forces no longer completely extend efficiently across their widths. Attractive nuclear forces in this region, as atomic mass increases, are nearly balanced by repellent electromagnetic forces between protons, as atomic number increases.
In elements heavier than xenon, there is a decrease in binding energy per nucleon as atomic number increases. In this region of nuclear size, electromagnetic repulsive forces are beginning to become bigger than the attractive strong nuclear force that exists between all nucleons.
Nickel-62 is the most tightly-bound nucleus (per nucleon), followed by iron-58 and iron-56. This is the basic reason why iron and nickel are very common metals in planetary cores, since they are produced as end products in supernovae and in the final stages of silicon burning in stars.
Extension – derivation of E = mc2 The Einstein equation (E = mc2) can be deduced from special relativity as follows:
m = mo/[1 - v2/c2]1/2 therefore m2c2 – m2v2 – moc2 = 0 where mo is the rest mass of the object.
Differentiating with respect to time gives: c2 2m[dm/dt] – 2mv[d(mv)/dt] = 0
Rearranging this equation gives: c2 2m[dm/dt] = 2mv[d(mv)/dt] and therefore c2[dm/dt] = v[d(mv)/dt]
However dE/dt = Fv and so dE/dt = Fv = v[d(mv)/dt] = c2dm/dt
And so E = mc2
Lesson 42 notes - Fission and FusionObjectives(e) describe the process of induced nuclear fission;(f) describe and explain the process of nuclear chain reaction;(g) describe the basic construction of a fission reactor and explain the role of the fuel rods, control rods and the moderator (HSW 6a and 7c);(h) describe the use of nuclear fission as an energy source (HSW 4 and 7c);(i) describe the peaceful and destructive uses of nuclear fission (HSW 4 and 7c);(j) describe the environmental effects of nuclear waste (HSW 4, 6a and b, 7c);(k) describe the process of nuclear fusion;(l) describe the conditions in the core of stars
OutcomesBe able to describe the process of induced nuclear fission;Be able to describe and explain the process of nuclear chain reaction;Be able to describe the basic construction of a fission reactor and explain the role of the fuel rods, control rods and the moderator (HSW 6a and 7c);Be able to describe the use of nuclear fission as an energy source (HSW 4 and 7c);Be able to describe the peaceful and destructive uses of nuclear fission (HSW 4 and 7c);Be able to describe the environmental effects of nuclear waste (HSW 4, 6a and b, 7c);Be able to describe the process of nuclear fusion;Be able to describe the conditions in the core of stars
FissionAn unstable nucleus may become stable and lose the extra energy in two ways: It can emit radiation (alpha, beta or gamma) or undergo fission. In nuclei there are two competing forces: The Strong Nuclear Force, which is an attractive force that acts between all nucleons; and the electrostatic force that repels the positive protons from each other.
Induced nuclear fission
The above diagram shows how in induced nuclear fission a slow moving neutron is fired at Uranium 235 so that it becomes uranium 236 which is unstable and so splits into to fission particles, 2-3 neutrons and Energy (in the form of Gamma and Heat).
Chain Reaction
The diagram shows how, because of the emission of 2-3 neutrons in each fission reaction, a chain reaction occurs.
Nuclear Fission Reactor
The fission reaction occurs in the reactor that is shielded with concrete. The fuel rods are of a material such as 235U. 235U is stable but can be made unstable by adding a neutron. A fast neutron from a neutron source (such as californium) must be slowed down (moderated) so that it can be absorbed. Graphite is used as a moderator by placing it in between the fuel rods so that as a neutron collides with the nuclei of the graphite it bounces about and
loses momentum so slowing it down. When the neutron is absorbed 236U is formed which is unstable and so it splits (fissions) into 2 daughter nuclei, 2 or 3 neutrons and lots of energy in the form of heat and gamma rays:
235U + 1n → 236U → 92Kr + 142Ba + 21n+ Energy
The heat boils up water which turns into steam, which turns turbines and generators and produces electricity.
If the electricity production needs to be slowed done, Boron rods are placed into the reactor that can absorb some of the stray neutrons and so reduce the amount of heat and therefore electricity.
Nuclear Fusion
IntroductionNuclear fusion is the process of binding nuclei together to form heavier nuclei with the release of energy. It is the process that powers the stars and our own Sun.
Fusing NucleiProtons won’t stick together easily because they are both positive and so repel each other. The strong force will win over very short distances but you need to give the protons a lot of energy to over come the repulsion before the can fuse. This means very high temperatures.
Extension – Estimating the temperature needed:The magnitude of the force between protons is given by:
F= (k q1 q2)/r2
where k=1/(4πε0) is the Coulomb constant = 8.988×109 m F-1.
The work done, U in moving the two protons together until they are attracted by the strong force is given by:
U= (k q1 q2)/r0.
The Coulomb barrier increases with increasing atomic number.U=(k Z1 Z2 q2)/r0.
Where Z1 and Z2 are the proton numbers of the nuclei being fused.
Using figures for 2 Deuterium nuclei, for which Z1,2=1, we obtain: U= (8.988 x 109 x 1 x 1 x (1.6 x 10-19) 2)/(1 x 10-15) = 2.298 x 10-13 J
The kinetic energy of the nuclei is related to the temperature by:(1/2) mv2 = (3/2) kBT
By equating the average thermal energy to the Coulomb barrier height and solving for T, gives a value for the temperature of around 1.1 x 1010 K.
Nuclear Fusion in StarsIn 1848, J.R. Mayer examined the then-popular theory of the Sun being composed of burning coal. He stated that if the Sun began burning 5000 years ago, corresponding to the Biblical age of the Earth, then by his calculations, it already would have burned out.
Lord Kelvin proposed several explanations on how the Sun might generate its energy including: mass contraction, which could allow the Sun to shine for up to 45 million years.
Charles Darwin was looking at the erosion of rocks and concluded that the Earth had to be at least 300 million years old. The Sun had to be at least as old as the Earth so clearly the solution had not been found.
Missing MassIt was discovered that the when nuclei of hydrogen combine to form Helium, the resulting mass was less than the sum of the mass entering the sum of the initial masses. The missing mass was converted into energy with an exchange rate of E=Δmc2 where Δm is the difference in the mass of the nuclear reaction and c is the speed of light (3 x 108 m s-1). The conversion of a relatively small amount of mass into energy would allow the Sun to shine for billions of years.
The fusion of Hydrogen into Helium in stars occurs in three stages:First, two ordinary Hydrogen (Hydrogen-1) nuclei, which are actually just single protons, fuse to form an isotope of Hydrogen called Deuterium (Hydrogen-2), which contains one proton and one neutron. A positron (a positively charged electron) and a neutrino (a neutral particle which travels nearly at the speed of light and has, perhaps, almost no mass) are also produced. The positron is very quickly annihilated in the collision with an electron and the neutrino travels right out of the Sun.
11H +1
1H = 21H+ 0
0e+ +00ν
The newly-formed Deuterium fuses with another regular Hydrogen to form an isotope of Helium, Helium-3 containing two protons and one neutron.
21H+ 1
1H = 31H
Next, two of the Helium-3 nuclei fuse to form a Helium-4 nucleus and two hydrogen-1 nuclei. Energy as gamma rays are produced in each step.
31H+3
1H = 42He + 21
0H
The resulting reaction cycle generates around 25 MeV of energy. The high density of the Sun allow the temperature for fusion to occur to be around 1.5 x 107 K
Nuclear Fusion on EarthMan-made nuclear fusion uses a different reaction to that which occurs in the stars. In stellar Hydrogen burning, the reaction of two Helium-1 nuclei requires a proton to change into a neutron. Fusion on Earth starts by fusing Deuterium and Tritium. (Tritium is an isotope of Hydrogren which contains 1 proton and 2 neutrons.) The reaction is as follows:
21H + 3
1H = 42He + 1
0n + 17.59 MeV.
An equally probable reaction can also occur between two Deuterium nuclei in one of two ways:
21H +2
1H = 32H + 1
0n + 3.27 MeVor alternatively,
21H +2
1H = 31H + 4.03 MeV
Nuclear Fusion Reactors
The TokamakThe tokamak uses magnetic fields to confine high temperature plasma to the interior of a torus shaped vacuum chamber. When the plasma has been heated to around 100 million K fusion occurs. Each fusion reaction results in the production of an alpha particle (4
2He nucleus) and a neutron. The neutrons carry most of the energy and can escape the confining fields. They are captured in the walls of the tokamak where they generate heat. A coolant circulating through the walls is passed through a heat exchanger to produce the steam to drive turbines, as in a conventional power station. The walls also double as a "breeding blanket" in which neutrons react with Lithium to produce further Tritium.
Figure 2. A Tokamak design of fusion reactor
From: www.splung.com/content/sid/5/page/fusion
Lesson 42 practice questions – Fission and Fusion
1 In nuclear fission, energy is released.a) Explain what is meant by nuclear fission.
…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… (1)
b) In a possible fission reaction Uranium 235 (atomic number 92) captures a neutron to become a compound nucleus before splitting into Ba 141 (atomic number 56) and Kr 92 (atomic number 36) releasing three neutrons.
Write down the nuclear reaction equation for this event.
……………………………………………………………………………………… (2)
c) The total mass of the compound nucleus Uranium 236 (atomic number 92) before fission is 236.053u. The total mass of the fission products is 235.867u. Use these data to calculate the energy released in the fission process. (1u=1.66054x10-
27kg)
energy = ……………………….. J (3)2 a) A possible fission reaction is
i) The two asterisks represent two numbers missing from the right hand side of the nuclear reaction.
Write down the missing nucleon (atomic mass) number …………………………..
and the missing proton (atomic) number …………………………………………..
ii) The total mass of the compound nucleus before fission is 236.053u. The total mass of the fission products is 235.840u. Show that the energy released in the fission process is about 3 x 10-11J.
(3)
3 One of the reactions that fuel the stars is the fusion of two protons to give deuterium. In turn the deuterium goes through a series of reactions, the end product being helium. This is also a process that releases energy. In this question you are asked to consider the energy that would be released if all the deuterium in the water contained in an electric kettle were to be converted by fusion into helium.
The kettle contains 1 litre of water. The data you need are listed below.
1 atomic mass unit (u) = 931 MeV
1 eV = 1.6 10–19 J
NA = 6.02 × 1023 mol–1
Particle Mass / u
H11 1.007 825
H21 2.014 102
3.016 030
1.008 665
a Two deuterium nuclei can fuse to give one nucleus of helium with the ejection of one other particle. Write down the balanced equation that represents this reaction.
b Calculate the mass change that occurs in this reaction.
c Convert this energy into joules.
This gives you the energy released when two deuterium nuclei fuse. The next steps take you through the calculation of the total energy released if all the deuterium in the kettle water were to fuse to make helium-3. The ratio of deuterium atoms to hydrogen in water is roughly 1 to 7000.
d What is the mass of 1 mole of water (H = 1 u; O = 16 u roughly)?
e How many moles of water are contained in the litre?
f How many molecules of water (H2O) are in the kettle?
g How many molecules of deuterium oxide (D2O) are in the kettle?
h Each heavy water molecule has two atoms of deuterium; what total energy is released if all the deuterium in the kettle is converted to helium-3?
Now to put this number in a new perspective. It requires 4200 J to increase the temperature of 1kg of water by 1K.
i How many litres of water could be heated through 100 K by the fusion energy you calculated in question 8?
