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Hazards of Radiation. Radioactivity. Learning Objectives. Recap properties of α , β , γ radiation. Describe and explain the relative hazards to humans when exposed to α , β , γ radiation. Recap the inverse square law of radiation.

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## Hazards of Radiation

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**Hazards of Radiation**Radioactivity**Learning Objectives**• Recap properties of α, β, γ radiation. • Describe and explain the relative hazards to humans when exposed to α, β, γ radiation. • Recap the inverse square law of radiation. • Describe how this applies to the safe handling of radioactive sources.**Quick Recap**• Complete the table (in pencil) that describes the properties of the three common radiations:-**Ionisation**• What is ionisation? • Which do you think is the most ionising?**Ionisation**• α - Intense, ionises about 10,000 atoms per alpha particle. (because strongly +ve and large) • β - Less intense than a, about 100 ionisations per particle. (because lower charge, but higher speed) • γ - Weak interaction about 1 ionisation per mm.**Hazards**• α – Once it gets in it is highly damaging to body tissue. • Luckily can’t pass through skin but could be inhaled or ingested. Considered less damaging than gamma rays or alpha particles. • β – Lower interaction rate means it is much less damaging to body tissue than alpha. Used as medical tracers. • γ - can be a dangerous form of radiation, as they are very penetrating. Needs intense or prolonged radiation to cause damage to cells.**Annoying Question (10 mins)**• Explain the dangers associated with radioactive sources (8 marks). • Write a mark scheme for this question i.e. assign 1 mark each to 8 short but relevant statements.**Possible Mark Scheme**• Alpha is intensely ionising. • Although short range and kept out by skin • Ingestion of alpha emitters can do immense damage to the cells. • Beta is less ionising • But can penetrate the body • Gamma is highly penetrating • But causes little ionisation. • Long term exposure leads to damage to DNA**Spare Question**• Alpha and beta particles lose about 5 × 10-18 J of kinetic energy in each collision they make with an air molecule. An alpha particle makes about 105 collisions per cm with air molecules, while a beta particle makes about 103 collisions. What is the range of an alpha particle and a beta particle if both start off with an energy of 4.8 × 10-13 J?**Spare Question Answer**• Both particles lose all their energy in 4.8 × 10-13 J ÷ 5 × 10-18 J (P) = 96 000 collisions (P) • The alpha particle has a range of 96000 ÷ 105 cm-1 = 0.96 cm (P) • The beta particle has a range of 96000 ÷ 1000 cm-1 = 96 cm. (P)**Inverse Square Law for γ**When taking a reading of intensity, I, at a distance x from the source, I0 is the intensity at the source and k is a constant. Intensity (number of photons per unit area) decreases by the square of the distance. So doubling distance from source means only a quarter of radiation reaches you – so keep them at a distance! -tongs • Can be written as:- • or:-**Learning Objectives**• Recap properties of α, β, γ radiation. • Describe and explain the relative hazards to humans when exposed to α, β, γ radiation. • Recap the inverse square law of radiation. • Describe how this applies to the safe handling of radioactive sources.**Nuclear Radius**Nuclear Physics**Homework**• Research and explain how electron diffraction can be used to determine the radius of the nucleus (6 Marks) • Past Paper Question on today’s material. • Complete both by Next Lesson, next Monday, Period 5.**Learning Objectives**• State and use the equation for dependence of radius on nucleon number. • Calculate nuclear density. • Recall the implications of the high nuclear density compared to atomic density.**Equation**• Dependence of radius on nucleon number:- [The term A1/3 means the cube root of A, the nucleon number. The term r0 is a constant with the value 1.4 × 10-15 m. R is the nuclear radius.] What physical quantity is r0? Rearrange in the form of A=. Try working out R for Gold (A =197 ) and Carbon (A=12)**Nuclear Density**• Radius of a carbon nucleus ~ 3.2 x 10-15m. • Radius of a gold nucleus ~ 8.1 x 10-15m. • Mass of a carbon nucleus ~ 2.00 x 10-26kg. • Mass of a gold nucleus ~ 3.27 x 10-25kg. • What are the densities of the nuclei?**Nuclear Density**• Density of carbon nucleus ~ 1.46 x 1017 kg m-3. • Density of gold nucleus ~ 1.47 x 1017 kg m-3. • Very high! One teaspoon = 500 million tonnes. • So pretty much the same, regardless of element. • Ext: Work out mass of neutron star based on this density. How does it compare to solar mass?**Nuclear Density**• Nuclear density >> Atomic Density • This implies:- • Most of an atom’s mass is in its nucleus. • The nucleus is small compared to the atom. • An atom must contain a lot of empty space.**Example Exam Questions**• Q1: • (a)If a carbon nucleus containing 12 nucleons has a radius of 3.2 x 10-15m, what is r0? • (b) Calculate the radius of a radium nucleus containing 226 nucleons. • (c) Calculate the density of a radium nucleus if its mass is 3.75 x 10-25 kg. • Q2: A sample of pure gold has a density of 19300 kg m-3. If the density of the gold nucleus is 1.47 x 1017kg m-3 discuss what this implies about the structure of a gold atom.**Learning Objectives**• State and use the equation for dependence of radius on nucleon number. • Calculate nuclear density. • Recall the implications of the high nuclear density compared to atomic density.**Nuclear Radius**Probing Matter**Homework**Why are spent fuel rods more radioactive after removal from the reactor? What are the different types of radioactive waste? How is each type stored/disposed?**Filling the Gaps**γ emitters can indeed be used as medical tracers as we thought. Cell membranes can also be destroyed as well as affecting DNA.**Learning Objectives**Explain how an estimate for the nuclear radius can be obtained from Rutherford’s Experiment.**Rutherford’s Alpha Scattering**At P, the point of closest approach, all of the initial kinetic energy of the alpha is converted to electrostatic potential energy.**Electrostatic Potential Energy**• The equation for Electrostatic Potential Energy is given by this equation:- • Where:- • EP is the Electrostatic Potential Energy • rc is the distance of closest approach • ε0 is the permittivity of free space (constant - see data booklet) • Q1Q2 are the charges of the two particles involved.**Little bit of Maths…**• Solving to find rC, what do we get? • Rearranging… • Remember: • Q1=2e = 2 × 1.60 × 10-19 C (α is He nucleus) • Q2=79e = 79 × 1.60 × 10-19 C (79 protons in Gold nucleus) • EP=7.68 MeV = 768 × 106 × 1.60 × 10-19 J (K.E. of α particles fired at the foil.) • ε0 = 8.85 × 10-12 Fm-1 (from the data booklet) • rc= 2.96 × 10-14 m(a bit large)**A couple of points…**The nucleus is treated as a point charge. At this level it is not. The alpha particles are stopped some distance away from the nucleus. It takes higher energy alpha particles to penetrate the nucleus. The values for the nuclear radius given by other particles such as protons, neutrons and electrons are slightly different. …so really it is only an upper limit on the nuclear radius.**Method 2**About 1 in 10,000 particles are deflected by more than 90º. For a thin foil (so that only one scattering) with n layers of atoms, the probability of being deflected is about 1 in 10,000n. This probability depends on effective cross section of nucleus to the atom:- Typically n=10,000 so d = D/10,000**Questions**A) For a metal foil which has layers of atoms, explain why the probability of an α particle being deflected by a given atom is therefore about 1 in 10,000n. (assume 1 in 10000 deflected by more than 90º) B) Assuming this probability is equal to the ratio of cross sectional area of the nucleus to that of the atom, estimate the diameter of a nucleus for atoms of diameter 0.5 nm in a metal foil of thickness 10 μm.**Safety Aspects**Nuclear Energy**Physics Workshop**Every Wednesday 3.40pm-4.40pm in O8 Is this time good for most people?**Learning Objectives**• Describe the range of safety features associated with a nuclear reactor. • Derive the equation relating half life and the decay constant. • Practice calculations on radioactive decay.**High Energy Electron Diffraction**• Any 6 from:- • A beam of high energy electrons is directed at a thin sheet of an element • and accelerated through a potential difference of about 108 volts (MeV/high energy) • A detector measures the number of electrons diffracted at a number of different angles. • Scattering effects occur due the charge of the nuclei and electron and this causes the count rate of the beam of electrons to decrease as angle increases. • The electrons are also diffracted by the nuclei in the sheet which causes minima and maxima to observed in the final pattern • as long the de Broglie wavelength of the electrons is of the same order as the size of the nucleus, which is about 10-15 m. • The diameter of the nucleus can be calculated using the angle to the first minimum, θmin • and the wavelength of the incoming beam λ using the diffraction equation R sin θmin = 0.61λ.**Safety Features**The reactor is a thick steel vessel designed to withstand the high pressure and temperature in the core. The core is in a building with very thick concrete walls which absorb the neutrons and gamma radiation Emergency shut down system – designed to insert the control rods fully into the core. Sealed fuel rods are inserted and removed using remote handling devices. Spent rods are more radioactive than before use.**How to Remember?**C - Concrete building. R - Remote Handling. E - Emergency shut down system. S - Steel vessel for reactor core. S - Spent fuel rods much more radioactive. Spells CRESS, helps us remember?**Dangers of Nuclear Power**Chernobyl (1986) Wanted to see if coolant pumps would keep operating if there was a loss of power. When they pushed control rods into reactor, caused loss of power and reversed direction of the rods!**Aftermath**Overheating caused decomposition of water into hydrogen and oxygen which gases collected at the top of the vessel. Ignited and blew the lid off the reactor and turned the vessel on its side. Nine tonnes of caesium-137 floated across Europe along with many other tons of radioactive material. Caesium-137 is water-soluble and extremely toxic in minute amounts (half life 30 years).**Okay let’s try it**Describe the range of safety features associated with a nuclear reactor (5 marks).**Specific Heat Capacity & Latent Heat - Theory**Thermal Physics Lesson 1**Learning Objectives**• Define specific heat capacity • Perform calculations using ∆Q=mc ∆θ • Describe how specific heat capacity can be measured in the lab**Homework**• Complete worksheet to practice calculations we will look at today.**Why do we care**• The anomalously large SHC of water is particularly important for the development and maintenance of life on Earth.**Why are we learning this?**• Historical Context • Images of Joule, Kelvin • Steam engines • Thermal physics in stars**Which is hotter? Bath water or a lit match?Which needed more**energy to heat it?**Heat vs. Temperature**• Heat as water and temperature as wetness analogy • ‘Heat’ is not an entity but a short hand name for a process (heating as oppose to working).**State vs. Phase**• Solids, liquids and gases are three of the different phases of matter (superfluids and plasmas are two others. Thus melting, boiling etc are changes of phase. • Each phase can exist in a variety of states depending upon e.g. the temperature and pressure.**Specific Heat Capacity**• Definition:- • The specific heat capacity (c) of a substance is the amount of energy needed to raise the temperature of 1kg of the substance by 1K (or 1 ºC)

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