PHYSICS A-LEVEL AUTHOR: KG/ ARCTIC KITTEN For Edexcel Physics 2018 CONTENTS Topic 2: Mechanics .............................................................................................................................................................................. 2 Topic 3: Electric Circuit ........................................................................................................................................................................ 4 Topic 4: Materials ................................................................................................................................................................................ 7 Topic 5: Waves and Particle nature of light ....................................................................................................................................... 10 Topic 6: Further Mechanics ............................................................................................................................................................... 16 Topic 7: Electric and Magnetic Field .................................................................................................................................................. 17 Topic 8: Nuclear and particles Physics .............................................................................................................................................. 22 Topic 9: Thermodynamics ................................................................................................................................................................. 27 Topic 10: Space ................................................................................................................................................................................. 29 Topic 11: Nuclear Radiation .............................................................................................................................................................. 33 Topic 12: Gravitation ......................................................................................................................................................................... 35 Topic 13: Oscillation .......................................................................................................................................................................... 37 TOPIC 2: MECHANICS Name Definition Formulae STATIC Centre of mass Centre of gravity The point from which all the mass of the object appears to act The point from which all the weight of the object appears to act KINEMATICS 1 π = π’π‘ + ππ‘ 2 2 π£ = π’ + ππ‘ Suvat equations π£ 2 = π’2 + 2ππ (π’ + π£)π‘ 2 NEWTON’S LAWS π = Newton’s First Law Newton’s Second Law An object will remain at rest, or in a state of uniform motion in a straight line, unless acted upon by a resultant force The resultant force is directly proportional to the rate of change of momentum and in the same direction of the momentum To every action, there’s an equal and opposite reaction Newton’s Third Law ο© ο© ο© ο© Act on different bodies Opposite direction Same magnitude Same kind MOMENT Moments The force multiplied by the perpendicular distance from the pivot to the line of action of the force Law of moments For any object in equilibrium the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moment about the same point π = πΉπ Unit: Nm WORK DONE, ENERGY & POWER The product of: Work Done F (magnitude of the force) π = πΉπ πππ π s (magnitude of displacement s of point of application of force) Unit: π½ = ππ cos ο± (ο± is the angle between the force and displacement vector) Energy Kinetic energy The property of object that gives it the ability to do work The work done to accelerate an object of mass m, from rest to a speed v βπΈπ = πΈπ′ − πΈπ = ππ· Potential energy The ability of an object to do work by virtue of its position or state Gravitational potential energy The energy an object has due to its position in a gravitational field 1 πΈπ = ππ£ 2 2 π2 = 2π πΈπ = ππβ The rate of doing work with respect to time Power Unit: π = π½ π π= βπ βπ‘ 1 watt is 1 J of E transferred in 1 sec Efficiency πππ = ππ πππ’π πΈ ππ’π‘ππ’π‘ πππ‘ππ πΈ ππππ’π‘ END π = πΉπ£ πππ π TOPIC 3: ELECTRIC CIRCUIT Name Definition Formulae ELECTRICITY Electron-volt Energy transferred when an electron moves through a potential difference of one volt Drift velocities Motion of electron in a wire Current π: πβππππ πππππππ ππ’ππππ ππππ ππ‘π¦ Different metals have different conductivity because different n emf π£ππππππ = 106 π⁄π π£πππππ‘ = 10−4 ππ⁄π The rate of flow of charged particle Potential difference 1ππ = 1.6 × 10−19 π½ The energy/charge transferred between two point WD per unit charge to move a charge around the circuit Emf = pd when πΌ = 0 because no energy/ pd lost on resistors πΌ= βπ βπ‘ πΌ = πππ£πππππ‘ π΄ π= π π π = π + πΌπ Plot V (across battery) against I Gradient = -r Y intercept is emf Resistance The opposition to the flow of electrical current Ohm’s Law A special case where πΌ ∝ π for constant temperature π = π πΌ Total resistance Resistivity Power Critical temperature Numerically equal to the resistant of a unit length and a unit area of wire π2 π The temperature below which its resistivity instantly drop to zero π = πΌπ = πΌ 2 π = ππ π΄ Unit: πΊπ π = CURRENT-POTENTIAL GRAPH Ohmic conductors Filament bulb Diodes Thermistor LDR Require a minimum driving V in the forward direction Temperature ο, more energy transfer to lattice ions Electrons gain energy from light Forward direction: low R ο current flow, Temperature ο Obeying Ohm’s Law Ions vibrates more, probability of collision ο, electron lose more energy Threshold voltage 0.6V Electrons to conduction band Backward direction: high R πΌ = πππ£π΄ ο so resistance ο Few charge carriers ο leakage charge carriers density ο, current ο Reverse pd high enough overcome E barrier π = πΌπ resistance ο Light intensity ο, electrons to conduction band Charge carriers density ο, current ο π = πΌπ resistance ο END TOPIC 4: MATERIALS Name Definition Formulae LIQUIDS πππ π Upthrust Terminal velocity The upthrust on an object in a fluid =the weight of the fluid displaced by the object Upthrust + drag = weight No resultant force so velocity is constant π = ππ π πππ π π = ππ ππ π π+π· =π Falling object: At first, π· = 0 π· ∝ π£ so drag force increases π − π − π· = πΉ so resultant force decreases When π· + π = π no resultant forces so N1L terminal velocity Fluid Streamline Path line Steady flow Substance that can flow A curve whose tangent at any point is along the direction of the velocity of the fluid particle at that point The path taken by a fluid particle as it moves Occurs when no aspect of the fluid motion change with time Fluid move with uniform lines in which velocity is constant over time Laminar flow • • • No mixing of layers Flows in layers/flowlines/streamlines No abrupt change in direction or speed of flow Mixing of layers Turbulent flow Contains eddies/vortices Abrupt/random changes in speed or direction For a spherical object of rad r Stokes’ Law Moving slowly through a fluid with speed v πΉπ = 6πο¨ ππ£ The flow of fluid is laminar The thickness of a fluid. Viscosity increase, rate of flow decrease (spread quicker) Viscosity Liquids, ο¨ ο² with temperature ο¨ Gasses, ο¨ ο± Drag for turbulent flow Hysteresis πΆπ :drag coef, no unit A: area of object facing fluid flow 1 πΉπ = πΆπ π΄ππ£ 2 2 The extension under a certain load will be different depending on its history of past load and extension HOOKE’S LAW Hooke’s Law The extension, e, is directly proportional to the applied force, if the limit of proportionality is not exceeded k: the stiffness of the spring/ the spring constant πΉ = ππ Outside the region that obeys Hooke’s law: Extension not proportional to force (greater extension for same force) Deform plastically, not return to original shape Elastic’s Potential Energy The ability of a deformed material to do work as it regains its original length Area under a force-extension graph 1 π = πΉπ 2 1 π = ππ 2 2 YOUNG MODULUS Stress The force per unit cross-sectional area perpendicular to the surface Strain Fractional change in length of the material The stress per unit strain Using thin long wire to measure Young modulus: Young’s Modulus Small extension is hard to measure and has high uncertainty πΉ π΄ Thin wire has smaller A hence larger P for a given F π= Long wire: greater extension for a given stress πΉ π΄ π π= πΏ π πΈ= π π= Name Definition Note GRAPH P/ Limit of The maximum extension that an object can proportionality exhibit, which still obeys Hooke’s Law The maximum extension or compression that a material can undergo and still return E/ Elastic limits to its original dimension when the force is removed Y/ Yield point The point after which a small increase in stress produces an appreciably greater increase in strain. UTS/ Ultimate Tensile Stress The maximum tensile stress the material can withstand before breaking PEYU If the mass exceeds maximum mass State maximum load The elastic limit is exceeded Spring deform permanently Spring constant change PROPERTIES OF MATERIALS Strength The maximum compressive stress applied before breaking Strong/ Weak Strong: High breaking stress (steel) Weak: Low breaking stress Stiff/ Flexible Stiff: High Young’s Modulus, large stress for small deformation Flexible: Low Young’s Modulus Tough/ Brittle Tough: large plastic deformation region on graphο°absorb lots of energy Brittle: little plastic deformation before breaking ο°absorb little energy Elastic/ Plastic Elastic: Regain their original shape when deforming force/stress is removed Plastic: Extend extensively and irreversibility for a small increase in stress beyond the yield point (copper, clay) Hardness Resistance to scratch on surface Hard/ Soft Hard: Not easy to scratch or indent Soft: Easy to scratch or indent Ductile/ Malleable Ductile: Undergo large plastic deformation under tension and hence can be made/ drawn into wires Malleable: Undergo large plastic deformation under compression and hence can be hammered into thin sheets END TOPIC 5: WAVES AND PARTICLE NATURE OF LIGHT Name Definition Note BASICS OF WAVE Mechanical wave Wave require medium to travel through Electromagnetic wave Require no medium to travel through Longitudinal Waves Compression Rarefaction Has oscillations that are parallel to the direction of movement of the wave energy (Vibrations of the particles parallel to the direction of propagation of the wave) Area in which particle oscillation put them closer than their equilibrium state Area in which the particle oscillation put them further apart than their eq state Transverse wave The oscillations are perpendicular to the direction of movement of the wave energy Displacement Distance and direction from the equilibrium position Amplitude The magnitude of maximum displacement from the equilibrium position Frequency The number of complete oscillations per unit time Period The time taken for one complete oscillation π£= Speed Wavelength ο¬ = πο¬ π π π£ = √ (π‘ππππ π£πππ π) π Minimum distance between two point on a wave with the same displacement STANDING/ STATIONARY WAVES No net transfer of energy Standing wave Nodes Antinodes 2π π Points where the amplitude of oscillation is 0 ο¬= Points where amplitude of oscillation is maximum Two waves with same amplitude and wavelength travelling at opposite direction/ Producing standing reflected off wave Principle of superposition give resultant displacement Nodes and antinodes produced Snapshot of the wave Constructive interference occurs when phase difference = 0 Destructive interference occur when phase difference = π π 2π π£ = √ ; π£ = πο¬; ο¬ = π π Harmonics ∴π= Name π π ×√ 2π π Definition Note PHASE Phase of oscillation The stage of a given point on a wave is through a complete cycle Phase difference The difference in phase angle between two parts of the same oscillation or between two oscillation In phase Antiphase Phase difference = ππ, n even Path difference = πο¬ Phase difference = ππ, n odd Path difference= πο¬/2 Wave front The line of a crest or trough of a transverse wave/ compression of rarefaction of longitudinal wave Coherence Waves with same frequency and constant phase difference π OP: 2π − 2 = OO’: 3π 2 OQ: π 3π 2 π 2 π π − = 2 2 − = π Monochromatic Same frequencies Interference The superposition outcomes of a combination of waves Constructive Interference Take place when the path difference is a whole number of wavelength Destructive Interference Path difference is 1/2, 3/2, 5/2… wavelength Produce interference: Superposition takes place Path difference = ππ in phase In phase: constructive interference Antiphase: destructive interference Antiphase amplitude = minimum = 0 Fringes Principle of superposition Pattern of light and dark band When two or more waves meet, the total displacement at any point is the sum of the displacements that each individual wave cause at that point Polarisation Orientation of the plane of oscillation of a transverse wave Polarised waves Oscillations occurs in only one plane or directions perpendicular to the direction of propagation of the wave Reflected light/ incident light is polarised Polarised light vibrates in one direction Polaroid only allow oscillation in one plane Polaroid When planes are parallel it allows plane through, the intensity is high Perpendicular block the light, intensity = 0 Each rotation by π will alternatively block and allow the light through Name Definition LENS Focal length Focus The distance from the centre of the length to the focal point The point where parallel incident rays be made to meet by the refraction of the lens 1 1 1 = + π π’ π£ Thin lens equation π= Magnification π= Power Combination Convex lens β1 π£ = β0 π’ 1 π π = π1 + π2 Cause light to converge, f is positive Converge parallel rays to a focus at the focal length from the lens Cause light to diverge, f is negative Concave lens Images Name Diverge parallel rays to appear to have come from a virtual focus at the focal length back from the lens π£ > 0: real, inverted π£ < 0: virtual, upright Definition Formulae Note REFRACTION – TOTAL INTERNAL REFLECTION Refraction The change in direction of a wave occurs when its speed change due to a change of medium Refractive index Relative refractive index π π£ π£1 π12 = π£2 1 π12 = π21 π= Wave travelling from medium 1 to medium 2 If π2 > π1 Then material 2 is optically denser than material 1. Absolute refractive index π ≥1 π£1 π2 π12 = π1 π1 = Light travelling in medium 1 π£2 < π1 π2 < π1 π1 sin π1 = π2 sin π2 Snell’s law Reflection The change in direction of a wave at an interface between two different media so that the wave returns into the medium from which it originated Critical angle The angle of incidence for which the angle of refraction is 900 Total Internal Reflection (TIR) When the angle of incidence is bigger than critical angle → light does not refract but bounces back at the interface 1 = sin πΆ π1 1 πΆ = sin−1 π1 DIFFRACTION Diffraction The spread out of the wave when it meets a solid obstacle For most of the light waves there is destructive interference Diffraction gratings Path difference = π sin π Constructive interference → πο¬ π sin π = πο¬ π = 1 ππ’ππππ ππ ππππ π: πππππ ππ’ππππ ππππ₯ ≤ Name π ο¬ Definition Notes WAVE THEORY Young’s double slit experiment Monochromatic coherent light passes through two parallel slits Light behaves light a wave The waves through the slits diffract, two diffracted waves overlaps Principle of superposition determines the resultant wave displacement at any point Constructive interference where in phase ο bright fringe Destructive interference where out of phase ο dark fringe The size of wavelength of photon is similar to the size of the slit PHOTOELECTRIC EFFECT Photon Photoelectron Experiment Results Packet of electromagnetic radiation Electron released from a metal surface as a result of its exposure to electromagnetic radiation Shine light on a metal surface, electron might be emitted Increase frequency Increase amplitude π < π0 No electron emitted Nothing happen π > π0 Max KE of electron increase as frequency ο Number of electron emitted/ s ο, max KE does not change Photon cause emission of electron from surface of metal Photon has energy πΈ = βπ 1 βπ = π + ππ£πππ₯ 2 2 One photon hit one electron Explanation If πΈ > π emission occurs 1 ππ£ 2 2 is KE of electron emitted Max because some energy lost to get to the metal surfaces Threshold frequencies Work function Minimum frequency that can cause electron emission If π < π0 no e- are emitted If π ≥ π0 e- are emitted Minimum energy needed to remove an electron from the metal surface The energy in waves theory depend only on amplitude not frequency Wave theory not explain photoelectric effect Increasing light intensity should increase maximum KE, max KE not depend of frequency Predict a delay between shining the light and emission of electron Cannot account for a threshold frequency of the metal/ emission occurs at all frequencies Increase the light intensity increase number of electron emitted/s Particle theory explain One photon release one electron Energy of photon depend on frequency not intensity πΈ = βπ Intensity determines number of electrons Wave-particle duality Quantum object sometimes have wave like properties & sometime have particle like properties depend on the experiment done on them ENERGY LEVELS/ QUANTISATION OF ENERGY Electron gains energy (become excite) and move to higher levels Line spectrum/ Photon emission Electron has fixed energy level Electrons fall to lower level, reduce energy by emitting photons Energy lost: π¬ = ππ Photon has specific energy hence form line spectrum Ground state The lowest energy level for a system Excitation The energy state that is higher energy than the ground state Energy level A specific quantity of energy an electron can/ is allowed to have inside an atom De Broglie wavelength The wavelength associated with a particle with a given momentum END ο¬= β β = π ππ£ TOPIC 6: FURTHER MECHANICS Name Definition Formulae MOMENTUM π = ππ£ Momentum Impulse Change in momentum πΌ = πΉβπ‘ = βπ π (π ) ππ‘ π‘ππ‘ππ π πΉπππ π’ππ‘πππ‘ = (ππ£) ππ‘ ππ£ πΉπππ π’ππ‘πππ‘ = π = ππ ππ‘ ππ πΉπππ π’ππ‘πππ‘ = π£ ππ‘ ππ£ ππ πΉπππ π’ππ‘πππ‘ = π +π£ ππ‘ ππ‘ ππππ π’ππ‘πππ‘ = 0 πΉπππ π’ππ‘πππ‘ = If π = ππππ π‘ Newton’s second Law If π ≠ ππππ π‘, π£ = ππππ π‘ If π ≠ ππππ π‘, π£ ≠ ππππ π‘ If πΈπ = 0 Conservation of momentum The total momentum before a collision is the same as total momentum after collision (provided that there is no external force acting on the system) Elastic collision KE is conserved (e.g. Identical masses move apart at 90O) Inelastic collision KE is not conserved (e.g. explosions, stick after collision) πππππ‘πππ = ππππππ CIRCULAR MOTION Angular velocity Centripetal acceleration Centripetal force A resultant force is required to produce and maintain circular motion No reaction forces Weightlessness For astronauts on ISS, gravitational force = centripetal force, hence no reaction force END π£ = ππ 2π π= π 2 π£ π= = π2 π π ππ£ 2 πΉ = ππ = = ππ2 π π TOPIC 7: ELECTRIC AND MAGNETIC FIELD Name Definition Formulae Notes ELECTROSTATICS Radial field πππ π2 1 π= 4ππ0 = 8.9 × 109 πΉ= Coulomb’s Law Forces between two charges obey an inverse Electric field A region where a charged particle experience a force Electric field strength The force per unit charge acting on a small positive charge πΈ= The work done against the electric field in moving the charge from infinity to that point in the field πΈππΈ = Electrical Potential Energy Electrical Potential π πΈ = πΌ π π πΈπ΄ = π0 π= πΉ ππ = 2 π π π= πππ π ππ π CAPACITOR Field strength are equal at all point Uniform field Arrows show the direction of a small (+) charges will move when placed in the electric field Equipotential surface The plates, always perpendicular to the electric field line Electric field strength d: distance from positive plate Capacitance Charge stored per unit p.d. Capacitor A device for storing charges π π π πΆ= π π΄π0 ππ = π πΈ= ππ : relative permittivity For air, ππ = 1 Energy stored by a capacitor Time constant Charging 1 π = ππ 2 1 2 = πΆπ 2 π2 = 2πΆ The area under the graph (triangle) 1 π = ππ = ππππ£πππππ 2 Time taken for the charge to fall to 0.37 of its initial value RC Shape of graph (current) exponential decay, current decrease by equal fraction in equal time interval The cell pushes charges through the circuit A current flows, charges are added to the || until πΌ = 0 πΌ = πΌ0 π ππ increases, ππ decreases, πΌ decrease π π = πΌπ + πΆ −π‘⁄ π πΆ πππΌ = πππΌ0 − π‘ π πΆ Discharging Capacitor pushes charges (opposite direction) through the resistor from negative plate to positive plate A current flow, charges are removed exponentially till 0 Changing AC to DC Smoothed DC, Exponential decay Rectified circuit Capacitor store charges Current change direction If RC > T of AC, the capacitor (Charge battery: without Normal circuit doesn’t fully discharge before being charged diode charges and discharge) π΄π0 π So as dοο , Cοο, Qοο, Iοο πΆ= Microphone condenser 1 π If π πΆ < , I vary with frequency f Root mean square Irms is equal to the direct current that give the same average power output π½πππ = π°πππ = Μ = π°ππππ πΉ π· Name Definition π½π √π π°π √π Formulae Note FLUX πΉ πΌπΏ The force per unit length per unit current on a long straight wire perpendicular to the magnetic field lines √π0 πΌ0 Flux The B*(the area perpendicular to the field lines) π = π΅π΄ cos π Unit: Wb Flux linkage For a coil of N turn Φ = ππ Unit: Wb or Wb turn Magnetic flux density π΅= 1 =π MAGNETIC FIELD Magnetic field The direction of magnetic field is the direction North pole of compass will point if placed in the field A moving charge create a magnetic field Field line are concentric circles The magnetic field gets weaker as the distance from the wire increase Magnetic field around a wire Right-hand grip rule tells the direction of the field All magnetic field are closed loops All magnetic field are created by a moving electrical charge Fleming’s left-hand rule give direction Two parallel wires carry current in the same direction attracts CURRENT CARRYING CONDUCTOR πΉ = π΅πΌπ sin π Equation The coil will rotate Speed of the motor depend on B, I, N, Area of the coil The dynamo effect The commutator ensures that the current always flow in the same direction around the loop so the loop rotate in the same direction. Magnetic flux goes from 0 to a maximum An alternating emf is produced CHARGED PARTICLE BEAMS Equation F perpendicular to v, v is constant hence centripetal force πΉ = π΅ππ£ sin π ELECTROMAGNETIC INDUCTION Faraday’s Law Magnitude of the induced emf is directly proportional to the rate of change of flux linkage Lenz’s Law The induced emf cause a current to flow as to oppose the change in flux linkage that creates it β= π(ππ ) ππ‘ β = −π(ππ ) ππ‘ As magnet move, there’s a change in flux Faraday’s law: induced emf proportional to the rate of change in flux Initial increase in emf as magnet get closer to the coil Magnet & coil When magnet is fully inside the coil there is no change in flux so no emf Changing direction of magnet, direction of emf change Magnitude of emf depends on the speed of magnet Same total flux so the areas of two graphs are equal Work done by magnet: Lenz’s law, induced current creates a B field to oppose motion Hence force in opposite direction to its motion Flux changeοinduced emf To create a current in the coil work must be done so there is a force ο induce B field in the coil oppose the change in B field π = πΉπ hence work is done Ways to create induced emf: Moving the magnet Changing the current (turn on off) Change into alternating current TRANSFORMER An electrical machine for converting an input AC PD into a different output AC PD ππ > ππ : Step up transformer ππ < ππ : Step down transformer ππ ππ πΌπ = = ππ ππ πΌπ Transformer effect The changing I in the primary coil create an changing B field in the iron core There is a changing in flux linked to the second coil Faraday’s law (β = π(ππ ) ππ‘ ) there’s an induced emf Ideal transformer: No flux loss π Since ππ < 1 Step down transformer so low emf across secondary coil Energy loss π Ohmic losses The primary and secondary coils get hot Flux losses Not all the flux stays in the iron core Hysteresis Magnetising and demagnetising the core losses produce heat Eddy current The changing flux in the iron core creates current in the core, which also generate heat, dissipate energy Make wire resistances small so heating losses are small Use soft iron core so the flux linkage is as large as possible & hysteresis losses are as small as possible Use laminated core, so the eddy current are as small as possible Power plant END TOPIC 8: NUCLEAR AND PARTICLES PHYSICS Name Definition Nucleon/ mass number Number of nucleons in the nucleus Proton/ atomic number Number of protons in the nucleus Notes A metal is heated Free electron gain KE Thermionic Emission KE > Φ the electron escape from the metal surface (how charged particles produced for use in particles accelerator) RUTHERFORD SCATTERING Rutherford’s Scattering Fire a beam of alpha particles at a very thin sheet of gold Count the number of α particles scattered at different angles Most go straight through θ ~ 0o Results Some α particles will be deflected by large angles (θ ~ 90o) A few α particles reflected/ go straight back (θ ~ 180o) Conclusion The atom is mostly empty All the positive charges and most of the mass is contained in a very small region Most does not get near enough to any matter to be affected Reasons Some came close enough to the charge to be affected A few deflected so nucleus must have mass much greater than the alpha particle mass to cause this deflection PARTICLE PHYSICS Particle Physics For every particle that is an identical particle with opposite electric charge called its antiparticles Antiparticles When a particle meets its own antiparticle, they annihilate, the energy released makes new particles De Broglie: π= Investigate Nucleons Structure π π To look at small distance λ must be small So p must be large So E must be large πΈ 2 = π2 π 2 + π2 π 4 If π β« ππ πΈ = ππ FUNDAMENTAL: not made out of other particles Leptons Electron Electron neutrino π− ππ Muon Muon neutrino π − Tau π − Up π’+ Quarks 2⁄ 3 Charm π+ 2⁄ 3 Top π‘+ 2⁄ 3 ππ Tau neutrino ππ Down π− 2nd gen Have a Leptons 3rd gen 1st gen 1⁄ 3 Strange π − 1st gen 2nd gen 1⁄ 3 Bottom π− 3rd gen 1⁄ 3 HADRON Baryons Mesons Proton Neutron Contains 3 quarks π+ π0 Baryon number π΅ = +1 Pions π+ π0 Contain π− 1 quark + 1 antiquark BOSON When particles interact, they are affected by one of 4 possible forces: Gauge Bosons ο¨ ο¨ ο¨ ο¨ Gravity (Graviton): act on energy Electromagnetism (Photon): charged particles Strong force (Gluons): quarks Weak force (W+, W-, Zo) log In Newtonian physics, we describe these forces using fields In quantum mechanics, the idea of fields is replaced by the transfer of particles called gauge bosons We then call these interactions, instead of forces Name Definition Notes PARTICLES ACCELERATOR When the next tube is positive the electron accelerates across the gap Inside each tube, the electron has constant v LINACS High-frequency supply ensure tube has the correct potential to accelerate the eAs particles are accelerated by the E field between the tube their speed increase The AC frequency is constant So the time inside each tube must be a constant = ½ period of the AC So the tube must be longer when vο The tube will increase in length until the speed reach the speed of light (constant) then the tube lengths become constant ππ π‘ 2ππ π= π΅π π΅π π= 2ππ π΅π = Cyclotron The e- accelerate across the gap end with speed v Inside the dee, the e- move in a semi-circle Time inside the dee π‘= So ππ π£ π ππ = π£ π π‘ π΅π = ππ π‘ E field produce a force Facing dee is always negative (for proton) Increases the KE of the particles across the gap βπΈπΎ = ππ B field causes the direction of the particles inside the Dees change Limitation: When vοc, cyclotron stop accelerating particle Newton’s Law of motion don’t apply when vοc Radius of orbit ο as energy ο but vο constant, so time inside dee ο so frequency ο Curvature Some particle tracks curve ‘clockwise’ others ‘anticlockwise’ Some have positive charge, some have negative charge Fleming’s left-hand rule tells us the sense of curvature Charge particles gain KE so p ο π ∝ π so rο The curvature decreases along the length Synchrotron Accelerate the particles with an electrical field Synchrotron vs Cyclotron Particle path is bent with a magnetic field The particles move in a circle Radius of path is constant As KEο, Bο to keep r constant As particle Eο, E field get stronger Because the particles are accelerating, they lose E by emitting radiation (synchrotron radiation) The magnetic field causes the track to bend Uncharged particles leave no track Electric field: Accelerate particle Direction of force indicates sign of charge Bubble π= Chamber πΈπ π Magnetic field: Circular motion Direction of curvature indicates sign of charge ππ£ π= π΅π Only moving charged particles leave a track Pion are charged so leave a track Pion interact with a stationary charged particle 2 neutral particles created (because no track) to conserve charge Track in different direction so momentum conserved Both particles decayed into opposite charged particle because charge is conserved At all collision momentum and charge are conserved Ad: lots of collision Dis: There’s momentum before collision so momentum after collision Fixed target Particles created must have KE So not all KE converted into mass Not many particles are created and their masses are not very big Ad: Final π = 0 so final KE is small Colliding beams All energy goes into making new particle ο can make new massive particles Dis: Not many collisions END TOPIC 9: THERMODYNAMICS Name Definition Formulae Specific heat capacity The amount of heat energy required to change the temperature Specific latent heat The amount of heat energy required per unit mass of a substance of a unit mass of a substance by a unit temperature to change the state from … to … at a constant temperature KINETIC THEORY Assumption: Kinetic model of gas Internal energy ο§ ο§ ο§ ο§ ο§ ο§ A gas is made of lots of particles Volume of particle << Volume of container Particles move at random Particles collide elastically (KE is conserved) No potential energy between particle (PE=0) The mean KE of particle is directly proportional to the temperature Random distribution of potential and kinetic energy among the molecules IDEAL GAS EQUATION Ideal gas equation 1 1 ππ = ππ < π 2 > = πππ = π < π 2 > 3 3 Initial x momentum: ππ£π₯ Final x momentum: −ππ£π₯ βπ = 2ππ£π₯ Time between collision: π‘= 2πΏ π£π₯ N2L, average force on wall πΉ= Derive ππ½ = βπΈ = ππβπ βπ 2ππ£π₯ ππ£π₯2 = = 2πΏ π‘ πΏ π£π₯ Total force on wall π π΅πππ π ∑ πππ ππππ‘πππππ ππ£π₯2 π = ∑ π£π₯2 πΏ πΏ Mean squared speed 2 < π >= Moving randomly: ∑ π£π₯2 ∑(π£π₯2 + π£π¦2 + π£π§2 ) π = ∑ π£π₯2 ∑ π£π¦2 = ∑ π£π§2 3 ∑ π£π₯2 π π < π2 > ∑ π£π₯2 = 3 < π2 > = βπΈ = πΏβπ Therefore, total force πΉ= π πΏ × π<π 2 > 3 Pressure π= πΉ ππ < π 2 > = π΄ 3πΏ3 But π = πΏ3 so 1 ππ = ππ < π 2 > 3 Internal energy = KE + PE Fr ideal gas, PE = 0 so Internal energy = KE Boyle’s Law Pressure Law Charles’ Law Absolute zero 1 1 π = ∑ π(π£π₯2 + π£π¦2 + π£π§2 ) = ππ < π 2 > 2 2 2 ∴ ππ = πππ = π 3 1 3 ∴ π = πππ 2 = πππ 2 2 For a fixed amount of an ideal gas at a constant temperature: π1 π1 = π2 π2 Its pressure is inversely proportional to its volume For a fixed amount of an ideal gas at a constant volume: Its pressure is directly proportional to its temperature For a fixed amount of an ideal gas at a constant pressure: Its volume is directly proportional to its temperature The temperature at which the pressure/ volume of a gas become zero π1 π2 = π1 π2 π1 π2 = π1 π2 1 3 πΎπΈ = π < π 2 >= ππ 2 2 BLACK BODIES MaxwellBoltzmann Distribution How many molecules will have a speed in a small range of speed At every temperature above 0K objects radiate energy as electromagnetic wave Black bodies radiator A blackbody absorbs all the radiation that falls on it Total energy radiated per second only depends on the surface area A and the absolute temperature T StefanBoltzmann Law The total amount of energy radiated per second is proportional to the surface area A and the absolute temperature πΈ = ππ΄π 4 ππππ₯ π = 2.898 × 103 = ππππ π‘πππ‘ Wein’s Law END TOPIC 10: SPACE Name Definition Note STELLAR PARALLAX Intensity/ Flux πΌ= πΏ 4ππ2 π= 1π΄π π The apparent shift in the position of a nearby star, relative to more distant ones, due to the moment of the Earth around the Sun. Stellar Parallax The star is viewed from two positions at 6month intervals The change in angular position of the star against backdrop of fixed stars is measured Use trigonometry parallax to calculate the distance Parsec The distance that a star would be if it had a parallax of 1 arcsec Elliptical orbit Over the course of one year the stars will trace out an elliptical path on the sky. Stars have orbits not perpendicular to earth will appear to have elliptical orbits because only see the projection of the diameter. Large distance When d is large θ is small so the fractional uncertainty is large, therefore there is a large fractional uncertainty in the calculated value of d. Since Mars is farther away from the Sun than the Earth, for a given parallax we can calculate a larger value of stellar distance. STANDARD CANDLES Standard candle An astronomical object whose luminosity is know Giant stars that become unstable and pulsate: their diameters oscillate and therefore they vary in luminosity Cepheid variables Cepheid variables out ward pressure P and inward gravity compression are out of sync so the star and temperature pulsates Determine distance to Cepheid Measuring period T πΏ = 4πππ 2 π 4 give luminosity πΏ Light flux can be determined πΌ = 4ππ2 π(π. π . ) = 1 π(′′) Inverse square law gives the distance The explosion of stars that have run out of fuel for nuclear fusion in their cores. Type 1a are standard candles Supernovae Type 1a supernovae are extremely luminous they can be seen from a very large distance The light curve must be calibrated by using Cepheid variables to determine the distance to a galaxy that contains a type 1a supernovae HR DIAGRAM HR diagram A Luminosity-Temperature diagram Main sequence Stars that convert Hydrogen into Helium via thermonuclear fusion in the core Blue giants Large mass, high temp and luminosity Red giants Low temp, high luminosity, converting He4 to C-12 and O-16 Core of a red giant star Do not have fusion reaction Radius is very small πΏ = 4πππ 2 π 4 so luminosity is low White dwarfs Surface temperature is high, πππππ is in UV spectrum Emits a lot of light in visible spectrum so appear white Stars are very good black bodies. The total radiations they emit per second only depend on the surface area and the absolute temperature. They obey Stefan’s law: Star πΏ = 4πππ 2 π 4 And Wein’s law: πππππ π = 2.898 × 103 A star position on the HR depend on its mass and its age Stars are large ball of gas (mostly hydrogen, helium) Life cycle of the stars Gravity cause a large cloud of gas and dust to collapse & heat up When neutral temperature reach ≈ 106 πΎ, nuclear reaction starts in the centre, H is converted to He A star is born, its life cycle of a star depends on its mass Young star groups have more red giant stars Planetary nebula Pulsar Gaseous nebula Shell of gas ejected from RG star on its way to becoming a WD Rotating neutron star with a very string magnetic field Pulsars beam radiation out along their magnetic axis Large cloud of gas & dust. They have very low temperature and density Name Definition Notes Doppler’s effect CMBR Cosmic Microwave Background Radiation: Come from all part of the sky Its intensity is almost the same in every direction Black body radiation produced in the hot Big Bang Whose wavelength have been stretched by the cosmological expansion The peak wavelength Is now in the microwave part of the spectrum It implies that the temperature after Big Bang was very high If the temperature was exactly uniform across the sky, the density of the universe would be exactly uniform Gravity would not be able to form structures such as galaxy, stars and planets Low temperature region has higher density and will collapse first to form galaxy Hubble’s Law The recession velocity of a galaxy is directly proportional its distance from our galaxy It implies that in the past the universe was smaller By extrapolating backward far enough, everything in the universe was at the same location: a point of infinite density and temperature, The Big Bang Hubble’s parameter The gradient of the Hubble’s law graph The present value is π»0 = 71 ππ/π /πππ Hubble’s constant not very accurate Because the distances to the galaxies are underestimated hence gradient is not as steep as in Hubble’s graph Dark matter Material that does not interact via the electromagnetic force. Its gravity may be responsible for explaining the rotation curves of galaxies and the stability of the galaxy clusters π£ = π»0 π Cosmological redshift The increase in wavelength of radiation from distant galaxies due to the expansion of the universe Redshift The fractional increase in wavelength of light emitted by a source and detected by an observer due to the relative motion between them Light from almost all galaxy are redshifted ππππ πππ£ππ > ππππ Due to Doppler effect galaxy are moving away from us Hubble’s law so distance between galaxy is increasing So, the universe is expanding Big Bang Nucleosynthesis The early universe was extremely hot and dense, the condition is suitable for thermonuclear fusion to occur DARK MATTER In order to account for the measured shape of the graph there has to be more mass than can be accounted for by the visible matter. This extra mass is called dark matter Dark matter does not emit electromagnetic radiation, but it has gravitational effects The dark matter affects the gravity of the universe, which affect the rate at which the universe expands, so it affects whether the universe is open, closed or flat Because the total density of the universe is uncertain, the future of the universe is uncertain π§= π0 ππ π£ = ππ π TOPIC 11: NUCLEAR RADIATION Name Definition Formulae NUCLEAR DECAY Nuclear decay Randomly: It is unpredictable which nucleus will decay next and when it decays Spontaneous: the rate of decay cannot be changed by changing the external conditions (temperature, pressure, etc.) Radioactive isotopes Isotope has an unstable nucleus, decay and emit radiation Alpha decay Alpha particles: ο§ ο§ ο§ ο§ Beta decay → 14 6πΆ 14 7π Beta decay ο Z by 1 Beta particles π: neutrino Moderately ionizing Range in air bout 1m Stopped by thin metal Deflected by magnetic fields (opposite direction to alpha) 234 90πβ + 42πΌ Alpha decay ο Z by 2, A by 4 Beta particles are high-speed electron emitted by the nucleus ο§ ο§ ο§ ο§ Gamma decay Strongly ionizing Short range in air Stopped by paper Deflected by magnetic fields 238 92π → + −10π½ + πΜ π ππ : electron neutrino πΜ π : anti-electron neutrino Gamma rays are high energy EM radiation (photon) Gamma rays: ο§ ο§ ο§ ο§ Weakly ionizing Obey inverse square law in air Stopped by 1m concrete Not deflected by magnetic fields HALF-LIFE π = π0 π −ππ‘ Half-life The time is taken for the number of radioactive nuclei to reduce into half of its initial value Decay constant The probability that a given nucleus will decay in one second π= The rate of decay of unstable nuclei π΄ = ππ Activity Unit: Bq (Becquerel) Rate of production The rate of production of C-14 (etc.) decrease The ratio was greater Ratio used is from current time not from the past ππ2 π1⁄ 2 π΄ = π΄0 π −ππ‘ So, the time is underestimated BACKGROUND RADIATION Background radiation Radioactive isotopes in the environment Sources of radiation: rocks, air, water, cosmic rays Background radiation may affect cancer rate, responsible for some mutations that drive evolution Before plotting activity graph the count rate must be corrected for background, otherwise π1⁄ will be overestimated 2 END Name Definition Formulae BINDING ENERGY Mass defect Free nucleons have more energy than when they’re trapped in the nucleus. 2 According to Einstein, πΈ = ππ so if the energy of the nucleus increases the mass must increase βπ = πππ π ππ ππ’ππππππ −πππ π ππ ππ’ππππ’π Since the mass of proton/neutron is constant, the mass of the nucleus < total mass of proton/neutron in it. Nuclear binding energy The energy needed to separate all nucleons in the nucleus π΅. πΈ. = βππ 2 πΉπ − 56 is the most stable isotope For A>56 the BE/nu decrease So required net energy input to undergo fusion So does not occur in massive stars FISSION Nuclear fission Split a large nucleus into small nuclei Release energy because the BE/nucleons of the fragment increase ο the energy is released in the reaction, provided that we do not pass the peak Number of neutrons always increase Chain reaction More than 1 neutron is produced in the reaction. Each neutron can induce further nuclei to fission The reaction grows exponentially Fissile Nucleus can be split by slow neutron π΄ππ‘ππ£ππ‘π¦ × πΈ πππ πππππ‘πππ Rate of energy radiation Rate of temperature Most KE released is carried by the alpha particles which escapes, so it does not heat the metal. ππ ππ = ππ ππ‘ ππ‘ increase So, rate of T is likely to be overestimated Radioactive waste Total activity is underestimated All isotopes produced in the decay will be radioactive, so they contribute to the total FUSION Nuclear reactor Pros: ο§ Lots of energy/kg of fuel ο§ No CO2 emission Cons: ο§ Radioactive waste must be stored for thousands of years ο§ Possibility of radiation escape during accident ο§ High cost of building reactors and decommissioning Nuclear fusion Joining 2 or more light nuclei into a heavier one and release energy Sustained fusion High energy/ temperature ο The particles have enough kinetic energy to overcome electrostatic repulsion ο They come close enough for fusion High density/ pressure ο Ensure that the reaction rate is high Fusion reactors Pros: ο§ Unlimited supply of fuel ο§ Little radioactive waste Cons: ο§ Very expensive, requires extremely high T, P ο Container problems ο§ Strong magnetic field required END TOPIC 12: GRAVITATION Name Definition Formulae GRAVITATIONAL FIELDS A gravitational field is caused by mass & it affects mass Gravitational fields Gravitational field strength Gravitational field lines show the direction that a positive mass will move in that field. The field line spacing tells us the field strength Force acting on unit mass in the field π= πΉ π β GPE Work done in moving a distance h in the field ∫ πΉππ₯ π = ππβ Total energy πΈ = πΎπΈ = − πΊππΈ 2 Gravitational potential π·=− βπ· = πβ βπ· π=− ππ₯ Gravitational Potential Change in GPE per unit mass Change in gravitational potential Distance R The attractive gravitational force between two point mass Newton’s law of universal gravitation Is directly proportional to the product of their mass And inversely proportional to the square of distance between them Escape velocities πΊπ π The speed of the object so that it just reach ∞ πΊπ π2 πΊππ πΉ= 2 π π= 2πΊπ π£ππ π = √ π KEPLER LAW OF PLANETARY ORBITS Kepler’s first law The planets orbit the sun in elliptical orbit with the sun at one focus of the ellipse Kepler’s second law The line joining the planet to the sun sweep out equal area in equal times Kepler’s third law If T is measured in year, d is measured in AU, π 2 = π3 π2 =( 4π 2 ) π3 πΊπππ’π SATELLITES π£2 = Near Earth orbit Geosynchronous πΊπ π Above equator, π = 24β → π = 4.2 × 107 BLACK HOLES Schwarzschild radius Radius of a black hole of mass M For Earth π π = 8.89 × 10−3 π END π π = 2πΊπ π2 TOPIC 13: OSCILLATION Name Definition Formulae SIMPLE HARMONIC MOTION Simple harmonic motion Occurs when there is a force always act toward equilibrium point and the force is directly proportional to the displacement from equilibrium πΉ = −ππ₯ π₯ = π΄ cos ππ‘ π£πππ₯ = π₯0 π π£ = −π΄π sin ππ‘ ππππ₯ = π₯0 π2 π = −π΄π2 cos ππ‘ = −π2 π₯ Equation of simple harmonic motion π= π2 π₯ π2 π₯ → 2 = −π2 π₯ ππ‘ 2 ππ‘ π₯ = π₯0 cos ππ‘ π₯ = π₯0 sin ππ‘ This equation has 2 solutions that tell us how x changes with time π₯0 is the maximum displacement from equilibrium = amplitude Angular frequency π π π=√ =√ π π RESONANCE Occur when the driving frequency is close to the natural frequency Resonance Maximum energy transferred from the driver to the oscillator The amplitude of oscillation increases rapidly/ the oscillation is amplified The amount of amplification ο as damping ο (the width of the curve ο) DAMPING A resistive force that opposes the natural motion of an oscillator Damping Energy is dissipated from the oscillation So, the amplitude of the oscillation decrease Light damping Heavy damping Critical Damping With air resistance, T does not change The amplitude decreases exponentially No oscillation The object returns to equilibrium point slowly The most efficient way of removing energy from an oscillator END π = π 0 π −ππ‘ cos ππ‘
