DO PHYSICS ONLINE IDEAS TO IMPLEMENTION QUESTIONS & PROBLEMS SOLUTION QUIDELINES How to answer a question: problem solving (t0_372.pdf) p1.23 Waves transfer energy without any material being transformed. At first cathode rays were though to be waves like X-rays which has been recently discovered. Cathode rays were deflected both by electric and magnetic fields indicating that cathode rays were a stream of negatively charged particles and J J Thompson measured the qe/me ratio providing compelling evidence that cathode rays were a beam of negatively charged electrons. p1.28 m = 10-30 kg q = +6×10-12 C a = 7.0×1021 m.s-2. p1.30 force on a moving charge in a magnetic field F force on charge particle [N] magnitude only B strength of uniform magnetic field [T tesla] magnitude only magnetic flux density q charge of particle [C coulomb] magnitude only v velocity of charged particle [m.s-1] magnitude only angle between magnetic field lines and current direction magnitude only = 0 F = 0 current direction & magnetic field
direction parallel = 90o F = Fmax current direction perpendicular to magnetic field direction p1.34 p1.44 B = 1.2310-3 T d = 22.5 m = 2.2510-3 m accelerating voltage Va = 10.5 kV = 10.5103 V (a) v = ? m.s-1 accelerating voltage does work on the electron increasing its kinetic energy (b) hand screw rule – magnetic force directed up the page (c) magnetic force & electric force must cancel – electric force directed down the page (d) Electric field between the plates – same direction as force on a positive charge – up the page (e) voltage across the plates – bottom plate is positive (electron attracted to bottom plate) (f) J.J Thompson’s e/m experiment From measurement of E (or V and d), B and Va the e/m can be found. Alternatively: Thomson adjusted the electric field strength and magnetic field strength to make the electrons pass straight through. This allowed him to calculate the velocity v of the electrons (cathode rays). He then passed the electrons through the magnetic field B alone, and from the radius R of their path calculated the q/m ratio p1.65 (a) striations (b) right hand side is the cathode – striations are nearer than anode (c) See notes Electrical discharges (d) Pattern will not significantly change but molecules of different gases emit radiation of different frequencies in such transitions. Therefore, the glow of the positive column has a characteristic colour for each gas. (e) The pressure of the gas is mainly responsible for the observed patterns.
C plates must be conductors plates must be conductors p1.94 (a) The shadow of the cross showed that the cathode rays travelled in straight lines. Without the cross there is no shadow. (b) The glass fluoresces because the energy carried by the electrons are absorbed by the glass molecules. The molecules of the glass become and excited and then loss energy by emitting photons of visible light. (c) The cathode rays are negatively charged electrons – negative charged particles moving in a magnetic field experience a force – electron beam deflected causing a distortion of the shadow of the cross. (d) Electrons are constituent of atoms, investigate properties of electrons, measure the qe/me ratio. (e) Need a high voltage source such as an induction coil ******************************************************************* p2.05 across the gap of the transmitter produced the induction coil emitted electromagnetic radiation which included radio waves and UV radiation. When a sheet of glass was placed between the transmitter and receiver, the spark in the receiver was weaker due to the absorption of the UV by the glass. Hertz noted the observation but did not follow it up. The UV absorbed by the metal receiver enabled electrons to be mitted more easily giving a bigger spark – photoelectric effect. p2.06 Planck energy of atomic oscillators is quantized E = h f Einstein particle model for light – light consisted of photons – energy of each photon hf An electron emitted immediately from the surface by the interaction with a photon Applied the Law of Conservation of Energy Energy of incident photon = Energy to release electron from surface + KE of ejected electron h f = ½ m v2 + = ½ m vmax2 + W W is the min energy needed to remove an electron from the surface – work function
electrons released from hot filament – thermionic emission large (bulky) hot – used lots of energy diodes – rectification triodes – amplifiers, switches state devices - miniaturisation diodes (pn junctions) transistors (npn, pnp) IC - integrated circuits consume little power Superconductor – zero resistance below critical temperature Applications – solid state devices switching amplification Applications – superconductors power transmission magnetic levitation p3.52 p3.55
m013 When the temperature of a metal is decreased its electrical resistance decreases because conduction electrons loss less energy in collisions with the lattice ions whose vibrations decrease by lowing the temperature. m020 Power transmission – see notes Magnetic levitation – trains – see notes m045 f = 102.8 MHz = 102.8106 Hz E = ? J = ? eV = ? m qe = 1.60210-19 C c = 3.0 m.s-1 h = 6.62610-34 J.s E = h f E(eV) = E(J)/qe c = f = c / f E = 6.810-26 J = 4.310-7 eV very small photon energy since in radio part of em spectrum = 2.9 m long wavelength because of small frequencies m047 (a) Cathode rays were deflected by electric and magnetic fields negatively charged particles. J.J. Thompson was able to measure qe/me ratio electrons (b) To produce an electrical discharge – need very high voltages > 10 000 V (c) make sure that your do not come into contact with the high voltage source m049 proton q = 1.610-19 C v = 6.00 m.s-1 B = 1.82 T R = 0.350 m m = ? kg m0 = ? kg Magnetic force = Centripetal force m050 Paddle spins when struck by cathode ray beam beam carries momentum m065 CRO step-up transformer Transistor circuit step-down transformer
Planck and Einstein good friends – supported each scientific research but their views on the role of science and scientific research were very different. Einstein Activist for world peace and was only one of four scientists to sign a partition for peace at the start of WW1. Maintained that scientific research should be removed from social and political forces and devoted to the pursuit of knowledge and understanding. Did support the development of the atomic bomb. WWII, totally opposed the research, development and use of nuclear weapons – warned of the devastating energy that could be released. Planck Staunch patriot – support the German government and the rise of the Nazi regime. Work for the the Nazi during WWII. Science to support a social and political agenda. Rapid progress in German science as a benefit to all Germans and a clear indication of their superior abilities.