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Thermionic Emission Properties of electrons ( Cathode Rays )1. They carry unit negative charge .2. They can be deflected by magnetic and electric fields ( they are charged ).3. They cause certain substances ( zinc sulphide ) to fluoresce. 4. High speed electrons produce X Rays when they strike a heavy metal target.Note: Discovered by JJ Thompson Named by GJ Stoney ( Irish ) Milikan calculated the charge on the electron in the oil drop experiment.Thermionic emission : Is the emission of electrons from a hot metal surface.The Cathode Ray Tube 9759955067301. The low voltage filament heats the cathode this results in thermionic emission.2. A vacuum tube has no gas molecules into which the electrons can collide and loose energy.3. There is a very high voltage between the anode ( + ) and the cathode ( - ) this accelerates the electrons. ( potential energy = eV ) 4. The anode focuses the beam of electrons on to the fluorescent screen.5. The X and Y plates move the beam of electrons in the vertical or horizontal planes using either electric or magnetic fields.Uses : TV SET, Oscilloscope, ECG. I V Characteristic In a Vacuum ( Electrons, produced by thermionic emission ) O to A : As the voltage is increased the electrons cross the tube and the current increases. A B A to B : No more electrons are released. I Further increasing the voltage causes no increase in the current Example : Cathode ray tube O V O VEnergy of electrons in an electric field The electrons gain potential energy from the high Voltage PE = e V As the electron moves through the tube all the potential energy is converted into kinetic energy KE = ? m v2Potential energy Kinetic energy e = the charge on the electron m = the mass of the electron eV ? m v2 V = applied voltage v = velocity of the electron To convert eV to Joules eV x Charge on the electron Force on a moving charge (q) in a magnetic field We know the force on a current carrying conductor ( tinfoil ) in a magnetic fieldF = B I L I = q Current = charge t time F = B q L v = L velocity = length t t time F = B q v q = a single charge if q = one electron ( e )F = B e vNote : An electron moving at a constant speed at right angles to a uniform magnetic field will move in a circular path.This movement can be described in terms of the force on the electron F = B e v Or it can be described in terms of centripetal force F = mv2 R From the above for an electron moving at right angles to a magnetic field Centipetal force = Force on an electron due to a magnetic field m v2 = B e v R m = mass of the electron v = velocity of the electron e = the charge on the electron R = radius of the orbit See page 332 Problem 7 and 2003 Q 9 2003 Q9PD = 4 x 103V e- = 1.9 x 10-19C me = 9.1 x 10-31 Kg(i) Find the Potential energy ( PE ) gained ( the electron gets it`s PE from the applied voltage) PE = ( charge on the electron ) x ( applied voltage ) PE = ( 1.6 x 10-19 ) x ( 4 x 103 ) = 6.4 X 10-16 Joules (ii) Find the speed v of the electron ( all the PE is converted into Kinetic energy ) loss of PE = gain in KE (6.4 x 10-16) = ? mev2 (6.4 x 10-16) = ? ( 9.1x10-31) x v2 2x (6.4 x 10-16) = v2 ( 9.1x10-31) 3.7x107 ms-1 = v ( velocity of the electron ) (iv) Find the force acting on the electron B = 5 x 10 -2 T e- = 1.6 x 10-19C v = 3.7x107 ms-1 me = ( 9.1x10-31 Kg) F = B e v F = ( 5 x 10 -2) x (1.6 x 10-19) x (3.7x107 ) F = 3 x 10 -13 N (V) Find the radius ( r ) of the circular path followed by the electron According to the left hand rule an electron travelling at right angles to a magnetic field follows a circular path .Therefore the force acting on the electron can be expressed as centripetal force F = mv2 r or as the force on the electron F = B e v Bev = mv2 r ( 3 x 10 -13 ) = ( 9.1x10-31) x ( 3.7x107 )2 r r = 4.3 x 10 – 3 m X Rays : Are high frequency electromagnetic radiation produced in a vacuum tube when high speed electrons strike a heavy metal target.Discovered by Roentegen ( 1895 )X Ray tube 12014202546351. At the cathode thermionic emission occurs producing a beam of electrons.2. The very high voltage accelerates the electrons to high speeds towards the anode.3. When the electrons strike the Tungsten target (i) 1% of the electrons produce X Rays (ii) 99% of the electron energy is turned into heat This is removed by the cooling fins. 4. The target must have a high melting point, this is why Tungsten ( a heavy metal ) is used. 5. The Lead shield does not allow X Rays to pass through it. Penetrating power of X Rays Penetration Frequency Frequency applied voltage Properties of X Rays 1. Unaffected by electric and magnetic fields. ( => they carry no charge ) 2. Affect photographic plates.3. Ionise materials they pass through ( causing cancer )4. They penetrate materials, the denser a substance the more it absorbs X Rays.Uses of X Rays: Medicine (photograph bones, destroy cancer cells ) Detection of cracks in metals.Note : In X Ray production High speed electrons are causing X Rays ( electromagnetic radiation ) to be released from a heavy metal. In the Photoelectric effect Electromagnetic radiation cause electrons to be released from a metal The Photoelectric effect : Is the emission of electrons from a metal surface when hit by electromagnetic radiation of a suitable frequency Hallwacks (1888)Photoelectric effect is the opposite of X ray production because Photoelectric effect: electromagnetic radiation hits the metal causing electrons to be released X Rays : High speed electrons hit the heavy metal causing the release of X Rays ( electromagnetic radiation )To demonstrate the photoelectric effect : (1) Put a negative charge on the electroscope and shine a UV light on the Zinc Result : The leaves fall Conclusion : UV light causes electrons to be emitted from the zinc (2) Put a sheet of glass between the UV light and the negative electroscope Result : The leaves do not fall Conclusion : The UV light cannot get through the glass (3) Put a positive charge on the electroscope and shine a UV light on the zinc Result : The leaves do not fall -19685195580 Conclusion : The electrons emitted are immediately attracted back by the positive charge on the zinc ( also: The loss of electrons increases the positive charge on the electroscope) -48260249555The Photocell : Conducts electricity when light of a suitable frequency falls on it (it acts like a switch) 1526540450215 No current flows unless light of a suitable frequency falls on the cathode Uses : Light meters in photography Galvanometer The Nature of light :1. Newton (1600) light is a stream of fast moving particles ( corpuscles ) ( Photons )2. Huygens (1600) light is a wave motion.( could not show diffraction and interference)3. Young 1802 showed the interference of light waves ( light is a wave motion )4. Einstein 1905 when trying to explain the photoelectric effect concluded light has a dual nature both a wave motion and particles ( photons ) rightcenterTo demonstrate the action of a photocell Method : 1. Fix the distance between the light and the cathode. 2. Change the filters (change the frequency of the light ) 2.Record the photocurrent. Result : For any metal there is a certain frequency (f0)below which photoemission (release of electrons)will not occur. Method : 1. Put in place one filter of suitable frequency (> fO) 2. Move the light closer to the photocell. 3. For each position of the light record the photocurrent. Photocurrent Result : Photocurrent 1/(distance)2 light intensity 1/(distance)2 Note: 1. The frequency of the light controls whether or not electrons are emitted. Below fO no electrons are released. f > fO will not increase the number of electrons released (size of the current ) it will increase the kinetic energy of the electrons 2. The intensity of the light controls the number of electrons emitted.( the current ) Einstein`s Photoelectric law Light consists of tiny bundles of energy called photons. The energy in each photon is : Energy = ( Planks constant ) x (frequency) E = h f Each photon hits one electron and it`s energy is transferred to the electron.If the frequency of the photon is the threshold frequency fO then all the energy is needed to remove the electron from the metal ( work function ) hfo = If the frequency f > fO hf = + KEmaxany extra energy the photon has is given to the electron as kinetic energy hf = + ? mv22004 Q 9 . = 240 x 10 -9m = 4.3 eV h = 6.6 x 10 -34J c = 3 x 108 ms-1 1eV = 1.6 x 10-19J(1) Find the threshold frequency of the Zinc Convert eV to Joules (4.3) x (1.6 x 10-19) = 6.9 x 10-19 J hfo = => fo = fo = 6.9 x 10-19 = 1.04 x 1015 Hz h 6.6x10-34 (2) The maximum kinetic energy of an emitted electron Find the frequency of the photon C = f f = c f = 3 x108 = 1.25 x1015 Hz 240 x 10 -9hf = + KE max = > hf - = KEmax (6.6 x 10 -34 ) ( 1.25 x 10 15 ) – (6.9 x 10-19) = 1.39 X 10 -19 J = KE max ................
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