A Level Physics Hidden Formulae

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  AQA A-Level Physics Component Code 7407 and 7408  Note foruser:This resources was created by an A Level Student (momondion TES) in2017 in preparation for upcoming A level Exams. As such if there are any errors, do contact the author of this resource. This is intended as a guide for useful formulae that can be used in exam questions and to help build understanding on the topics studied at AQA Resources referred to making this resource: CGP A Level Year 1  – The Complete Course for AQA CGP ALevel Year 2  – The Complete Course for AQA Advanced Physics For YouAQA SpecificationAQA A Level Physics formula sheet  Key Blue equation  – Given formulae Red equation  – Not given formulae ℎ =   () ℎ =       = ℎ = ℎ    ℎ =  This is the charge to mass ratio for a particle/nucleus/ion. ( − ) 1 = 1.6×10 −  ℎ = 6.63×10 −   Planck’s constant  = ℎ = ℎ The energy of a photon of EM radiation is directly proportional to its frequency  = ℎ  The work function ,  is the minimum amount of energy needed to release an electron from the surface of a metal The threshold frequency ,    is the minimum frequency of EM radiation needed to release an electron from the surface of a metal The kinetic energy of photoelectrons is up to a maximum because: ã Photons are absorbed by electrons on a one to one basis ã Some electrons are deeper in the metal than others ã This means that they require more energy than the work function to be released, so their kinetic energy will be lower than those from the surfaceParticles with a larger momentum will have a smaller de Broglie wavelength.    ()  =   Stopping potential,   is the potential difference needed to stop the fastest moving electron in a photocell experimentThe de-excitation of an electron between two energy levels will emit a photon of EM radiation.An electron must gain a specific amount of energy in order to be excited to a higher energy level    =  2  Pair Production: The minimum energy needed for pair production is equal to double the rest energy of one particle produced    =    Annihilation The minimum energy of one photon produced is equal to the rest energy of one of the particles which have been annihilated  Key Blue equation  – Given formulae Red equation  – Not given formulae    =    = sin =   sin   =   sin    = 12  = 1 =  = sin   =        =         ∝ ()  ℎ  = 2  =   The frequency of the first harmonic on a stringWhere  is the total mass of the string divided by its total length (mass per unit length,  − ) .The frequency of the 2 nd , 3 rd , 4 th harmonics etc will be a multiple of the first harmonicYoung Double Slit Experiment ã   is the fringe spacing ã   is the distance from the screen to the slit ã   is the slit separationDiffraction Gratings ã   stands for the order, the position of a maxima ã For a fixed slit width,  , light with a longer wavelength will be diffracted more Intensity is the energy transferred per second over an area of 1  ( − ) The refractive index must always be higher than 1! The higher the number, the more optically dense the medium is   is the more dense medium in this caseThe critical angle can only occurs when light goes from a more dense to a less dense medium.This means that   must be less dense in this case Note: If you get math error, the refractive indexes are the wrong way round Snell’s law   is the angle of incidence   is the angle of refraction
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