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•Set up the circuit to supply an opposing Voltage to measure the (maximum) kinetic energy of the electrons •Adjust the variable voltage supply until the current is zero. This voltage is called the cutoff voltage. •Note how the cell is arranged – where it is ‘pushing’ electrons •Measure the cut-off voltage for different light frequencies and plot data on a graph. •Convert Voltage to Kinetic energy Ek V= Energy / charge so E = qV Ek = eV where e is the charge on 1 electron 1 electron volt =1 eV = 1.6 x 10-19 J V f f Kinetic energy vs. frequency: Ek Ek hf -ϕ Gradient = fo h f Terms & Definitions: Threshold frequency (fo) the minimum frequency of light that results in the emission of photoelectrons Work function (ϕ) the minimum amount of work needed to eject an electron from the surface ϕ = h fo Stopping potential (V) equivalent to the maximum kinetic energy of the ejected photoelectrons (Ek) What is the gradient? A beam of 640nm red light shines onto a polished copper disc. 1. Calculate the energy of each photon E=hf=hc/=3.11x10-19J 2.Write this energy in eV 1eV=1.6x10-19J 1.9eV The work function of copper is 1.28x10-18J 3.Figure out whether an electron will be emitted Φ = 1.28 x 10-18J. Ei = 3.11x10-19J is about ¼ of what's needed NO electron emitted 4.What about 160 nm light? 1.24x10-18J. Still not quiet enough 5.What is 160 nm Light? UV Blue light wavelength 429nm is shone on a metal surface with threshold frequency 3.0x1014Hz. What is the maximum velocity of the electrons (mass=9.1x10-31kg)? Step one find the frequency of the light. Step two calc the photon energy 4.6x10-19J Use hfo=Φ, Ek=hf- Φ, Ek=½mv2 to get v v=7.6x105ms-1 7.0x1014Hz A sodium surface is illuminated with light of wavelength 300nm. The work function for sodium metal is 2.46 eV. Find: a) The Maximum kinetic energy of the ejected photoelectrons b) The threshold frequency (& wavelength) for sodium Solution a) The energy of the photons in the illuminating light beam is E = hf = hc/ E = (6.626 x 10-34 J.s)(3.00 x 108 m/s) / 300 x 10-9 m = 6.63 x 10-19 = 6.63 x 10-19 J / 1.60 x 10-19 J/eV = 4.14 eV Using equation Ek max = hf - Φ we have Ek max = 4.14 -2.46 = 1.68 eV = 2.69 x 10-19 J Solution b) At the Threshold frequency Ek max = 0. So Φ = hf o since Φ = 2.46 eV = (2.46eV) (1.60 x 10-19 J/eV) = 3.94 x 10-19 J Then fo = Φ /h = 3.94 x 10-19 J / 6.626 x 10-34 J.s = 5.95 x 10 14Hz = 505 nm This is in the green region of the visible spectrum What did you take in? Try and answer these without looking back at your notes 1) What is a photon? 2) Which theory of light does the photo electric effect support? 3) What quantity dictates the amount of energy a photo electron contains? 4) What is an electron-volt? 5) Does dull light produce photo-electrons? 6) What happens when the same light is brightened up?