13.1 Photoelectric Effect.notebook March 11, 2015

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2 13.1 Photoelectric Effect.notebook March 11, Quantum Physics Quanta, Photons & the Photoelectric Effect Robert Millikan Early Quantum mechanics demonstrated that the charge iof an electron is comes in packets or "quanta", of 1.6x10 19C and all the charges were found to be integral multiples of a basic charge and no smaller. In the early 1900's Max Planck analyzed the spectrum of hot gases & solids. He theorized a "black box" capable of absorbing (cool) or emitting (hot) all frequencies of the electromagnetic spectrum (light) which follows from Stefan Boltzman's Law that says that energy radiated in time, Power, P, is proportional to temperature in Kelvin, T 4, (P α T4), The frequency of the elecromagnetic spectrum emitted is described by Planck as being given in the form of packets or "quanta" of energy. The intensity of the spectrum at 5000 K is shown in blue, the electrons in their shells are falling and giving off energy is discreet quanta, unlike the black line which is what we would expect if energy was continuous. therefore E = nhf where h is Planck's constant (6.63x10 34Js) and n is the number of packets. photons of light electrons ejected (photoelectrons) metal object (zinc) When light falls on a metal surface like zinc, which is attached to an ammeter, a current can be observed to flow. Einstein deduced that this was due to the photons of light having the right amount of energy to transfer to the electons in the metal to eject the electrons from the metal. This is known as the Photoelectric Effect (he won Nobel Prize in 1921 for this not relativity). Einstein theorized that energy in the form of packets (quanta) was responsible for this effect and that the right number of packets would eject the photoelectrons. He able to show that this energy was also equal to E = nhf. 2

3 Photoelectricity was first discovered in 1887 by Heinrich Hertz during investigations into radio waves using a spark gap. Radio waves are produced when a high voltage is supplied across two electrodes causing a spark in the gap. Hertz found that if ultraviolet light was shone on the electrodes, the sparks were much stronger and thicker. 3

4 Powerful red laser No electrons released Powerful blue laser electrons released 4

5 What role does wavelength play in the photoelectric effect? 5

6 Photoelectric Effect The kinetic energy of ejected electrons from a metal (called photoelectrons), is not determined by the intensity or frequency of the light. A threshold frequency is needed for electrons to have the energy to be ejected, so E=hf. Each metal needs its own distictive amount of energy to liberate electrons called a Work Function, φ, an electron absorbing a photon of whose energy is E, greater than φ, will have a kinetic energy E φ. K max = hf φ If f < φ/h then there is not energy to eject electrons. f o = φ/h is the threshold frequency. Using electronvolts instead of Joules is benefical due to the small energies involved: 1eV = 1.6 x J therefore h = 4.14 x evs The maximum kinetic energy of the photoelectrons can be determined a a stopping potential V = W/q = KE max /e KE max = ev 6

7 Robert Milliken verified Einstein's photoelectric equation, and made the first direct photoelectric determination of Planck's constant h by using a stopping potential. 7

8 The work funtion φ, for aluminum is 4.08 ev. a) What is the threshold frequency required to produce photoelectrons from aluminium? The minimum energy needed is 4.08 ev so the threshold frequency is f o = φ/h = 4.08eV/4.14 x evs = 9.86 x Hz b) Classify the elecrtromagnetic radiation that can produce photoelectrons. This is in the ultraviolet region of the EM spectrum. c) If light of frequency f = 4.00x10 15 Hz is used to illuminate a piece of aluminum, i) what is K max, the maximum kinetic energy of ejected photoelectroons? K max = hf φ = (4.14 x evs)(4.00x10 15 Hz) (4.08 ev) = 12.5 ev ii) what's the maximum speed of the photoelectrons? (electron mass 9.11x10 31 kg)? KE = 1/2 mv 2 so v 2 = 2KE/m e = 2(12.5 ev)(1.6 x J/eV)/9.11 x 10 6 kg so v = 2.1 x 10 6 m/s d) If the light described in part b were increased by a factor of 2 in intensity, what would happen to the value of K max? More intensity means more photos striking the metal which will eject more photoelectrons, however the kinetic energy will remain the same nothing will happen! the only way to increase the KE is to increase the frequency, not the intensity. 8

9 Photoelectric experiment Take measurements of stopping potenal and wavelength to determine Planck s constant and the threshold frequency Plot a graph of stopping potential versus frequency From your graph, determine values of: Planck s constant (h), Threshold frequency & Work function 9

10 Hamper HL page 231 Q s

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