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Grade12, U9-L2 Photoelectric Effect

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Photoelectric effect

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Grade12, U9-L2 Photoelectric Effect

  1. 1. The Quantum Theory of Light and The Photoelectric Effect Nelson Reference Pages:Nelson Reference Pages: Page 620 -628Page 620 -628
  2. 2. Blackbody Radiation- An Introduction – Do not copy If a piece of steel is heated with a torch it will initiallyIf a piece of steel is heated with a torch it will initially not change colour. But, it will give off heat, which ischange colour. But, it will give off heat, which is invisible because it consists ofinvisible because it consists of infrared radiation.radiation. Further heating, resulting in a temperature increaseFurther heating, resulting in a temperature increase will cause the steel to glow red. As the temperaturewill cause the steel to glow red. As the temperature is further increased, the steel will glowis further increased, the steel will glow yellow andand further heating will result in the steel glowingfurther heating will result in the steel glowing white.. A further temperature increase will result inA further temperature increase will result in UV radiation (UV is not visible).radiation (UV is not visible). As the steel is heated, it initially will emit low frequencyAs the steel is heated, it initially will emit low frequency (large(large λλ)) electromagnetic radiation (EM) and as theelectromagnetic radiation (EM) and as the temperature increases so does the frequency of EMtemperature increases so does the frequency of EM radiation. White light indicates that all frequencies ofradiation. White light indicates that all frequencies of the visible spectrum are given off.the visible spectrum are given off.
  3. 3. Gustav Kirchhoff (1824 - 1887)Gustav Kirchhoff (1824 - 1887) defined thethe properties of aproperties of a blackbody from his studies on thefrom his studies on the emission and absorption spectra for gases. Heemission and absorption spectra for gases. He found that when a gas is heated tofound that when a gas is heated to incandescence (glows) it will give off specificincandescence (glows) it will give off specific frequencies of light. And conversely, when whitefrequencies of light. And conversely, when white light is shone on the same gas, it will absorb thelight is shone on the same gas, it will absorb the same frequencies as it gave off. Kirchhoffsame frequencies as it gave off. Kirchhoff proposed thatproposed that all objects will absorb the sameobjects will absorb the same characteristic frequencies of EM radiation thatcharacteristic frequencies of EM radiation that they give off. He then defined a blackbody asthey give off. He then defined a blackbody as the “the “perfect radiator” as it could emit the entire” as it could emit the entire EM spectrum. Since the blackbodyEM spectrum. Since the blackbody must alsoalso absorb the entire spectrum, Kirchhoff showedabsorb the entire spectrum, Kirchhoff showed how this could easily be done using a block withhow this could easily be done using a block with a small hole and an internal cavity.a small hole and an internal cavity.
  4. 4. The imperfect diagram of aThe imperfect diagram of a blackbody radiator isis given below. It shows a block with a hole, ofgiven below. It shows a block with a hole, of very small diameter, leading to an inner cavity.very small diameter, leading to an inner cavity. Any EM radiation entering the cavity through theAny EM radiation entering the cavity through the hole will not be able to leave. If the inner cavityhole will not be able to leave. If the inner cavity is continuously heated, the entire EM spectrumis continuously heated, the entire EM spectrum will be emitted through the tiny hole.will be emitted through the tiny hole.
  5. 5. Max Planck’s “Quantum” of Energy - Start Planck (1858-1947) proposed that the allowed energy of an oscillator (the e - had not yet been discovered) of a specific material could only exist as a integer (n) multiple of hf, here h was a constant (Planck’s constant = 6.63 x 10 -34 Js) and f was the frequency. So, E = nhf represented the amount (quantum) of energy that could be absorbed or given off. {This had to do with either absorbing, or giving off a photon – although this aspect of EM radiation was not yet known.}
  6. 6. The Photoelectric Effect In 1902 Philipp Lenard, a physicist, performedIn 1902 Philipp Lenard, a physicist, performed experiments using the setup shown below forexperiments using the setup shown below for various frequencies of light. Initially, Lenard usedvarious frequencies of light. Initially, Lenard used UV light and found that a current was present. ThisUV light and found that a current was present. This indicated that the light was able to freeindicated that the light was able to free e- whichwhich could then travel through the vacuum chamber. Ascould then travel through the vacuum chamber. As thethe intensity of light was increased the ______ alsoof light was increased the ______ also increased. The ejectedincreased. The ejected e- were called photoelectrons. A V a c u u m c h a m b e r w i t h m e t a l l i c p l a t e s a t e a c h e n d
  7. 7. Lenard modified his experiment by reversing theLenard modified his experiment by reversing the terminals of theterminals of the variable voltage power supply. By. By creating an electric field that opposed the motion ofcreating an electric field that opposed the motion of the electrons, Lenard was able to determine Ek ofthe electrons, Lenard was able to determine Ek of the photoelectrons by adjusting the voltage until thethe photoelectrons by adjusting the voltage until the current completely stopped. This was called thecurrent completely stopped. This was called the stopping PD (depended on the light frequency and type of metal plate used).. A V a c u u m c h a m b e r w i t h m e t a l l i c p l a t e s a t e a c h e n d - +
  8. 8. Lenard also used a prism to breakup whiteLenard also used a prism to breakup white light and then used different colours of light.light and then used different colours of light. He found that the stopping PD was greater forHe found that the stopping PD was greater for higher frequency light (smallerfrequency light (smaller λλ) and that) and that higher intensity light of a lower frequency could not compensate for a stopping potential set for a higher frequency light.. TheThe conclusion was that higher frequency lightwas that higher frequency light would produce photoelectrons with greaterwould produce photoelectrons with greater Ek and light intensity wasEk and light intensity was not a factor ina factor in achieving theachieving the maximum Ek for the photo efor the photo e-- ..
  9. 9. Einstein and the Photoelectric Effect (PE) In 1905 Einstein showed the relationship between Planck’s quantum of energy and the PE. Einstein proposed the following: 1. Light must be emitted and absorbed as a quantum of energy. These quanta, or packets of energy, were later called photons. 2. Planck’s unit of energy, hf, was the energy of a _________ . 3. When a photon strikes a metal surface, all of the energy of the photon is absorbed by one e- in one event. 4. Since the energy of a photon is related to its frequency, a higher frequency photon will transfer more energy to an e- then a lower frequency photon.
  10. 10. Einstein also proposed answers to other issues related to the PE:  Some photo e- had more Ek than others. This was related to how “deep” the e- was buried. Those e- buried deeper required more energy to be freed from the metal and thus had less Ek energy. The e- closest to the surface were the easiest to remove and would have max Ek. Einstein named the minimum amount of energy to remove an e- as the work function (W). The value of W depends on the type of metal (values pg 621).
  11. 11. The maximum Ek for a photoThe maximum Ek for a photo e- is: Ek = hf – W, {recall 1.0 eV = 1.6 x 10recall 1.0 eV = 1.6 x 10-19-19 JJ} Einstein did receive the Nobel Prize in 1921 for his contributions to the understanding of the photoelectric effect. The following video summarizes the PE.The following video summarizes the PE.  http://www.youtube.com/watch?v=0qKrOF-gJZ4http://www.youtube.com/watch?v=0qKrOF-gJZ4 The next video shows the effect of light color on currentThe next video shows the effect of light color on current  http://www.youtube.com/watch?v=kcSYV8bJox8 Threshold Frequency (f0) Using Ek = hf – W, the lowest frequency to free an e- is called the threshold frequency and does depend on the type of metal. This freed photo e- will have no Ek and will be re-absorbed by the metal.
  12. 12. Robert Millikan and the Photoelectric Effect Millikan performed experiments using different metals to find the Ek of the photoelectrons. The vertical axis is a measure of Ek and the horizontal axis indicates the frequency of light used. The intersection of each line with the frequency axis represents the _____ _____ . If the line is continued below the f axis it would intersect the Ek axis at the value of the ______ _______ . {See pg 623} Which metal requires the most energy to remove an e- ? Ans. Na K N a f0 ,K f0 ,N a f0 E k v s f E k
  13. 13. Momentum of a Photon Although a photon canAlthough a photon can not have a rest mass,not have a rest mass, it can be shownit can be shown mathematically that amathematically that a photon does possessphoton does possess momentum. Themomentum. The equation on the rightequation on the right equates Einstein’sequates Einstein’s famous E = mcfamous E = mc22 to theto the energy of a photon.energy of a photon.  E = mcE = mc22 and E = hfand E = hf  So, hf = mcSo, hf = mc22  But p = mv = mcBut p = mv = mc  So, hf = p cSo, hf = p c  p = hf/cp = hf/c  ButBut λ = c/fλ = c/f  So,So, p = h/p = h/λλ
  14. 14. Practice Questions  Nelson Textbook  Page 624 #1, 2 Page 626 #1-3Page 624 #1, 2 Page 626 #1-3  Page 631#1 - 6Page 631#1 - 6 From McGraw-Hill Textbook The eThe e-- emitted from a surface illuminated byemitted from a surface illuminated by light of λ = 460 nm has a max speed oflight of λ = 460 nm has a max speed of 4.2 x 104.2 x 10 55 m/s. Calculate W (in eV) of them/s. Calculate W (in eV) of the surface material. (Ans 2.2 eV)surface material. (Ans 2.2 eV)  Workbook: Page 54 #1-5

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