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 In the last lesson you found out about Planck's hypothesis that radiant energy came in discrete packets called quanta, and that for each frequency or.

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Presentation on theme: " In the last lesson you found out about Planck's hypothesis that radiant energy came in discrete packets called quanta, and that for each frequency or."— Presentation transcript:

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2  In the last lesson you found out about Planck's hypothesis that radiant energy came in discrete packets called quanta, and that for each frequency or wavelength of radiant energy  Using the relationship Topic 7.1 Extended B – Photons and the Photoelectric Effect E n = nhf, for n = 1,2,3,... where h = 6.63  J  s Planck's HypothesisPlanck's Constant c = f Relation between and f for radiant energy where c = 3.00  10 8 m/s Speed of Light we can restate Planck's hypothesis in terms of wavelength: E n = nhc Planck's Hypothesis

3  Planck's hypothesis was made to make theoretical calculations agree with experimental observations. Topic 7.1 Extended B – Photons and the Photoelectric Effect  In 1905 Elbert Einstein published a paper on the photoelectric effect, in which he postulated that energy quantization is a fundamental property of electromagnetic waves (including visible light and heat).  He called the energy packet a photon, each photon having an energy given by E = hf Energy of a Photon  In his paper Einstein said...the radiant energy from a point source is not distributed continuously throughout an increasingly larger region, but, instead, this energy consists of a finite number of spatially localized energy quanta which, moving without subdividing, can only be absorbed and created in whole units. FYI: In short, light has particle-like properties. FYI: But light also has wave-like properties since frequency is a property of a wave. FYI: The ramifications of the particle-wave duality of light will be explored later.

4 T HE P HOTOELECTRIC E FFECT Topic 7.1 Extended B – Photons and the Photoelectric Effect  Einstein based his reasoning on an experiment he conducted in 1905, describing the photoelectric effect.  Certain metals are photosensitive - meaning that when they are struck by radiant energy, they emit electrons from their surface.  In order for this to happen, the light must have done work on the electron.

5 T HE P HOTOELECTRIC E FFECT Topic 7.1 Extended B – Photons and the Photoelectric Effect  We can enhance this process if we add another metal electrode, and apply a voltage like so: + -  In fact, you can read the ammeter to determine the current of the emitted photoelectrons.  As with all positively and negatively charged plates, we have the anode and the cathode. (+) anode (-) cathode FYI: If we reverse the voltage (and it is big enough), we can stop the photoelectrons from making it to the opposite plate. In other words, we can stop the photocurrent. + - A

6 T HE P HOTOELECTRIC E FFECT Topic 7.1 Extended B – Photons and the Photoelectric Effect  We can enclose our anode and cathode in an evacuated glass envelope, and hook it up as shown: + - A V  If we adjust the voltage on the phototube using the variable resistor and we limit the radiant energy to a single frequency (monochromatic light) and intensity, we get the following I vs. V graph: IpIp + -  If we reverse the polarity of the voltage, our graph looks like this: IpIp V -V0-V0  If we increase the intensity of the same monochromatic light, we get this graph: FYI: The negative voltage necessary to stop the photocurrent is called the STOPPING POTENTIAL.  Two important observations can be made regarding the graphs: (1) The photoelectric current is proportional to the intensity. (2) The cutoff voltage is independent of the intensity. FYI: Point (1) is predicted by classical theory. The more intense the barrage of light, the more electrons will be emitted. FYI: Point (2) is NOT predicted by classical theory. According to classical theory, the more intense the barrage of light, the faster electrons will be emitted, and therefore the larger the required stopping potential. E X P E C T E D N O T E X P E C T E D

7 T HE P HOTOELECTRIC E FFECT Topic 7.1 Extended B – Photons and the Photoelectric Effect  The stopping potential V 0 is related to the maximum kinetic energy of the emitted electrons by K max = eV 0  If we change the frequency of our monochromatic light, we observe that the maximum kinetic energy of the emitted electrons increases linearly with the frequency: K f f0f0  We also observe that below a certain frequency no more electrons are emitted no matter how intense the light.  We call this lower limit the cutoff frequency. cutoff frequency  We also observe that above this frequency the electron emissions begin instantaneously, even at extremely low intensities.

8 T HE P HOTOELECTRIC E FFECT Topic 7.1 Extended B – Photons and the Photoelectric Effect  The following table summarizes the problems classical theory has with the photoelectric effect: Photoelectric Effect and Classical Theory CharacteristicsClassical Prediction? The photocurrent is proportional to the intensity of the light. YES The maximum kinetic energy of the emitted electrons is dependent on the frequency of the light but not on its intensity. NO No photoemission occurs for light with a frequency below a certain cutoff frequency f 0 regardless of its intensity. NO A photocurrent is observed immediately when the light frequency is greater than f 0 even if the light intensity is low. NO

9 T HE P HOTOELECTRIC E FFECT Topic 7.1 Extended B – Photons and the Photoelectric Effect  The last effect is a very serious problem classically because classical wave theory predicts that at low intensities, times of the order of minutes are required to dislodge an electron. But Einstein's observation was that "A photocurrent is observed immediately when the light frequency is greater than f 0 even if the light intensity is low."  Thus, if light were treated as a wave, current theory failed to predict this result.  Einstein thus suggested that light was a particle and he called it a photon. It was a particle having an associated energy of E = hf.

10 T HE P HOTOELECTRIC E FFECT Topic 7.1 Extended B – Photons and the Photoelectric Effect  Einstein also stated that an electron was held in its photomaterial by "cohesive forces" that needed a minimum amount of work to be overcome. He called the energy needed to dislodge the electron the work function  0.  Thus, the photon must overcome the work function in order to cause the electron to become free, or a photoelectron. The following relationship follows: hf = K max +  0 The Work Function incident photon maximum kinetic energy of dislodged electron minimum work needed to dislodge electron FYI: This relationship satisfactorily addresses all of the failures of classical theory regarding the photoelectric effect.

11 T HE P HOTOELECTRIC E FFECT Topic 7.1 Extended B – Photons and the Photoelectric Effect Suppose the work function for a particular metal is 1.00 eV. (a) If the metal is illuminated with a monochromatic light having a wavelength of 600 nm, what will be the maximum kinetic energy of the emitted electrons? Since c = f we can easily find f: f = c / = 3  10 8 / 600  = 5  Hz Furthermore, since  0 = 1.00 eV = 1.6  J we have hf = K max +  0 K max = hf -  0 K max = 6.63  (5  )  K max = 1.72  J

12 T HE P HOTOELECTRIC E FFECT Topic 7.1 Extended B – Photons and the Photoelectric Effect Suppose the work function for a particular metal is 1.00 eV. (b) What, then, is the maximum speed of the emitted electron? Since m = 9.11  kg and we have K max = mv K max m v max = 2(1.72  ) 9.11  v max = v max = 6.14  10 5 m s -1 (c) What is the cutoff voltage for this metal? K max = eV  = (1.6  )V 0 V 0 = V

13 T HE P HOTOELECTRIC E FFECT Topic 7.1 Extended B – Photons and the Photoelectric Effect Suppose the work function for a particular metal is 1.00 eV. (d) What is the cutoff frequency f 0 for this metal? From hf = K max +  0 we have hf 0 = K max +  0 f 0 = 0 0h0h Cutoff Frequency FYI: This relationship gives you the minimum frequency that will dislodge electrons from a metal. f 0 = 1.6   f 0 = 2.41  Hz FYI: This corresponds to a wavelength given by = c / f = 3  10 8 / 2.41  = 1243 nm.


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