chapter 38 solutions - Chapter 38 Solutions to Exercises 12...

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Unformatted text preview: Chapter 38 Solutions to Exercises 12 th Ed. 38.1.I DENTIFY and S ET U P : The stopping potential V is related to the frequency of the light by h V f e e =- . The slope of V versus f is h/e . The value f th of f when V = is related to by th hf = . E XECUTE : (a) From the graph, 15 th 1.25 10 Hz f = . Therefore, with the value of h from part (b), th 4.8 eV hf = = . (b) From the graph, the slope is 15 3.8 10 V s- . 16 15 34 ( )(slope) (1.60 10 C)(3.8 10 V s) 6.1 10 J s h e--- = = = (c) No photoelectrons are produced for th f f < . (d) For a different metal f th and are different. The slope is h/e so would be the same, but the graph would be shifted right or left so it has a different intercept with the horizontal axis. E VALUATE : As the frequency f of the light is increased above f th the energy of the photons in the light increases and more energetic photons are produced. The work function we calculated is similar to that for gold or nickel. 38.2. I DENTIFY and S ET U P : c f = relates frequency and wavelength and E hf = relates energy and frequency for a photon. 8 3.00 10 m/s c = . 16 1 eV 1.60 10 J- = . E XECUTE : (a) 8 14 9 3.00 10 m/s 5.94 10 Hz 505 10 m c f - = = = (b) 34 14 19 (6.626 10 J s)(5.94 10 Hz) 3.94 10 J 2.46 eV E hf-- = = = = (c) 2 1 2 K mv = so 19 15 2 2(3.94 10 J) 9.1 mm/s 9.5 10 kg K v m-- = = = 38.3. 8 14 7 3.00 10 m s 5.77 10 Hz 5.20 10 m c f- = = = 34 27 7 27 8 19 6.63 10 J s 1.28 10 kg m s 5.20 10 m (1.28 10 kg m s) (3.00 10 m s) 3.84 10 J 2.40 eV. h p E pc----- = = = = = = = 38.4. I DENTIFY and S ET U P : av energy P t = . 19 1 eV 1.60 10 J- = . For a photon, hc E hf = = . 34 6.63 10 J s h- = . E XECUTE : (a) 3 2 16 av energy (0.600 W)(20.0 10 s) 1.20 10 J 7.5 10 eV P t-- = = = = (b) 34 8 19 9 (6.63 10 J s)(3.00 10 m/s) 3.05 10 J 1.91 eV 652 10 m hc E --- = = = = (c) The number of photons is the total energy in a pulse divided by the energy of one photon: 2 16 19 1.20 10 J 3.93 10 photons 3.05 10 J/photon-- = . E VALUATE : The number of photons in each pulse is very large. 38.5. I DENTIFY and S ET U P : Eq.(38.2) relates the photon energy and wavelength. c f = relates speed, frequency and wavelength for an electromagnetic wave. E XECUTE : (a) E hf = so 6 19 20 34 (2.45 10 eV)(1.602 10 J/1 eV) 5.92 10 Hz 6.626 10 J s E f h-- = = = (b) c f = so 8 13 20 2.998 10 m/s 5.06 10 m 5.92 10 Hz c f - = = = (c) E VALUATE : is comparable to a nuclear radius. Note that in doing the calculation the energy in MeV was converted to the SI unit of Joules. 38.6. I DENTIFY and S ET U P : th 272 nm = . c f = . 2 max 1 2 mv hf =- . At the threshold frequency, th f , max v h . 15 4.136 10 eV s h- = . E XECUTE : (a) 8 15 th 9 th 3.00 10 m/s 1.10 10 Hz 272 10 m c f - = = = . (b) 15 15 th (4.136 10 eV s)(1.10 10 Hz) 4.55 eV hf - = = = ....
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This note was uploaded on 04/14/2008 for the course PHYS 2303 taught by Professor Hoffman during the Fall '07 term at University of Texas at Dallas, Richardson.

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chapter 38 solutions - Chapter 38 Solutions to Exercises 12...

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