28.5. Set Up: In 1.00 s the energy output is 12.0 J. The energy of a photon is Solve: The number of photons emitted per second is Reflect: A large number of photons are emitted each second. 28.6. Set Up: The source emits of energy as visible light each second. with Solve: (a) (b) The number of photons emitted per second is (c) No. The frequency of the light depends on the energy of each photon. The number of photons emitted per second is proportional to the power output of the source. 28.7. Set Up: Solve: the threshold frequency, when 28.8. Set Up: For nickel, Solve: 28.9. Set Up: At the threshold frequency, Solve: (a) (b) (c) Reflect: The threshold wavelength depends on the work function for the surface. 28.10. Set Up: The minimum corresponds to the minimum Solve: 28.11. Set Up: Solve: Use the data for 400.0 nm to calculate Then for 300.0 nm, f5 hc l 2 1 2 m v max 2 5 1 4.136 3 10 2 15 eV # s 21 3.00 3 10 8 m / s 2 400.0 3 10 2 9 m 2 1.10 eV 5 3.10 eV 2 1.10 eV 5 2.00 eV. f
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.