# chap34 - PARTICLES AND WAVES 34 EXERCISES Section 34.2...

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34.2 Chapter 34 20. From Equation 34.6, which equals (a) (b) 2.07 eV, and (c) 12.4 keV for the given frequencies. , Eh f γ = 15 9 (4.136 10 eV s)(1 MHz) 4.14 10 eV, −− ×⋅ = × 21. INTERPRET This problem is about the relationship between the wavelength and energy of a photon. DEVELOP Using / cf λ = and Equation 34.6, the wavelength of a photon can be written as ch c f E == EVALUATE With and 6.5 eV E = 1240 eV nm, hc = we find the wavelength to be 1240 eV nm 191 nm 6.5 eV hc E = This photon is in the ultraviolet region of the electromagnetic spectrum.
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## This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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chap34 - PARTICLES AND WAVES 34 EXERCISES Section 34.2...

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