Unformatted text preview: kind depicted in Â±ig. 4115. Thus, using the result oF problem 3 in Chapter 39, eV = hc Î» min = 1240 eV Â· nm . 10 nm = 1 . 24 Ã— 10 4 eV . ThereFore, the accelerating potential diference is V = 1 . 24 Ã— 10 4 V = 12 . 4 kV....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy, Kinetic Energy, Photon

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