p41_036 - kind depicted in ±ig 41-15 Thus using the result...

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36. The kinetic energy gained by the electron is eV ,wh e re V is the accelerating potential diference. A photon with the minimum wavelength (which, because oF E = hc/λ
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Unformatted text preview: kind depicted in ±ig. 41-15. Thus, using the result oF problem 3 in Chapter 39, eV = hc λ min = 1240 eV · nm . 10 nm = 1 . 24 × 10 4 eV . ThereFore, the accelerating potential diference is V = 1 . 24 × 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.

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