Daily_007.SOLN - = hc 2 Kmc 2 (2) we have 2 = ( hc ) 2 2...

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Phys 274 – c ± Eric B. Szarmes 2 February 2011 Daily Homework 7 SOLUTIONS Q4B.1 The electrons in this problem have a kinetic energy of K = 25 eV, and thus are non-relativistic since K ² mc 2 , where mc 2 = 511 , 000 eV for electrons. We therefore may use λ = hc 2 Kmc 2 = 1240 eV · nm p 2(25 eV)(511 , 000 eV) = 0 . 245 nm . (1) Q4B.3 Let’s use the same formula as in the first problem; the only difference is that we’ll need to verify at the end of the calculation that the electrons are non-relativistic. Since
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Unformatted text preview: = hc 2 Kmc 2 (2) we have 2 = ( hc ) 2 2 Kmc 2 , (3) and therefore K = ( hc ) 2 2 2 mc 2 = (1240 eV nm) 2 2(1 . 0 nm) 2 (511 , 000 eV) = 1 . 50 eV . (4) Since this value for K clearly satises K mc 2 , the electrons are indeed non-relativistic, and we were justied in using the non-relativistic formula. The corresponding potential dierence between the two plates of the electron gun is 1.50 V. 1...
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This note was uploaded on 03/07/2011 for the course PHYS 274 taught by Professor Ericszarmes during the Spring '11 term at University of Hawaii, Manoa.

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