1132_Physics ProblemsTechnical Physics

1132_Physics ProblemsTechnical Physics - Chapter 40 *P40.27...

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Chapter 40 473 *P40.27 The electron’s kinetic energy is Km v == × × = × −− 1 2 1 2 9 11 10 2 16 10 23 1 2 18 .. kg 2.18 10 m s J 6 ej . This is the energy lost by the photon, hf hf 0 −′ hc hc λλ 0 18 216 10 . J. We also have ′− = = × ×× −° =+ × θ 0 34 31 0 13 1 663 10 11 7 4 111 10 h mc e cos . cos . . af Js s 9.11 10 kg 3 10 m m 8 (a) Combining the equations by substitution, 0111 109 10 0 111 1 09 10 1 21 10 1 09 10 1 21 10 1 11 10 0 1 21 10 10 4 1 09 10 1 11 10 2 00 18 34 7 0 2 0 7 7 0 26 0 7 0 1 3 0 66 2 71 3 λ + = × +− + + × ×+ × × = = −× ± × × . . . . . . . . . pm J s 6.63 10 Js 3 10 m m pm pm m pm m m m m 1.21 m m 8 0 2 2 bg e j e j 7 . × only the positive answer is physical: 0 10 101 10 . m . (b) Then −−− 1 01 10 1 11 10 1 01 10 10 13 10 ...
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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