47. Let the distance in question be r. The initial kinetic energy of the electron is Kmviei=122,where vi= 3.2 ×105m/s. As the speed doubles, Kbecomes 4Ki. Thus ∆∆UerKKKKmviiiei=−=−−2244332π0ε(),or 22219
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.