78.(a) Using the result of problem 3,E=hcλ=1240 nm·eV10.0×10−3nm= 124 keV.(b) The kinetic energy gained by the electron is equal to the energy decrease of the photon:∆E=∆µhcλ¶=hcµ1λ−1λ+∆λ¶=µhcλ¶µ∆λλλ¶=E1+λ∆λ=EλλC(1−cosφ)=124 keV10.0pm(2.43 pm)(1−cos 180◦)=4
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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.