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 Homework Statement:
 A photon with wavelength 0.1365 nm scatters from an electron that is initially at rest. What must be the angle between the direction of propagation of the incident and scattered photons if the speed of the electron immediately after the collision is 8.20×106 m/s?
 Relevant Equations:

K=1/2 *m*v^2
E=hc/(wl)
(wl)' = (wl) + (h/mc)(1cos(theta))
K= 1/2 (9.11*10^31)(8.2*^6)^2
(wl)' = hc/k = 6.5 nm
(wl)'  (wl) = 6.36 nm
cos(theta) = 1 (((6.36nm)(9.11*10^31)(3*10^8))/(6.63*10^34)) = 2620.7
theta = error
(wl)' = hc/k = 6.5 nm
(wl)'  (wl) = 6.36 nm
cos(theta) = 1 (((6.36nm)(9.11*10^31)(3*10^8))/(6.63*10^34)) = 2620.7
theta = error