24. A photon having wavelength lambda scatters off a free electron at A (Fig. P40.24), producing a second photon having wavelength lambda-prime. This photon then scatters off another free electron at B, producing a third photon having wavelength lambda double prime and moving in a direction directly opposite the original photon as shown in the figure. Determine the value of delta lambda= lambda double prime-lambda.

Question

roadjester · Answer

attachment