Example 2.9 in Serway's Modern Physics, 2nd ed., asks Why are x-ray photons used in the Compton experiment, rather than visible light photons? To answer this question, we shall first calculate the Compton shift for scattering at 90o from graphite for the following cases: (1) very high energy γrays from cobalt, λ = 0.0106A; (2) x-rays from molybdenum, λ = 0.712A; and (3) green light from a mercury lamp, λ = 5461A. (A = angstrom units) They calculate the wavelength shift in each case using the Compton shift formula, then calculate the fractional change in wavelength in each case, showing that: for x-rays from cobalt, Δλ/λ0 = 2.29 for x-rays from molybdenum, Δλ/λ0 = .0341 for visible light from mercury, Δλ/λ0 = 4.45 x 10-6 and conclude that, "Because both incident and scattered wavelengths are simultaneously present in the beam, they can be easily resolved only if Δλ/λ0 is a few percent or if λ0 ≤ 1A." FINALLY my question: the computations are straightforward and the conclusion about resolution makes perfect sense, but... Why are both incident and scattered wavelengths present simultaneously in the beam? This beam is at a 90o angle to the original direction of the beam from the x-ray source. So didn't all the photons in this beam get there by scattering? How do photons of the original, incident wavelength get there? Same question, another way: on pg. 81 of the same text, there are graphs of the distribution of intensity vs. wavelength for x-rays scattered at 0, 45, 90 and 135 degrees. All except the 0-degree graph show two peaks in the distribution: at λ0 and at λ'. Why not just λ'?