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Question

Gamma rays of photon energy 0.511MeV are directed onto an aluminum target and are scattered in various directions by loosely bound electrons there. (a) What is the wavelength of the incident gamma rays?(b) What is the wavelength of gamma rays scattered at 90.0 to the incident beam? (c) What is the photon energy of the rays scattered in this direction?

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Solution

(a) Using the value hc=1240eVnm, we find
λ=hcE=1240nmeV0.511MeV=2.43×103nm=2.43pm
(b) Now, Eq. 38-11 leads to
λ=λ+Δλ=λ+hmeC(1cosϕ)=2.43pm+(2.43pm)(1cos90.0)=4.86pm
(c) The scattered photons have energy equal to
E=E(λλ)=(0.511MeV)(2.43pm4.86pm)=0.255MeV

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