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Thermal Radiation. Quantum Nature of Light.

Problem 5.286

A photon with energy ℏω=250keV\hbar \omega=250 \mathrm{keV} is scattered at an angle θ=120∘\theta=120^{\circ} by a stationary free electron. Find the energy of the scattered photon.

Reveal Answer
ℏω′=ℏω1+2(ℏω/mc2)sin⁡(θ/2)=0.144MeV\hbar \omega^{\prime}=\frac{\hbar \omega}{1+2\left(\hbar \omega / m c^{2}\right) \sin (\theta / 2)}=0.144 \mathrm{MeV}
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