The Compton equation determines the shift of the wave-length of the scattered photon as a function of the ##cos\theta## of the scattered photon. It does not depend on the energy of the incident photon. With some manipulation using the equations of conservation of the momentum and energy, one gets another equation describing the relation between the scattering angles, energy of the incident photon and that of the scattered electron.(adsbygoogle = window.adsbygoogle || []).push({});

Still, how does the scattering angle vary with the energy of the incident photon? Intuitively, the higher the energy the smaller the angle, but is it really true? I mean what makes the nature prefer a small angle when we increase the incident photon energy.

I thought of the following as a possible solution: At any angle, the higher the angle value, the lower the energy of the scattered photon by Compton equation. The difference between the incident and the scattered energy is equal to the kinetic energy of the scattered electrone by law of energy conservation. But, because the kinetric energy of the electron can not increase infinitely, this means a higher left-over energy of the scattered photon when the incident energy is high. But from Compton equation, the higher energy the lower wavelength and this can be only achieved at lower scattered angle.

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# I Is the Compton scattering angle related to the incident energy?

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