What fraction of the photon's energy is transmitted to the recoil electron in the Compton effect? The energy of X-ray photons prior to scattering was E. Do The calcula-tions for a photon energy of 10 keV and for a scattering angle of 60.

Answer

`Deltaepsilon=(2epsilon^2sin^2(theta//2))/(epsilon_0+2epsilonsin^2(theta//2));1%`

Explanation

The shift in the wavelength in the Compton effect is `Deltalamda = lamda.-lamda=2ldma_(c)sin^2(theta//2)` where `theta` is the photon scattering angle.
Substituting . `lamda=hc//epsilon and lamda_C=h//m_ec` we obtain `(hc)/(epsilon.)-(hc)/epsilon=(2h)/(m_ec)sin^2""(theta)/2, " or " 1/epsilon,-1/epsilon=(2sin^2(theta//2))/(epsilon_0)`
where `epsilon_0=mec^2` is the rest energy of the electron. In the course of the collision of the photon with the electron the latter acquires an energy
`Deltaepsilon=epsilon-epsilon.=(2epsilon^2sin^2(theta//2))/(epsilon_0+2epsilonsin^2(theta//2))`