Can anyone help me in understanding the selection rules for the optical transitions in case of optical and e-beam excitations? A photon is capable of producing an electron-hole pair however the electron beam can produce many such e-h pairs. And because of high energy, the electron will be excited high in the conduction band. Optical transitions are different and hence the selection rules should also be different. Can someone help me in understanding those selection rules?
 A: Your question is not very precise, but there are still a few things one can clarify.
There is one fundamental difference for an (optical) photon induced e-h pair production and the one induced by an electron (beam):
One single optical photon can excite exactly one e-h pair. Thus, if the energy of the photon is E (=h*nu), and the binding energy of the electron in the material is Eb, then the kinetic energy of the produced e-h pair is E-Eb. This is basically the photoelectric effect (https://en.wikipedia.org/wiki/Photoelectric_effect), for which Einstein got his Nobel price.
On the other hand, a single high-energy electron from an electron-beam will NOT produce a single e-h pair. It will not transfer all its energy (and mass/identity) onto a formed e-h pair, which is actually impossible because of conservation laws. A high-energy electron will electromagnetically interact with the many electrons in the material and "ionize" many of them on its way (just by knocking them away from their atoms). Ionization is described by the Bethe-Bloch formula (https://en.wikipedia.org/wiki/Bethe_formula), which describes the almost continuous energy transfer from the high-energy particle on the many many electrons of a material.
