As it is known, the potential tube voltage of the CT scanner is equivalent to the maximum energy of the produced photons, but it is not equivalent to the effective energy of the photons beam. Theoretically, effective energy = 1/2 to 1/3 of the maximum energy (or kVp). Furthermore, it varies from one CT modality to another and differs when exposure parameters are changed.
So, how could the effective energy be determined practically and precisely?
Thanks