Abstract
A method of solving the transport equation to study the transport of secondary pions due to the interaction of high-energy protons is described. The pion decays into a muon and a neutrino during its flight. This decay probability is considered during the flight of the pion. Towards the end of the range, a negative pion is absorbed by the nucleus, which results in the release of its rest energy of 140 MeV. About 30 MeV of the energy, released in the form of charged particles, is absorbed locally. When a positive pion is stopped, about 34 MeV energy, (4.2 MeV due to the muon and 30 MeV due to the positron) is deposited locally. The method described is used to calculate the depth-dose distribution of pions, which are produced due to high-energy protons incident normally on a 30 cm thick tissue slab. The results thus obtained are compared with the Monte Carlo values.
Sign-in/Register to access full text options 
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have
Schedule a callDisclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.
