The isotope 512B having a mass 12.014 u undergoes β-decay to 612C. 612C Has an excited state of the nucleus 612C* at 4.041 MeV above its ground state. If 512B decays to 612C*, the maximum kinetic energy of the β particle in units of MeV is (1 u=931.5 MeV/C2, where c is the speed of light in vacuum).
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 9.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Energy released E=[Δm]931.5=[12.014−12]931.5=13.041MeV Energy of β particle =13.041−4.041=9MeV