The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.
The isotope $$_{5}^{12}B$$ having a mass 12.014 u undergoes $$\beta$$- decay to $$_{6}^{12}C$$, $$_{6}^{12}C$$ has an excited state of the nucleus ($$_{6}^{12}C^*$$) at 4.041 MeV above its ground state. If $$_{5}^{12}B$$ decays to $$_{6}^{12}C^*$$, the maximumkinetic energy of the $$\beta$$- particle in units of MeV is
(1 u = 931.5 MeV/c$$^2$$, where c is the speed of light in vacuum).
Correct Answer: e
Create a FREE account and get: