The isotope ¹²₅B of mass 12.014 u undergoes β⁻ decay to an excited state ¹²₆C* lying 4.041 MeV above the ¹²₆C ground state. Given m(¹²₆C)=12.000 u and 1u=931.5 MeV c⁻², find the maximum β-particle kinetic energy in MeV.
Physics · Nuclear physics
JEE Advanced 2016

Question

The isotope ¹²₅B of mass 12.014 u undergoes β⁻ decay to an excited state ¹²₆C* lying 4.041 MeV above the ¹²₆C ground state. Given m(¹²₆C)=12.000 u and 1u=931.5 MeV c⁻², find the maximum β-particle kinetic energy in MeV.

Physics · Nuclear physics

JEE Advanced 2016

G1 Physics · Accepted Answer

The β⁻ Q-value to the excited state is [m(¹²B)−m(¹²C)]c²−4.041 MeV. Using 12.014u−12.000u gives about 13.041−4.041≈9 MeV. Neglecting tiny recoil, Kβ,max≈9 MeV.