242/96 Qr aftergoing through beta minus alpha decay
Assuming that the parent nucleus before decay was 242/96 Cm (Cm-242) and it underwent beta-minus (β-) decay followed by alpha (α) decay:
1) In beta-minus decay, a neutron in the nucleus is converted to a proton, emitting an electron and an antineutrino:
242/96 Cm → 242/97 Bk + β- + ν¯
So the daughter nucleus after β- decay is 242/97 Bk (Bk-242)
2) In alpha decay, the nucleus emits an alpha particle (two protons and two neutrons), resulting in a new nucleus with atomic number two less and mass number four less than the original nucleus:
242/97 Bk → 238/95 Am + α
So the final product nucleus after going through beta-minus and alpha decay is 238/95 Am (Am-238).
Therefore, the QR (Q-value) for this decay chain can be calculated by subtracting the mass of the parent nucleus from the mass of the final product nucleus, and converting the mass difference to energy using Einstein's famous equation E=mc²:
Q = [m(Cm-242) - m(Am-238)]c²
Using atomic mass values from the periodic table, we get:
Q = [(242.05955 u x 931.5 MeV/u) - (238.05079 u x 931.5 MeV/u)] = 149.2 MeV (rounded to one decimal place)
So the Q-value for this decay chain is 149.2 MeV, which represents the maximum energy that can be released through this decay process.