Correct option: (4) 5.57 MeV
Explanation:
The reaction
Represents the decay process of Pu238
Here Q is representing the energy of nuclear reaction
Q = m (Pu238) – [m (U234) + m (2He4)]
Q = 238.04954 – [234.04096 + 4.002603] = 0.00598 amu = 0.00598×931 = 5.57 MeV
This energy would have to shared by U234 and α particle. However U234 is very massive as compared to the mass of the α particles. Hence this energy of 5.57 MeV can be taken as the kinetic energy of the emitted α particle.