Question

Calculate the energy released if U²³⁸ –nucleus emits an –particle. 

Calculate the energy released in MeV in the following nuclear reaction

Given Atomic mass of ²³⁸U = 238.05079 u

Atomic mass of ²³⁴Th = 234.04363 u 

Atomic mass of alpha particle = 4.00260 u

Is the decay spontaneous, give reason. 

Answer 1

The process is

The energy released

Q = (M_U – M_TH — M_He) c²

= (238.05079 – 234.04363 – 4.00260)u × c² 

= (0.00456 u) × c² 

= 0.00456 × (931.5 mev/c²)c²  = 4.25 MeV

Yes, the decay is spontaneous (since Q is positive)