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The nucleus of Pu238 decays to U234 by emitting a α particle. The kinetic energy of the emitted α particle is given by [you are given that atomic mass of Pu238 is 238.04954 amu that of U234 is 234.040969 amu].

(1) 0059 MeV

(2) 5.57 amu

(3) 2.76 MeV

(4) 5.57 MeV

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Best answer

(4) 5.57 MeV

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.

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