1) Calculate the energy released in the fission reaction,

Uranium(235,92) + neutron--->[Uranium(235,92)]*--->Neodymium(143,60) + Zirconium(90,40)+ 3 neutrons + 8 electrons +8 antineutrino

__Solution:__

If atomic rest masses are used in calculating the Q value, the term 8 e- may be dropped.Let M(U),M(Nd),M(Zr) and M(n) be the atomic masses of Uranium, Neodymium,Zirconium and neutron respectively.

Hence,

Q=[M(U) - M(Nd) - M(Zr) - {3-1}M(n)]c^2

Q= 197.6 MeV

This is the solution given in Schaums book on Modern Physics.

Just because atomic rest masses are used, how can we neglect the 8 e- term?