Consider the following problem:
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?

 

Because it is so small when compared to the masses of everything else. The rest mass of an electron is about 0.0005 the mass of a proton. So you do the math.

 

well can u give me the masses used for the calculations?
i believe neglecting 8 e wud mean abt 4.05 Mev of non negligible energy error,
u will have to think of it like;
in a beta(-) decay a neutron is converted into proton;
hence the energy in that case is calculated by
Q=(m(a,z) - M(a,z+1))*931.5Mev;
here the mass of M(a,z+1) has i electron extra in its orbit hence taking atomic mass of m(a,z+1) accounts for the electron mass automatically no need for adding it separately.
similarly,
look carefully in the example problem;
8 neutrons have been converted into 8 protons;
hence electron masses are already accounted for by just subtracting atomic masses,
getting my point? (the atomic masses are such that u dont have to worry abt 8 e )

example:
say a beta emission takes Br(87,35)---Beta(-)--->Kr(87,36) +e-
now if mass of Kr is taken as atomic mass mass of 36 electrons are added (the confusion arises as the above Kr is not a atom but a nucleus so electron evolved is shown separately )
understand it this way (it might be wrong but good for understanding)
atomic mass of Kr = nucleus mass + electrons mass = mass of neutrons + mass of protons + mass of electrons
=(A-Z)(n) +Z(hydrogen)

 

Thats cool!thanx.