1)Determine the maximum possible speed of the daughter nucleus in negative beta decay of a He(6,2) nucleus which is initially at rest.
I solved in the following way:
He(6,2) --> Li(6,3) + electron + antineutrino
The daughter nucleus will have maximum speed when the electron and antineutrino has zero kinetic energy. Applying mass-energy equivalence,
M(He)c^2 = M(Li)c^2 + K(Li)
Where M(He)=atomic mass of helium isotope, M(Li)= atomic mass of Lithium isotope,
K(Li)=Kinetic energy of daughter nucleus
On simplification I get,
K(Li)=3.52 MeV
(1/2)M(Li)v^2=3.52 MeV
On simplification I get,
v=1x10^7 m/sec
But the answer given in my book is 1x10^5 m/sec. Could someone please tell me where I have gone wrong?

.... you're just a factor of 100 out ...... did you input c wrong? as it's quoted in both m/s & km/s (1000 fold difference) .... how's about that for starters?

You've just forgotten that the total impulse must be kept...
So the electron flies away much faster than the lithium and takes most of the energy.
Li will get its maximum energy when the neutrino takes none, this is correct.