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?
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?
