. × MeV or . × J . i) Ra Rn+ He ii) Pu U+ He iii) – P S+e + iv) – B Po+e + v) + C B+e + vi) + Tc Mo+e + vii) + Xe+e I+ .
(a) T years (b) . T years . years . .
MeV (b) Q = . MeV, E ( = . MeV . + C B+e + + Q C – B – N N e Q c , where the masses used are those of nuclei and not of atoms.
If we use atomic masses, we have to add m e in case of C and m e in case of B. Hence C – B – e Q c (Note m e has been doubled) Using given masses, Q = . MeV. Q = E d + E e + E + The daughter nucleus is too heavy compared to e + and + , so it carries negligible energy ( E d ).
If the kinetic energy ( E + ) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q ; hence maximum E e Q ). . – Ne Na +e + Q ; Q = Ne – Na – N N e c , where the masses used are masses of nuclei and not of atoms as in Exercise . .
Using atomic masses Ne – Na Q c . Note m e has been cancelled. Using given masses, Q = . MeV.
As in Exercise . , maximum kinetic energy of the electron (max E e ) = Q = . MeV. .
(i) Q = – . MeV; endothermic (ii) Q = . MeV; exothermic . Q = Fe – Al = .
MeV; not possible. . . × MeV .
Energy generated per gram of U = . J g