`A._(7)^(14)N` nucleus, when bombarded by `._(2)^(4)He`, converts into `._(8)^(17)O` accoring to the following reaction.
`._(7)^(14)N+._(2)^(4)Herarr._(8)^(17)O+._(1)^(1)H`
Atomic masses are:
`.^(14)N:14.003u,.^(4)He:4.003u,.^(17)O:16.999u,.^(1)H:1.008u`
(a) Find the Q value of the reactions
(b) Assume that a `.^(4)He` nucleus collides with a free `.^(14)N` nucleus originally at rest. Calculate the minimum kinetic energy `(k_(0))` that `.^(4)He` must have so as to cause this reaction.
(c) If the `.^(17)O` nucleus has an excitation energy of 1.0 MeV, find the minimum kinetic energy `(k_(0))` that `.^(4)He` must have.
[Take `1u=930MeV//c^(2)`]

Answer

(a) 0.93MeV (b) `k_(0)=1.20MeV` (c) `k_(0)=2.20MeV`