Find the energy of the reaction `.^(14)N(alpha, p) .^(17)O`, if the kinetic energy of the incoming `alpha`-particle is `T_(alpha) = 4.0 MeV` & the proton outgoing at an angle `theta = 60^(@)` to the motion direction of the `alpha`-particle has a kinetic energy `T_(p) = 2.09 MeV`.

Answer

`[Q = (1-eta_(p))T_(p) - (1-eta_(alpha))T_(alpha)-2 sqrt(eta_(p)eta_(alpha)T_(p)T_(alpha)) cos theta =-12 MeV, "where" n_(p) = m_(p)//m_(o), n_(alpha) = (m_(alpha))/(m_(0))]`