1.

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»


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))]`


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