Question

Calculate the energy released in the reaction in which an alpha particle bombarded on nucleus with an initial energy of 7.7 mega electron volt the reaction is 2he4+7n14 -------8o17+1h1

Answer 1

1 amu = 931.4 MeV zX220 → z-2Y216 + 2He4. From energy and momentum conservation we can derive the equation for K.E. of alpha particle as (K.E.)α = [(A−4)/A] Q Q = (K.E.)α A/(A−4) = 5.4 x 220/216 = 5.5 MeV K.E. of alpha particle = 5.4 MeV K.E. of daughter nucleus = Q − K.E. of alpha particle = 5.5 − 5.4 = 0.1 MeV.

Answer 2

The answer is 1.21MeV

Mass on side of reactant = (14.00307 + 4.00260)amu

Mass on side of product = (16.99914 + 1.007783)amu

So mass defect is 1.3⁻³MeV

Energy of α particle = 931.5 × 1.3 × 10⁻³MeV

                                 = 1.21MeV