A radioactive nucleus X of mass number 220 undergoes alpha decay to form a daughter nucleus Y . The binding energy per nucleon of X is 7.4 MeV , that of Y is 7.6 MeV , and that of the alpha particle is 7.0 MeV . The energy released in this decay process is:

Step-by-step solution

The alpha decay equation is given by: X²²⁰ Y²¹⁶ + ⁴ The energy released ( Q -value) in the decay process is the difference between the total binding energy of the products and the total binding energy of the reactant. Total binding energy of X is: BE_X = 220 7.4 = 1628 MeV Total binding energy of Y is: BE_Y = 216 7.6 = 1641.6 MeV Total binding energy of the alpha particle is: BE_ = 4 7.0 = 28.0 MeV The energy released is: Q = BE_Y + BE_ - BE_X Q = 1641.6 + 28.0 - 1628 Q = 1669.6 - 1628 = 41.6 MeV Answer: 41.6 MeV