A stationary nucleus X of mass number 220 undergoes alpha decay to form a daughter nucleus Y of mass number 216 . The rest mass energy of X is 205000.0 MeV , the rest mass energy of Y is 201270.0 MeV , and the rest mass energy of the emitted alpha particle is 3724.5 MeV . The kinetic energy of the alpha particle is:

Step-by-step solution

First, calculate the Q -value of the reaction, which is the difference between the rest mass energy of the reactant and the products: Q = E_ X - (E_ Y + E_ ) Q = 205000.0 - (201270.0 + 3724.5) = 205000.0 - 204994.5 = 5.5 MeV This released energy is shared as kinetic energy between the daughter nucleus Y and the alpha particle. Since the initial momentum is zero, by conservation of linear momentum, the magnitudes of their momenta are equal ( p_ Y = p_ ). The kinetic energy is inversely proportional to mass ( K = p^2