Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous -decay with same Q value of 1 MeV. The ratio of energies of -rays produced by A and B is ________.

Step-by-step solution

In an -decay, the kinetic energy of the emitted -particle is given by the conservation of momentum and energy as: K_ = A - 4 A Q where A is the mass number of the parent nucleus. For substance A ( A = 200 ): K_ , A = 200 - 4 200 Q = 196 200 Q For substance B ( A = 212 ): K_ , B = 212 - 4 212 Q = 208 212 Q The ratio of the energies of the -rays produced by A and B is: K_ , A K_ , B = 196 200 Q 208 212 Q = 196 200 212 208 Simplifying the fractions: K_ , A K_ , B = 49 50 53 52 = 2597 2600 Answer: 2597 2600