In an α-decay, the kinetic energy of the emitted α-particle is given by the conservation of momentum and energy as:
Kα=AA−4Q
where A is the mass number of the parent nucleus.
For substance A (A=200):
Kα,A=200200−4Q=200196Q
For substance B (A=212):
Kα,B=212212−4Q=212208Q
The ratio of the energies of the α-rays produced by A and B is:
Kα,BKα,A=212208Q200196Q=200196×208212
Simplifying the fractions:
Kα,BKα,A=5049×5253=26002597
Answer: 26002597