Two radioactive nuclei A and B have mass numbers 200 and 212, respectively. Both undergo spontaneous α-decay with the same Q-value of 1 MeV. What is the ratio of the kinetic energies of the emitted α-particles, K(A) ÷ K(B)?
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Question

Two radioactive nuclei A and B have mass numbers 200 and 212, respectively. Both undergo spontaneous α-decay with the same Q-value of 1 MeV. What is the ratio of the kinetic energies of the emitted α-particles, K(A) ÷ K(B)?

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G1 Physics · Accepted Answer

For a two-body α-decay of a parent nucleus of mass number A, parent → daughter of mass number (A-4) + α. Momentum conservation gives equal momentum magnitudes to the daughter and α-particle. Therefore the Q-value divides inversely in proportion to the product masses, and the α-particle receives the fraction K(α) = [(A-4) ÷ A]Q. For nucleus A with parent mass number 200: K(A) = 196 ÷ 200 × Q. For nucleus B with parent mass number 212: K(B) = 208 ÷ 212 × Q. Since Q is the same, K(A) ÷ K(B) = [196…