Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous α -decay with same Q value of 1 MeV. The ratio of energies of α -rays produced by A and B is ______. (a) (2548)/(2650) (b) (2706)/(2646) (c) (2597)/(2600) (d) (2862)/(2499)

Answer: (c). In α decay from rest, Kα = Q (A - 4)/(A). KA = 1 × (196)/(200), KB = 1 × (208)/(212) (KA)/(KB) = (196 × 212)/(200 × 208) = (41552)/(41600) = (2597)/(2600) So the answer is (c) (2597)/(2600).

Two radioactive substances A and B of mass numbers 200 and 212… | JEE Main 2026 (8 Apr, Shift 2)

Q4JEE Main 2026 · 8 APR · SHIFT 2Medium

Nuclear Reactions & Q-value

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

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Answer: (c)

In $\alpha$ decay from rest, $K_\alpha = Q\,\dfrac{A - 4}{A}$ .
$K_A = 1 \times \dfrac{196}{200}$ , $K_B = 1 \times \dfrac{208}{212}$
$\dfrac{K_A}{K_B} = \dfrac{196 \times 212}{200 \times 208} = \dfrac{41552}{41600} = \dfrac{2597}{2600}$
So the answer is (c) $\dfrac{2597}{2600}$ .

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