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

Physics · Nuclear Physics

JEE Main 2026 — 8 April, Evening Shift — Question 15

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

Option A:
$\frac{2548}{2650}$
Option B:
$\frac{2706}{2646}$
Option C:
$\frac{2597}{2600}$
Correct
Option D:
$\frac{2862}{2499}$

Answer: C

Step-by-step solution

Kinetic energy of $\alpha$ -particle $k_\alpha = \frac{A-4}{A}Q$ . Ratio $\frac{k_{A}}{k_{B}} = \frac{196/200}{208/212} = \frac{196\times212}{200\times208} = \frac{41552}{41600} = \frac{2597}{2600}$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 8 April, Evening Shift
Subject: Physics
Chapter: Nuclear Physics
Topic: Laws of Radioactive Decay