JEE Main 2026 · 8 Apr · Shift 2 · Physics Q44

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PhysicsSingle correct+4 / -1Question 44 of 75

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 $\alpha$ -rays produced by A and B is ________.

\dfrac{2548}{2650}

\dfrac{2706}{2646}

\dfrac{2597}{2600}

\dfrac{2862}{2499}

Answer and worked solution

Official answer: C

α energy. $K_\alpha=Q\dfrac{A-4}A$ (momentum shared with the daughter).

Ratio.
$\frac{196/200}{208/212}=\frac{196\times212}{200\times208}=\frac{2597}{2600}.$

(C)

Topics:Radioactive decay Conservation of linear momentum

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JEE Main 2026 · 8 Apr · Shift 2, Physics question 44