JEE Main 2026 Apr 8 Shift 2, Physics Q19: Radioactivity

JEE Main2026Apr 8, Shift 2Physics

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

Solution

In $α$ -decay of a parent at rest, momentum conservation gives the $α$ particle a kinetic-energy fraction of the total $Q$ value:

$K_{α} = (\frac{A - 4}{A}) Q$

where $A$ is the parent's mass number (so $A - 4$ is the daughter's mass number, which carries the recoil momentum).

For substance $A$ ( $A = 200$ ): $K_{α, 1} = \frac{196}{200} Q$ .

For substance $B$ ( $A = 212$ ): $K_{α, 2} = \frac{208}{212} Q$ .

Ratio: $\frac{K _{α, 1}}{K _{α, 2}} = \frac{196}{200} \cdot \frac{212}{208} = \frac{196 \times 212}{200 \times 208} = \frac{41552}{41600} = \frac{2597}{2600}$ .

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JEE Main 2026 Apr 8 Shift 2, Physics Q19: Radioactivity

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