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JEE MAIN_2026
08-04-26_S2
Question
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 $\_\_\_\_$
Select the correct option:
A
$\frac{2548}{2650}$
B
$\frac{2706}{2646}$
C
$\frac{2597}{2600}$
D
$\frac{2862}{2499}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \mathrm{k}_\alpha=\left(\frac{\mathrm{A}-4}{\mathrm{~A}}\right) \mathrm{Q} \\ & \therefore \frac{\mathrm{k}_{\alpha_1}}{\mathrm{k}_{\alpha_2}}=\frac{196}{200} \times \frac{212}{208}=\frac{2597}{2600}\end{aligned}$
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