Q69 / 75 · JEE Main 2026 (08 April Shift 2) · Physics
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 ________.
- A$\dfrac{2548}{2650}$
- B$\dfrac{2706}{2646}$
- Correct answerC$\dfrac{2597}{2600}$
- D$\dfrac{2862}{2499}$
Solution
In an $\alpha$-decay, the kinetic energy of the emitted $\alpha$-particle is given by the conservation of momentum and energy as:
$K_{\alpha} = \dfrac{A - 4}{A} Q$
where $A$ is the mass number of the parent nucleus.
For substance A ($A = 200$):
$K_{\alpha, A} = \dfrac{200 - 4}{200} Q = \dfrac{196}{200} Q$
For substance B ($A = 212$):
$K_{\alpha, B} = \dfrac{212 - 4}{212} Q = \dfrac{208}{212} Q$
The ratio of the energies of the $\alpha$-rays produced by A and B is:
$\dfrac{K_{\alpha, A}}{K_{\alpha, B}} = \dfrac{\dfrac{196}{200} Q}{\dfrac{208}{212} Q} = \dfrac{196}{200} \times \dfrac{212}{208}$
Simplifying the fractions:
$\dfrac{K_{\alpha, A}}{K_{\alpha, B}} = \dfrac{49}{50} \times \dfrac{53}{52} = \dfrac{2597}{2600}$
Answer: $\dfrac{2597}{2600}$
$K_{\alpha} = \dfrac{A - 4}{A} Q$
where $A$ is the mass number of the parent nucleus.
For substance A ($A = 200$):
$K_{\alpha, A} = \dfrac{200 - 4}{200} Q = \dfrac{196}{200} Q$
For substance B ($A = 212$):
$K_{\alpha, B} = \dfrac{212 - 4}{212} Q = \dfrac{208}{212} Q$
The ratio of the energies of the $\alpha$-rays produced by A and B is:
$\dfrac{K_{\alpha, A}}{K_{\alpha, B}} = \dfrac{\dfrac{196}{200} Q}{\dfrac{208}{212} Q} = \dfrac{196}{200} \times \dfrac{212}{208}$
Simplifying the fractions:
$\dfrac{K_{\alpha, A}}{K_{\alpha, B}} = \dfrac{49}{50} \times \dfrac{53}{52} = \dfrac{2597}{2600}$
Answer: $\dfrac{2597}{2600}$
Hence, the correct answer is option: 'C'.