Physics Q44 – JEE-MAIN 2026

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 _______.

Clue 1 of 3: Energy Conservation in Alpha Decay

In alpha decay, the total energy released (Q-value) is shared between the kinetic energy of the alpha particle and the recoiling daughter nucleus.

🧠 Full Solution Path

Step 1: Relate Alpha Particle Energy to Q-value and Mass Number✦ Active

In alpha decay, a parent nucleus with mass number $A$ decays into a daughter nucleus with mass number $A - 4$ and an alpha particle with mass number 4. By conservation of energy and momentum, the kinetic energy of the emitted alpha particle ( $E_{α}$ ) is given by the formula:

E_{α} = Q (\frac{A - 4}{A})

where $Q$ is the disintegration energy (Q-value) and $A$ is the mass number of the parent nucleus.

Step 2: Calculate Alpha Particle Energies for A and B○ Expand

For substance A, with mass number $A_{A} = 200$ :

E_{α, A} = Q (\frac{200 - 4}{200}) = Q (\frac{196}{200})

For substance B, with mass number $A_{B} = 212$ :

E_{α, B} = Q (\frac{212 - 4}{212}) = Q (\frac{208}{212})

The Q-value is given as $1$ MeV for both, but it will cancel out in the ratio.

Step 3: Determine the Ratio of Energies○ Expand

The ratio of energies of $α$ -rays produced by A and B is:

\frac{E_{α, A}}{E_{α, B}} = \frac{Q (\frac{196}{200})}{Q (\frac{208}{212})} = \frac{196}{200} \times \frac{212}{208} = \frac{49 \times 4}{50 \times 4} \times \frac{53 \times 4}{52 \times 4} = \frac{49 \times 53}{50 \times 52} = \frac{2597}{2600}

💡 Teacher's Secret Hint

Ensure careful simplification of the fractions to arrive at the correct ratio.

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Physics Question 44 – JEE-MAIN 2026

Clue 1 of 3: Energy Conservation in Alpha Decay

🧠 Full Solution Path