What is the ratio E_A/E_B?

2597/2600. From 196×212/(200×208)=41552/41600=2597/2600.

Alpha decay A=200 and A=212 same Q=1MeV — ratio of alpha energies

Q: What fraction of Q goes to alpha in alpha-decay from mass A?

E_α=Q(A-4)/A. For A=200: 196/200. For A=212: 208/212.

JEERANKERS

PhysicsNuclear Physics

QMCQNuclear Physics

Substances A ($A=200$) and B ($A=212$) show $\alpha$-decay with same $Q=1$ MeV. Ratio of $\alpha$-ray energies $E_A/E_B$ is:

A) $2548/2650$ B) $2706/2646$ C) $2597/2600$ D) $2862/2499$

✅ Correct Answer

C) 2597/2600

✓

Solution

Energy partition in α-decay

Parent at rest → momentum conservation: $m_\alpha v_\alpha=m_d v_d$.

$E_\alpha=Q\times\dfrac{m_d}{m_\alpha+m_d}=Q\times\dfrac{A-4}{A}$

Compute ratio

$E_A=\dfrac{196}{200}\text{ MeV},\quad E_B=\dfrac{208}{212}\text{ MeV}$

$\dfrac{E_A}{E_B}=\dfrac{196\times212}{200\times208}=\dfrac{41552}{41600}=\dfrac{2597}{2600}$

📘 Alpha Particle Energy: E_α = Q×(A-4)/A

From momentum conservation in $\alpha$-decay: $E_\alpha=Q(A-4)/A$. Heavier parent → more energy to alpha. GCD(41552,41600)=16 gives $2597/2600$.

💡

Important Concepts

Momentum Conservationm_α×v_α=m_d×v_d. KE_α/KE_d=m_d/m_α=(A−4)/4.

Energy FormulaKE_α=Q×(A−4)/A. For A=200: 196/200. For A=212: 208/212.

Computing Ratio196×212=41552, 200×208=41600. GCD=16. Ratio=2597/2600.

Why heavier parent gives more energy to alpha?Larger A-4 ratio. Also heavier daughter takes less recoil energy (momentum conservation).

FAQs

Why E_α=Q(A-4)/A?

⌄

KE_α/KE_d=(A-4)/4. And KE_α+KE_d=Q. Solving: KE_α=Q(A-4)/A.

What is GCD(41552,41600)?

⌄

41600-41552=48. GCD(41552,48): 41552=865×48+32, GCD(48,32)=16. So GCD=16.

41552/16?

⌄

41552÷16=2597. 41600÷16=2600. ✓

What Q value is used?

⌄

Q=1MeV for both. The ratio depends only on A, not Q (since Q cancels).

Is this from JEE Main 2026?

⌄

Yes, this appeared in JEE Main 2026.

🔗