Question

Two nuclei \(A(200\,amu)\) and \(B(212\,amu)\) undergoes \(\alpha\)-decay and Q-value is same and equal to \(1\,MeV\). Find ratio of KE of \(\alpha\)-particle.

• A. \( \dfrac{2597}{2600} \)
• B. \( \dfrac{2600}{2597} \)
• C. \( \dfrac{5200}{2597} \)
• D. \( \dfrac{2597}{5200} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In an $\alpha$-decay process, the parent nucleus is initially at rest.
The Q-value is shared between the emitted $\alpha$-particle and the recoiling daughter nucleus to conserve linear momentum.
Step 2: Key Formula or Approach:
Based on conservation of momentum, the kinetic energy of the $\alpha$-particle is $K_{\alpha} = Q \cdot \left(\frac{M-4}{M}\right)$, where $M$ is the mass of the parent nucleus.
Step 3: Detailed Explanation:
For nucleus A ($M_A = 200 \text{ amu}$):
\[ K_{\alpha A} = Q \times \left(\frac{200 - 4}{200}\right) = Q \times \frac{196}{200} \text{ MeV} \]
For nucleus B ($M_B = 212 \text{ amu}$):
\[ K_{\alpha B} = Q \times \left(\frac{212 - 4}{212}\right) = Q \times \frac{208}{212} \text{ MeV} \]
Taking the ratio of the kinetic energies:
\[ \frac{K_{\alpha A}}{K_{\alpha B}} = \frac{196/200}{208/212} = \frac{196 \times 212}{200 \times 208} \]
Simplifying the ratio by dividing numbers:
\[ \frac{K_{\alpha A}}{K_{\alpha B}} = \left(\frac{196}{200}\right) \times \left(\frac{212}{208}\right) = \left(\frac{49}{50}\right) \times \left(\frac{53}{52}\right) \]
\[ \frac{K_{\alpha A}}{K_{\alpha B}} = \frac{2597}{2600} \]
Step 4: Final Answer:
The ratio is $\frac{2597}{2600}$.