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

Concept: During \(\alpha\)-decay, \[ M \rightarrow (M-4) + \alpha \] Due to conservation of momentum: \[ p_{\alpha} = p_{\text{daughter}} \] Thus kinetic energy distribution becomes \[ \frac{KE_{\text{daughter}}}{KE_{\alpha}} = \frac{m_\alpha}{m_d} \] Energy of the \(\alpha\)-particle is \[ KE_\alpha = \frac{M-4}{M}Q \]
Step 1:Write expression for kinetic energy. For nucleus \(M\), \[ KE_\alpha = \frac{M-4}{M}Q \]
Step 2:Write ratio for the two nuclei. \[ \frac{(KE_\alpha)_1}{(KE_\alpha)_2} = \frac{\frac{200-4}{200}}{\frac{212-4}{212}} \]
Step 3:Substitute values. \[ = \frac{\frac{196}{200}}{\frac{208}{212}} \] \[ = \frac{196}{200}\times\frac{212}{208} \] \[ = \frac{2597}{2600} \] \[ \boxed{\frac{2597}{2600}} \]