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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA stationary radium nucleus (\( ^{226}\text{Ra} \)) undergoes alpha decay into a radon nucleus (\( ^{222}\text{Rn} \)) and an alpha particle (\( ^{4}\text{He} \)). If the total kinetic energy released in the decay is \( Q \), what fraction of the total kinetic energy is carried away by the radon nucleus?
Correct answer · Option 2
\( \frac{2}{113} \)
Explanation
By conservation of momentum, \( p_{\text{Rn}} = p_\alpha \). Since \( K = \frac{p^2}{2m} \), \( \frac{K_{\text{Rn}}}{K_\alpha} = \frac{m_\alpha}{m_{\text{Rn}}} = \frac{4}{222} = \frac{2}{111} \). Thus, \( \frac{K_{\text{Rn}}}{Q} = \frac{4}{222+4} = \frac{4}{226} = \frac{2}{113} \).
