Kinetic energy of the emitted \( \alpha \)-particle in the \( \alpha \)-decay of \( {}^{226}_{88}\mathrm{Ra} \) will be
\( m\left({}^{226}_{88}\mathrm{Ra}\right) = 226.02540\,\mathrm{u} \)
\( m\left({}^{4}_{2}\mathrm{He}\right) = 4.002603\,\mathrm{u} \)
\( m\left({}^{222}_{86}\mathrm{Rn}\right) = 222.01750\,\mathrm{u} \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4.85 MeV
The energy released (Q-value) is the mass defect times 931.5 MeV/u. It is shared between the \( \alpha \)-particle and the recoiling daughter nucleus, and conservation of momentum decides the share.
\( {}^{226}_{88}\mathrm{Ra} \rightarrow {}^{222}_{86}\mathrm{Rn} + {}^{4}_{2}\mathrm{He} \)
\( \Delta m = 226.02540 - 222.01750 - 4.002603 = 0.005297\,\mathrm{u} \)
\( Q = 0.005297\,\mathrm{u} \times 931.5\,\mathrm{MeV/u} = 4.934\,\mathrm{MeV} \)
The Ra nucleus is at rest, so the \( \alpha \)-particle and the Rn nucleus move off with equal and opposite momenta \( p \). With \( K = p^2/2m \), the kinetic energies are in the inverse ratio of the masses:
\( \displaystyle K_\alpha = Q\,\frac{m_{\mathrm{Rn}}}{m_{\mathrm{Rn}} + m_\alpha} \approx Q\,\frac{222}{226} \)
\( \displaystyle K_\alpha = 4.934\,\mathrm{MeV} \times \frac{222}{226} = 4.85\,\mathrm{MeV} \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Exercises – …
