A nucleus of uranium decays at rest into nuclei of thorium and helium. Then:

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Accepted Answer

1. **Conservation of Momentum:** The uranium nucleus is initially at rest, so the initial momentum is zero. According to the law of conservation of momentum, the final total momentum must also be zero .
   $$ \vec{P}_{initial} = 0 $$
   $$ \vec{P}_{final} = \vec{P}_{Th} + \vec{P}_{He} = 0 \implies \vec{P}_{Th} = -\vec{P}_{He} $$
   This means the momenta of the thorium nucleus and the helium nucleus (alpha particle) are equal in magnitude and opposite in direction ($P_{Th} = P_{He} = P$). Thus, options 2 and 3 are incorrect.
2. **Kinetic Energy Relation:** Kinetic energy ($K$) is related to momentum ($P$) and mass ($m$) by the equation $K = \frac{P^2}{2m}$ .
3. **Comparison:** Since the magnitudes of momentum ($P$) are the same for both particles, the kinetic energy is inversely proportional to the mass ($K \propto \frac{1}{m}$).
   - The mass of the helium nucleus ($m_{He} \approx 4$ u) is much less than the mass of the thorium nucleus ($m_{Th} \approx 234$ u).
   - Therefore, the lighter helium nucleus acquires significantly more kinetic energy than the heavier thorium nucleus ($K_{He} > K_{Th}$).

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