An U^{238} nucleus is initially at rest. It d | vedclass

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(C) According to the law of conservation of linear momentum,the initial momentum of the system is equal to the final momentum.Initially,the $U^{238}$

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$\frac{4u}{234}$

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(C) According to the law of conservation of linear momentum,the initial momentum of the system is equal to the final momentum.Initially,the $U^{238}$ nucleus is at rest,so the initial momentum is $0$.The decay process is: $U^{238} 	o Th^{234} + He^{4}$.Let $M_D = 234$ be the mass of the daughter nucleus and $M_{\alpha} = 4$ be the mass of the $\alpha$-particle.Let $v$ be the recoil speed of the daughter nucleus and $u$ be the speed of the $\alpha$-particle.Applying conservation of momentum: $0 = M_D \cdot v + M_{\alpha} \cdot u$.$0 = 234 \cdot v + 4 \cdot u$.Solving for $v$: $v = -\frac{4u}{234}$.The magnitude of the recoil speed is $\frac{4u}{234}$.

An $U^{238}$ nucleus is initially at rest. It decays by emitting an $\alpha$-particle with speed $u$. Find the recoil speed of the daughter nucleus.

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