When a U²³⁸ nucleus originally at rest, decays by emitting an alpha particle having a speed ′u′, the recoil speed of the residual nucleus is (a) 4u/238 (b) -4u/234 (c) 4u/234 (d) -(4μ)/(238)

Answer: (c). U²³⁸ → Th²³⁴ + α, starting from rest: 4u = 234 v. v = 4u/234 (opposite to the alpha particle). So the answer is (c) 4u/234.

When a U²³⁸ nucleus originally at rest, decays by emitting an alpha… | JEE Main 2003 (AIEEE)

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Nuclear Reactions & Q-value

When a $U^{238}$ nucleus originally at rest, decays by emitting an alpha particle having a speed ${}^{\prime}u^{\prime},$ the recoil speed of the residual nucleus is

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Answer: (c)

$U^{238} \to Th^{234} + \alpha$ , starting from rest: $4u = 234\,v$ .
$v = \dfrac{4u}{234}$ (opposite to the alpha particle).
So the answer is (c) $\dfrac{4u}{234}$ .

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