A ^{238}U nucleus decays by emitting an alpha | vedclass

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(A) Initially,the $^{238}U$ nucleus is at rest. After the decay,the $\alpha$ particle and the residual nucleus move in opposite directions.According t

Vedclass · Accepted Answer

$-4v/234$

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(A) Initially,the $^{238}U$ nucleus is at rest. After the decay,the $\alpha$ particle and the residual nucleus move in opposite directions.According to the law of conservation of linear momentum:$P_{	ext{initial}} = P_{	ext{final}}$$0 = m_{\alpha}v_{\alpha} + m_{	ext{residual}}V$Here,$m_{\alpha} = 4$ units,$v_{\alpha} = -v$ (assuming the $\alpha$ particle moves in the negative direction),and $m_{	ext{residual}} = 234$ units.$0 = 4(-v) + 234V$$234V = 4v$$V = \frac{4v}{234}$However,since the question asks for the recoil velocity vector relative to the $\alpha$ particle's velocity direction,the recoil velocity is $V = -\frac{4v}{234} \ m/s$.

$A$ $^{238}U$ nucleus decays by emitting an $\alpha$ particle of speed $v \ m/s$. The recoil velocity of the residual nucleus is (in $m/s$):

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