When a U^{238} nucleus,initially at rest,deca | vedclass

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(D) According to the law of conservation of linear momentum,the initial momentum of the system is zero because the nucleus is at rest.$\vec{P}_i = \ve

Vedclass · Accepted Answer

$-\frac{4u}{234}$

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(D) According to the law of conservation of linear momentum,the initial momentum of the system is zero because the nucleus is at rest.$\vec{P}_i = \vec{P}_f = 0$Let $m_{\alpha}$ be the mass of the alpha particle and $m_R$ be the mass of the residual nucleus $(Th^{234})$.$m_{\alpha} = 4$ units and $m_R = 234$ units.$m_{\alpha}\vec{v}_{\alpha} + m_R\vec{v}_R = 0$Given the speed of the alpha particle is $u$,so $\vec{v}_{\alpha} = u$.$4u + 234\vec{v}_R = 0$$\vec{v}_R = -\frac{4u}{234}$The negative sign indicates that the residual nucleus moves in the opposite direction to the alpha particle.

When a $U^{238}$ nucleus,initially at rest,decays by emitting an alpha particle with speed $u$,what will be the recoil velocity of the residual nucleus?

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