A nucleus of _{84}^{210}Po originally at rest | vedclass

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(A) The decay process is given by: $_{84}^{210}Po \rightarrow _{82}^{206}X + _{2}^{4}He$.According to the law of conservation of linear momentum,the i

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$\frac{4v}{206}$

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(A) The decay process is given by: $_{84}^{210}Po \rightarrow _{82}^{206}X + _{2}^{4}He$.According to the law of conservation of linear momentum,the initial momentum of the system is zero because the nucleus is at rest.Let $M_d$ be the mass of the daughter nucleus ($206$ amu) and $M_{\alpha}$ be the mass of the $\alpha$ particle ($4$ amu).Let $v'$ be the recoil speed of the daughter nucleus and $v$ be the speed of the $\alpha$ particle.$M_d v' + M_{\alpha} v = 0$$206 v' + 4v = 0$$v' = -\frac{4v}{206}$The magnitude of the recoil speed is $|v'| = \frac{4v}{206}$.

$A$ nucleus of $_{84}^{210}Po$ originally at rest emits an $\alpha$ particle with speed $v$. What will be the recoil speed of the daughter nucleus?

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