A nucleus of mass 218, amu in a free state de | vedclass

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(D) The decay process is given by: $X^{218} \rightarrow Y^{214} + \alpha^{4}$.According to the law of conservation of linear momentum,the magnitude of

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$0.125$

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(D) The decay process is given by: $X^{218} ightarrow Y^{214} + \alpha^{4}$.According to the law of conservation of linear momentum,the magnitude of the momentum of the daughter nucleus $(P_d)$ must be equal to the magnitude of the momentum of the $\alpha$-particle $(P_{\alpha})$,i.e.,$P_d = P_{\alpha}$.The kinetic energy $E$ is related to momentum $P$ and mass $m$ by the formula $E = \frac{P^2}{2m}$,which implies $P = \sqrt{2mE}$.Thus,$\sqrt{2m_d E_d} = \sqrt{2m_{\alpha} E_{\alpha}}$.Squaring both sides,we get $m_d E_d = m_{\alpha} E_{\alpha}$.Therefore,the recoil energy of the daughter nucleus is $E_d = \frac{m_{\alpha} E_{\alpha}}{m_d}$.Given: $m_{\alpha} = 4\, amu$,$m_d = 218 - 4 = 214\, amu$,and $E_{\alpha} = 6.7\, MeV$.Substituting the values: $E_d = \frac{4 	imes 6.7}{214} = \frac{26.8}{214} = 0.125\, MeV$.

$A$ nucleus of mass $218\, amu$ in a free state decays to emit an $\alpha$-particle. The kinetic energy of the emitted $\alpha$-particle is $6.7\, MeV$. The recoil energy (in $MeV$) of the daughter nucleus is:

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