A nucleus with mass number 220 initially at r | vedclass

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(B) The $Q$-value of the reaction is the total kinetic energy released,so $K_{\alpha} + K_{D} = Q = 5.5\, MeV$,where $K_{\alpha}$ is the kinetic energ

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$5.4$

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(B) The $Q$-value of the reaction is the total kinetic energy released,so $K_{\alpha} + K_{D} = Q = 5.5\, MeV$,where $K_{\alpha}$ is the kinetic energy of the $\alpha$-particle and $K_{D}$ is the kinetic energy of the daughter nucleus.By the law of conservation of linear momentum,the magnitudes of the momenta of the $\alpha$-particle and the daughter nucleus must be equal: $p_{\alpha} = p_{D}$.Using the relation $K = \frac{p^2}{2m}$,we have $p = \sqrt{2mK}$.Thus,$\sqrt{2 m_{\alpha} K_{\alpha}} = \sqrt{2 m_{D} K_{D}}$.Squaring both sides: $m_{\alpha} K_{\alpha} = m_{D} K_{D}$.Given $m_{\alpha} = 4$ and $m_{D} = 220 - 4 = 216$,we get $4 K_{\alpha} = 216 K_{D}$,which simplifies to $K_{D} = \frac{4}{216} K_{\alpha} = \frac{1}{54} K_{\alpha}$.Substituting this into the energy equation: $K_{\alpha} + \frac{1}{54} K_{\alpha} = 5.5\, MeV$.$\frac{55}{54} K_{\alpha} = 5.5\, MeV$.$K_{\alpha} = 5.5 	imes \frac{54}{55} = 0.1 	imes 54 = 5.4\, MeV$.

$A$ nucleus with mass number $220$ initially at rest emits an $\alpha$-particle. If the $Q$ value of the reaction is $5.5\, MeV$,calculate the kinetic energy of the $\alpha$-particle in $MeV$.

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