A nucleus with mass number 220 initially at r | vedclass

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(B) Given: $Q = 5.5 \ MeV$,Mass number of parent nucleus $A = 220$.The reaction is: ${}^{220}Y \longrightarrow {}^{216}X + {}^{4}_{2}\alpha$.By the la

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$5.4$

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(B) Given: $Q = 5.5 \ MeV$,Mass number of parent nucleus $A = 220$.The reaction is: ${}^{220}Y \longrightarrow {}^{216}X + {}^{4}_{2}\alpha$.By the law of conservation of linear momentum,the magnitude of momentum of the alpha particle $(p_{\alpha})$ must be equal to the magnitude of momentum of the daughter nucleus $(p_{X})$: $p_{\alpha} = p_{X}$.The kinetic energy $K$ is related to momentum $p$ by $K = \frac{p^2}{2m}$.Since $p_{\alpha} = p_{X}$,we have $K_{\alpha} = \frac{p^2}{2M_{\alpha}}$ and $K_{X} = \frac{p^2}{2M_{X}}$.Thus,$\frac{K_{\alpha}}{K_{X}} = \frac{M_{X}}{M_{\alpha}} = \frac{216}{4} = 54$.Therefore,$K_{X} = \frac{K_{\alpha}}{54}$.The total $Q$ value is the sum of kinetic energies: $Q = K_{\alpha} + K_{X}$.$5.5 = K_{\alpha} + \frac{K_{\alpha}}{54} = K_{\alpha} \left( 1 + \frac{1}{54} ight) = K_{\alpha} \left( \frac{55}{54} ight)$.$K_{\alpha} = 5.5 	imes \frac{54}{55} = 0.1 	imes 54 = 5.4 \ MeV$.

$A$ 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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