A nucleus with mass number 220 initially at r | vedclass

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Question

(A) The total energy available for the kinetic energy of the products is the $Q$ value minus the energy of the emitted photon.$E_{	ext{total}} = Q -

Vedclass · Accepted Answer

$6.48$

Vedclass · Answer

(A) The total energy available for the kinetic energy of the products is the $Q$ value minus the energy of the emitted photon.$E_{	ext{total}} = Q - E_{	ext{photon}} = 7.8 \, MeV - 1.2 \, MeV = 6.6 \, MeV$.Let the mass number of the parent nucleus be $A = 220$. The $\alpha -$ particle has mass number $4$,and the daughter nucleus has mass number $A - 4 = 216$.By conservation of linear momentum,the kinetic energy of the $\alpha -$ particle $(KE_{\alpha})$ is given by the formula:$KE_{\alpha} = E_{	ext{total}} 	imes \left( \frac{A - 4}{A} ight)$.Substituting the values:$KE_{\alpha} = 6.6 	imes \left( \frac{216}{220} ight) = 6.6 	imes \frac{54}{55} = 6.48 \, MeV$.

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