A nuclear fission process is given by A^{240} | vedclass

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(C) The nuclear fission reaction is given by: $A^{240} \rightarrow B^{100} + C^{140} + Q$. The energy released $Q$ in a nuclear reaction is equal to t

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) The nuclear fission reaction is given by: $A^{240} ightarrow B^{100} + C^{140} + Q$. The energy released $Q$ in a nuclear reaction is equal to the difference between the total binding energy of the products and the total binding energy of the reactants. Total binding energy of reactant $A = 240 	imes 7.6 \, MeV = 1824 \, MeV$. Total binding energy of products $B$ and $C = (100 	imes 8.1 \, MeV) + (140 	imes 8.1 \, MeV) = 810 \, MeV + 1134 \, MeV = 1944 \, MeV$. The energy released $Q = 	ext{Total Binding Energy of Products} - 	ext{Total Binding Energy of Reactants}$. $Q = 1944 \, MeV - 1824 \, MeV = 120 \, MeV$.

$A$ nuclear fission process is given by $A^{240} \rightarrow B^{100} + C^{140} + Q$ (energy). If the binding energy per nucleon for nuclei $A, B,$ and $C$ are $7.6 \, MeV, 8.1 \, MeV,$ and $8.1 \, MeV$ respectively,then the energy $Q$ released is approximately $...... \, MeV$.

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