The binding energy per nucleon versus mass nu | vedclass

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(C) Energy is released in a nuclear process when the total binding energy $(B.E.)$ of the products is greater than the total binding energy of the rea

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$W 	o 2Y$

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(C) Energy is released in a nuclear process when the total binding energy $(B.E.)$ of the products is greater than the total binding energy of the reactants.From the graph, we can identify the mass number $(A)$ and binding energy per nucleon $(B.E./A)$ for each nucleus:For $W: A = 120, B.E./A = 7.5 \, MeV \implies 	ext{Total } B.E. = 120 	imes 7.5 = 900 \, MeV$For $Y: A = 60, B.E./A = 8.5 \, MeV \implies 	ext{Total } B.E. = 60 	imes 8.5 = 510 \, MeV$For $X: A = 90, B.E./A = 8.0 \, MeV \implies 	ext{Total } B.E. = 90 	imes 8.0 = 720 \, MeV$For $Z: A = 30, B.E./A = 5.0 \, MeV \implies 	ext{Total } B.E. = 30 	imes 5.0 = 150 \, MeV$Now, let us check the options:$(A)$ $Y 	o 2Z$: Reactant $B.E. = 510 \, MeV$, Product $B.E. = 2 	imes 150 = 300 \, MeV$. (Energy absorbed)$(B)$ $W 	o X + Z$: Reactant $B.E. = 900 \, MeV$, Product $B.E. = 720 + 150 = 870 \, MeV$. (Energy absorbed)$(C)$ $W 	o 2Y$: Reactant $B.E. = 900 \, MeV$, Product $B.E. = 2 	imes 510 = 1020 \, MeV$. (Energy released as $1020 > 900$)$(D)$ $X 	o Y + Z$: Reactant $B.E. = 720 \, MeV$, Product $B.E. = 510 + 150 = 660 \, MeV$. (Energy absorbed)Therefore, the correct process is $W 	o 2Y$.

The binding energy per nucleon versus mass number curve for nuclei is shown in the figure. $W, X, Y$ and $Z$ are four nuclei indicated on the curve. The process that would release energy is

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