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 $X$: $A = 90, B.E./A = 8.0 \, MeV \implies 	ext{Total } B.E. = 90 	imes 8.0 = 720 \, MeV$For $Y$: $A = 60, B.E./A = 8.5 \, MeV \implies 	ext{Total } B.E. = 60 	imes 8.5 = 510 \, MeV$For $Z$: $A = 30, B.E./A = 5.0 \, MeV \implies 	ext{Total } B.E. = 30 	imes 5.0 = 150 \, MeV$Now, checking option $(c)$: $W 	o 2Y$Total $B.E.$ of reactants $= 900 \, MeV$Total $B.E.$ of products $= 2 	imes (510) = 1020 \, MeV$Since the total $B.E.$ of products $(1020 \, MeV)$ is greater than the total $B.E.$ of reactants $(900 \, MeV)$, energy is released in this process.

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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In the nuclear process $n \to p + e^- + \bar{\nu}$,if the masses of proton,neutron,and electron are $1.6725 \times 10^{-27} \ kg$,$1.6747 \times 10^{-27} \ kg$,and $9 \times 10^{-31} \ kg$ respectively,then the energy released is ...... $MeV$.

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