The binding energy per nucleon v/s mass numbe | vedclass

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(C) Energy is released in a nuclear reaction if the total binding energy of the products is greater than the total binding energy of the reactants. Th

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$w \rightarrow 2 y$

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(C) Energy is released in a nuclear reaction if the total binding energy of the products is greater than the total binding energy of the reactants. This is equivalent to saying that the binding energy per nucleon of the product nucleus must be greater than that of the reactant nucleus.Let us analyze the given options:$A$) $y \longrightarrow 2 z$: Binding energy per nucleon of $y$ is $8.5 	ext{ MeV}$ and that of $z$ is $5 	ext{ MeV}$. Since $5 $B$) $w ightarrow x+z$: Binding energy per nucleon of $w$ is $7.5 	ext{ MeV}$,$x$ is $8 	ext{ MeV}$,and $z$ is $5 	ext{ MeV}$. The average binding energy per nucleon of the products $(x+z)$ is roughly $\frac{90 	imes 8 + 30 	imes 5}{120} = \frac{720 + 150}{120} = \frac{870}{120} = 7.25 	ext{ MeV}$. Since $7.25 $C$) $w ightarrow 2 y$: This is not possible as mass number is not conserved ($120 
eq 2 	imes 60$ is fine,but $w$ is $120$ and $y$ is $60$,so $w ightarrow 2y$ is physically possible). Binding energy per nucleon of $w$ is $7.5 	ext{ MeV}$ and $y$ is $8.5 	ext{ MeV}$. Since $8.5 > 7.5$,the binding energy per nucleon increases,so energy is released.$D$) $x ightarrow y+z$: Binding energy per nucleon of $x$ is $8 	ext{ MeV}$,$y$ is $8.5 	ext{ MeV}$,and $z$ is $5 	ext{ MeV}$. The average binding energy per nucleon of the products is $\frac{60 	imes 8.5 + 30 	imes 5}{90} = \frac{510 + 150}{90} = \frac{660}{90} \approx 7.33 	ext{ MeV}$. Since $7.33 Therefore,the correct option is $C$.

The binding energy per nucleon $v/s$ 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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