If the binding energy per nucleon of _1H^2 is | vedclass

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(B) The reaction is $_1H^2 + _1H^2 	o _2He^4$. Total binding energy of reactants = $(2 	imes 2 	imes x) = 4x$. Total binding energy of product = $(

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$24$

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(B) The reaction is $_1H^2 + _1H^2 	o _2He^4$. Total binding energy of reactants = $(2 	imes 2 	imes x) = 4x$. Total binding energy of product = $(4 	imes 7x) = 28x$. Energy liberated $(Q)$ = $(	ext{Binding energy of product} - 	ext{Binding energy of reactants}) = 28x - 4x = 24x$. The ratio of energy liberated to the binding energy per nucleon of $_1H^2$ is $\frac{24x}{x} = 24$.

If the binding energy per nucleon of $_1H^2$ is $x$ and that of $_2He^4$ is $7x$,then the ratio of the energy liberated in the reaction $_1H^2 + _1H^2 \to _2He^4$ to the binding energy per nucleon of $_1H^2$ is:

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