The binding energy per nucleon of {}_3^7Li an | vedclass

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(D) The binding energy of a nucleus is given by $BE = (	ext{number of nucleons}) 	imes (	ext{binding energy per nucleon})$.For ${}_3^7Li$: $BE_1 =

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$17.3$

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(D) The binding energy of a nucleus is given by $BE = (	ext{number of nucleons}) 	imes (	ext{binding energy per nucleon})$.For ${}_3^7Li$: $BE_1 = 7 	imes 5.60\,MeV = 39.2\,MeV$.For ${}_1^1H$: The binding energy is $0\,MeV$ as it is a single proton.For $2{}_2^4He$: $BE_2 = 2 	imes (4 	imes 7.06\,MeV) = 2 	imes 28.24\,MeV = 56.48\,MeV$.The energy released $Q$ is the difference between the total binding energy of the products and the reactants:$Q = BE_{	ext{products}} - BE_{	ext{reactants}}$$Q = 56.48\,MeV - (39.2\,MeV + 0\,MeV)$$Q = 17.28\,MeV \approx 17.3\,MeV$.

The binding energy per nucleon of ${}_3^7Li$ and ${}_2^4He$ nuclei are $5.60\,MeV$ and $7.06\,MeV$ respectively. In the nuclear reaction ${}_3^7Li + {}_1^1H \to 2{}_2^4He + Q$, the value of energy $Q$ released is.........$MeV$.

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