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

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(D) The binding energy $(BE)$ of a nucleus is calculated as: $BE = (	ext{number of nucleons}) 	imes (	ext{binding energy per nucleon})$.For ${}_2^4

Vedclass · Accepted Answer

$17.28$

Vedclass · Answer

(D) The binding energy $(BE)$ of a nucleus is calculated as: $BE = (	ext{number of nucleons}) 	imes (	ext{binding energy per nucleon})$.For ${}_2^4He$: $BE = 4 	imes 7.06 \, MeV = 28.24 \, MeV$.For ${}_3^7Li$: $BE = 7 	imes 5.60 \, MeV = 39.20 \, MeV$.The binding energy of ${}_1^1H$ is $0 \, MeV$ as it is a single proton.The energy released $Q$ in the reaction is given by the difference in total binding energy of products and reactants:$Q = [2 	imes BE({}_2^4He)] - [BE({}_3^7Li) + BE({}_1^1H)]$$Q = [2 	imes 28.24] - [39.20 + 0]$$Q = 56.48 - 39.20 = 17.28 \, 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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