If the binding energy per nucleon in _3^7Li a | vedclass

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(C) The binding energy of a nucleus is given by the product of the number of nucleons and the binding energy per nucleon.For $_3^7Li$,the total bindin

Vedclass · Accepted Answer

$17.28$

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(C) The binding energy of a nucleus is given by the product of the number of nucleons and the binding energy per nucleon.For $_3^7Li$,the total binding energy is $7 	imes 5.60 \, MeV = 39.20 \, MeV$.For $_2^4He$,the total binding energy of one nucleus is $4 	imes 7.06 \, MeV = 28.24 \, MeV$.In the reaction $p + {}_3^7Li 	o 2 {}_2^4He$,the total energy must be conserved.The energy of the proton $(E_p)$ plus the binding energy of the reactant nucleus must equal the total binding energy of the product nuclei.$E_p + 39.20 \, MeV = 2 	imes (28.24 \, MeV)$.$E_p + 39.20 = 56.48$.$E_p = 56.48 - 39.20 = 17.28 \, MeV$.

If the binding energy per nucleon in $_3^7Li$ and $_2^4He$ nuclei are $5.60 \, MeV$ and $7.06 \, MeV$ respectively,then in the reaction $p + {}_3^7Li \to 2 {}_2^4He$,the energy of the proton must be ........... $MeV$.

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