If the binding energy per nucleon for ,_3^7,L | vedclass

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Question

(B) The binding energy of $\,_3^7\,Li\,\,$ nucleus $= 7 	imes 5.60 = 39.2 \,MeV$.The binding energy of $\,_2^4\,He\,\,$ nucleus $= 4 	imes 7.06 = 28

Vedclass · Accepted Answer

$17.28$

Vedclass · Answer

(B) The binding energy of $\,_3^7\,Li\,\,$ nucleus $= 7 	imes 5.60 = 39.2 \,MeV$.The binding energy of $\,_2^4\,He\,\,$ nucleus $= 4 	imes 7.06 = 28.24 \,MeV$.The reaction is $p + \,_3^7\,Li 	o 2\,_2^4\,He$.The energy released ($Q$-value) in the reaction is given by the difference in total binding energies of the products and the reactants.$Q = [2 	imes (	ext{Binding Energy of } \,_2^4\,He)] - [(	ext{Binding Energy of } \,_3^7\,Li) + (	ext{Binding Energy of } p)]$.Since the binding energy of a free proton is $0$, we have:$Q = (2 	imes 28.24) - 39.2 = 56.48 - 39.2 = 17.28 \,MeV$.

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

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