The binding energies per nucleon of $Li^{7}$ and ${He}^{4}$ are 5.6 MeV and 7.06 MeV respectively, then the energy of the reaction $Li^{7} + p = 2\lbrack\ _{2}He^{4}\rbrack$ will be

Question

Dr. Abubacker · Accepted Answer

(a) The reaction is $\ _{3}Li^{7} +_{1}P^{1} ightarrow 2\left( \ _{2}He^{4}) ight.\ $
$	herefore\ \ E_{p} = 2E\left( \ _{2}He^{4}) ight.\ E_{(Li)}$
$= 2(4 	imes 7.06) - 7 	imes 5.6$
$= 56.48 - 39.2 = 17.28\ MeV$

Q. The binding energies per nucleon of $Li^{7}$ and ${He}^{4}$ are 5.6 MeV and 7.06 MeV respectively, then the energy of the reaction $Li^{7} + p = 2\lbrack\ _{2}He^{4}\rbrack$ will be

Explanation

Reviewed by Dr. Abubacker

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