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If the binding energy per nucleon in \mathrm{Li^{7}}and \mathrm{He^{4}} nuclei are respectively 5.60 MeV and 7.06 MeV, then energy of reaction \mathrm{Li^{7}+p\rightarrow 2\, \: _{2}He^{4}} is

Option: 1

19.6 MeV


Option: 2

2.4 MeV


Option: 3

8.4 MeV


Option: 4

17.3 MeV


Answers (1)

best_answer

\mathrm{B.E.of\, \mathrm{Li}^7=39.20 \mathrm{MeV}\: and\: \mathrm{He}^4=28.24 \mathrm{MeV}}

Hence binding energy of \mathrm{\mathrm{He}^4=56.48 \mathrm{MeV}}

Energy of reaction \mathrm{=56.48-39.20=17.28 \mathrm{MeV}}.

Posted by

Kshitij

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