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The binding energy/nucleon of deuteron (""_(2)H^(4)) and the helium atom (He) are 1.1 MeV and 7 Mev respectively. If the two deuteron atoms fuse to form a single helium atom, then the energy released is : |
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Answer» <P>26.9 MeV and BE/nucleon for `""_(2)He^(4)=7MeV` and `""_(1)H^(2) +""_(1)H^(2) to ""_(2)He^(4)+Q` `Q=BE_(p)-BE_(R)` `=4 XX 7-4 xx 1.1=23.6MeV` |
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