1.

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 :

Answer»

<P>26.9 MeV
25.8 MeV
23.6 MeV
12.9 MeV.

SOLUTION :Given, BE/nucleon for `""_(1)H^(2)=1.1MeV`
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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