1.

The B.E. per nucleon of ""_(1)H^(2) and ""_(2)He^(4) are 1.1 MeV and 7 MeV respectively. If two deuteron (""_(1)H^(2)) nuclei react to form single helium nucleus, the energy released

Answer»

13.9 MEV
23.6 MeV
26.9 MeV
19.2 MeV.

Solution :`2 ""_(1)H^(2)=""_(2)He^(4)+E`
`E="energy of "H_(2)^(4)-"energy of "2_(1)H^2`
`=(4 xx 7-2 xx 1.1 xx 2)MeV=(28-4.4)MeV`
`E=23.6 MeV`


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