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 :

Options

A. 26.9 MeV
B. 25.8 MeV
C. 23.6 MeV
D. 12.9 MeV.

Answer

C

Explanation

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`